Engineering-economics-by-j-tiong-pdf.pdf [z0xj2yzxooln] (2023)

CONTENTs Ques

Question Bank .l

2 Question Bank 3 Question Bank 4 Question Bank 5 Question Bank 6 Question Bank 7 Question Bank g Question Bank g

present

Question Bank

1-24

Simple lnterest and Discount Compound lnterest

25_40 41 _ 102

Annuity 103

Uniform Gradient

A bookstore should this b

194

195 _ 212

A. P 2o0.oo B. P 3oo.oo

Bonds 213 _ 220

Depreciation and Depletion

10 Objective

Theories and Formulas Glossary

Economy

and AnnualCost sis

euestions

tion

221

_240

241

_260

261

_ 280

281

_ 302

c. P 4oo.oo

P 5oo.o0

Letx=sellingprice

303 324

Note: The profit of 30o/

th"-r!'rri"g;.t""'"^ffi

Profit =

323

0.30x

::t#:Tf

Eq.

X?i#^"";1;innor#"or,n"capitaror3oyoor

Also,

337

5ffi :1'[%']3?;"*,'*r-

= F'rotrt = 0.90x _

;:1f

200

i%:'#lo;111

_+ Eq.

discount

and 2;

200 = g.5x

x=400

;

The selting price of the book is p 400.00

A businessma tncome from a what minimum to be justified?

A. B.o2 % B. 12.07 c. 10.89 %%

.08

0/o

Question Bank

Present Economy 3

Engineering Economics by Jaime R. 'l'iong

,%*Let x = capital y

Dalisay Corporation's gross margin is 45o/o of sales. Operating expenses such as sales and administration are 15oh of sales. Dalisay Corporation is in 40% tax bracket. What percent of sales is their profit after taxes?

gross income

Projected earnings = 0.07x

A. 21% B. 20 c. 19% D. 18

Taxableincome=y-x

tax

= 0.42 (y

o/o

- x)

o/o

,(*rt,"..

Net earnings = Taxable income - tax Net earnings = (y - x)- o.+z(v - *) Net earnings = 0.58(y

Let x = sales of Dalisay Corporation

- x)

Gross margin = 0.45x Operating expenses = 0.15x

Projected income = net earnings

0.07x=0.58(y-x) 0.07x:0.58y-0.58x

Net income before taxes = 0.45x = 0.30x

0.65x = 0.58y \r-_ 0.65x o.s8

Net profit = 0.60(0.30x) Net profit = 0.18x

=1.1207x

.'. Rate of return before payment of taxes = 0.1207 or

12.07o/o

of the capital

A manufacturing flrm maintains one product assembly line to produce signal generators. Weekly demand for the generators is 35 units. The line operates for 7 hours per day, 5 days per week. What is the maximum production time per unit in hours required of the line to meet the demand?

.'. Net profit is 18% of the sales

ln determining the cost involved in fabricating sub-assembly B within a company, the following data have been gathered:

.... P 0.30 per unit .... P 0.50 per unit .... P 300.00 per set up

A. t hour B. t hour and 10 minutes C. t hour and 15 minutes D. t hour and 20 minutes

It is decided to subcontract the manufacturing of assembly B to an outside company. For an order of 100 units, what is the cost per unit that is acceptable to the company?

,%/,*

A. B. c.

Let t = maximum production time per unit ') . _ (, z nours )(s oays)( l week ' day jt *""k ,1.35 ,",tr.l

[

t = t hour per unit .'. The maximum production time per unit is

0.15x

Since tax is 40%, net profit after taxes will be:

t hour.

P 3.80 P 4.00 P 4.10

Present Economy 5

Question []ank

[!ngint,t'rirrg l,)r'onomir:s by,]airnu ll,. 'l'iong

Cost per unit = 0.30 * O.so * 1-09 100

Cost per unit = P 3.80

drr;-_.'8 h;rr/day work, *itn so men workins-

.'. The cost per unit is P 3.80.

By selling balut at P 5 per dozen, a vendor gains 20%. The cost of the eggs rises by 12.Soh.lf he sells at the same price as before, find his new gain in %.

A. B. C. D.

in 50 An equipment installation job in the completion stage can be complete.d

6.89 % 6.67 % 6.58 o/o 6.12 o/o

yll

t!9,P1tr.::t-"llil11s,i40

job, Jri", in" mechanical engineer contractor decided to add 15 men on the overtime not being Permitted.

are paid P150 lf the liquidated damages is P 5000 per day of delay, and the men workers? additional the re with sa he p"r OrV, how much m6ney would

A. B. c.

P 43,450 P 43,750

P 44.250

S/.k"o...

,c/7,.l,t,,..

With 50 men only:

Let x = original cost of a dozen of balut

No. of days delaYed = 50

cost+profit=5

Liquidated damages = 5000(10) = 59999 Salary of 50 men = 50(50)150) = 375000

x+0.20x=5 1.2Ox =5

40 = 10 days

Total exPenses = 50000 + 375000 Total exPenses = 425000

x = 4.17

Cost per dozen of balut = P4.17

With 65 men (an additional of 15 men):

Let y = new cost of dozen of balut

Solving for the man-day to finish the job' N:

y = 1.'125x y =1.125(.17)

y = 4.69125

(50 men)(50 daYs)

2500 man - day

(a total of 65) Let x = number of days to finish the job with 15 more men

cost+Profit=5 4.69125 + (% gain)(4.6912s) = 5 ohgain 0.0658 = o/o

gain: 6.58%

.'. The new gain of the balot vendor is 6.58 %.

(50+15)x=2500 x = 38.46

Sayx=39daYs paid by the Since number of days is 39 < 40, thus no penalty will be workers: the of salary the on solely Will contractor. So expenses Salary of 65 men = 65(150X39) = 380250

.'. The amount saved = 425000

PROPERTY TAG

21432 n DO NOT REMOVE

380250 = P 44,750'00

Quest,ion

lllrnk

It)rrgirrct,rirrg l,l,orrornir:s by Jainro Ii,. 'l'iong

Present Econorny 7

ln a eertain department store, the monthly salary of a saleslady is parfly constant and partly varies as the value of her sales for the month. When the value of her sales for the month is P 10,000.00, her salary for the month is P g00.00. When her monthly sales go up to P 12,000.00, her monthly salary goes up to P1,000.00. What must be the value of her sales for the month so that her salary for the month will be P 2,000.00?

A. B. c. D.

P 30,000 P 31,000 P 32,000 P 33,000

Jojo bought a second-hand Betamax VCR and then sold it to Rudy at a profit of 40%. Rudy then sold the VCR to Noel at a profit ol20%. lf Noel paid P 2,856 more than it cost Jojo, how much did Jojo pay for the unit?

A. B. c. D. '

P 4,100

P 3,900 P 4,000 P 4,200

/ /,/,t,,,

Let x = amount Jojo paid for the VCR unit

,'/./,r/rz"r..

Amount Noel paid for the VCR unit = x +

Let x=fixedsalary

900

-+ Eq. 1

-->

000

Amount Rudy paid for the VCR unit = x + 0.40x = 1.40x Amount Noel paid for the VCR unit =

Eq. 2

(

.+Ox)(t

.20)= 1 .68x -+ Eq.

Equating equations 1 and2'.

Subtract Eq. 1 from Eq.2:

x+2856=1.68x 0.68x = 2856 x= 4200

2000y = 1 00 100 v=' 2000 Y

-+ Eq. 1

k = constant of proportionality of salesgirl's variable salary z = sales for salesgirl to earn P2,000.00 x + k(10000) = x + k(12000) =

2856

.'.

Jojo paid

P 4,200 for the VCR unit.

= 0'05

Substitute the value of y in Eq. x + 0.05(1 0000) = 900

x=400 To earn a salary of P 2,000.00:

x+y(z)=2000 400+0.052=2000 z=32000

The selling price of a TV set is double that of its net cost. lf the W set is sold to a customer at a profit of 25Yo of the net cost, how much discount was given to the customer?

A. B. c. D.

37.5 37.9

o/o o/o

38.2% 38.5 %

Let x=netcost Selling price = 2x Profit = 0.25x Profit = Selling price

net cost

discount

(,lrrt,sl

i.lr llrrrrk l,),grrrr,r,ri,g l,l.rrr.r.r,.:,

0.25x 2x x O.25x -.

I)rcsent Economy 9

lr1,,lr1rrrrr, lf 'l'iorrg

disr:ounl

discount

discount = 0.75x

The quarrying cost of marble and granite blocks plus delivery cost to the processing plant each is P 2,400.00 per cubic meter. Processing cost of marble lnto tile is P 200.00 per square meter and that of granite into tiles also is P 600.00 per square meter.

Discount is75% of the net cost.

Solving for discount in terms of percent of selling price; o'75x discount 2x discount = 0.375 discount = 375% .'.

lf marble has a net yield of 40 square meters of tiles per cubic meter of block and eells at P 400 per square meter, and granite gives a net yield of 50 square meters of tiles per cubic meter of block and sells at P 1000 per square meter, Considering all other costs to be the same, the granite is more profitable than the marble by how much?

The discount given to the customer is 3T.S %.

A. B. C. D.

A Mechanical Engineer who was awarded a p 450,000 contract to install the

machineries of an oil mill failed to finish the work on time. As nrovirtecr far in rho contract, hE ha enalty t to on er day for the

first

per

per day for eve lf the t days was the completion of the contract

A. B. G.

ty was ?

P 12,000 per cubic P 13,000 per cubic P 14,000 per cubic P 15,000 per cubic

meter

meter meter meter

/ /./,t., For Marble (Per cubic meter): Quarrying cost & delivery cost to processing plant = 2400

26 days

processing cost into tit"" =

27 days

200

[( sq.m ]1+o ,q. *1= aooo

J' Total production cost =2400 + 8000 = 10400

28 days

Total income = 40(400) = 16000 Daily penatty for the first 1 0 days =

Total penalty for the first

days

(O.OrX+SOOO0) =

12s

= 1125(10) = 1 12SO

Profit = Total income

Total production cost

Profit=16000-10400 Profit = 5600

Daily penalty for the next 10 days = (O.OO5X45OO00) 2259 =

For Granite (Per cubic meter):

Total penalty for the next 10 days 22SO(1O) = 22SOO =

Quarrying cost & delivery cost to processing plant = 2400

Daily penalty for the succeeding days = (0.01)(450,000) = 45gO solving for the number of days (x) beyond 20 days from end of contract that would amount the penalty to 60750.0b 4500x + 22500 +11250 :60750

x=6 .'. The completion of the contract was delayed for 26 days.

processins cost into tir" =

[!99)@o

sq. m)= soooo

Total production cost = 2400 + 30000 = 32400 Total income = 50(1000) = 50000 Profit = Total income

Profit=50000-32400 Profit = 17600

Total production cost

(luestion Br.ruJt_ l,)rrgirroor.rrrg lrlcononrir:s by Jairrro Il 'l'iong

I)resent Economy

Solving for the difference in profit per cubic meter. Difference = 1 7600 _ 5600 Difference = '12,000

.'. Granite is the more than marble by p,|2,000 per cubic meter.

A man would like to invest p 50,000 in government bonds and stocks that will give an overall annual return of about 5%. The money to be invested in government bonds will give an annual return of 4.5o/o and the stocks of about 6%. The investments are in units of p 100.00 each. lf he desires to klep r,isito"t investment to minimum in order to reduce his rrsk, determine now riany stocts should be purchased.

A. B. c.

165 166 167 168

A22OV 2 hp motor has an efficiency of 80%. lf power costsP3.00 per.kw-hr for the first 50 kw-hr, P 2.90 per kw-hr for the second 50 kw-hr, P 2.80 for the third kw-hr and so on until a minimum of P 2.50 per kw-hr is reached. How much does It cost to run this rhotor continuously for 7 days?

A. P 800 B. P 820 c. P 840 D. P 860

Solving for the input power of motor: outDut

' =Linput

Eftrclencv

o.8o

2 =input

input = 2.5 hP

Let x = number of government bonds

input=rttr[,ffit)

y = number ofstocks

input = 1.865 kw

'100x+100y=5gggg

x+y=500 0.0a5(1 00x) + 0.06(1 00y)

-+ Eq

0.os(so0oo)

4.5x+6Y=2500

-+ Eq.2

Let P = total power consumption for 7 days p=(1

865**Xro*.{{$t)

P = 313.32 kw - hr

Multiply Eq. 1 by 6:

6x+6Y=3000

Table of Power Cost: Eq. 1a

Subtract Eq. 2 from Eq. 1a: 1.5x = 500 x = 333.33 Substitute value of x in Eq.

1,:

333.33{y=SOO Y

= 166'67

since there was a desire to keep stock investment minimum, y must be

Total Power Cost

P 860.00

,'. lt will cost P 860 to run the motor continuously for 7 days'

166 instead ot 167.

TIP ABLEEUILIBBARY

Quest,ion tsank , l,)ngirrccrirrg lrl<:onomics by Jairno lt. Tiong

Present, Economy 13

Cost of Project = Total cost + contingency Cost of Project = 1050000 + 157500 Cost of Project = P 1,207'500

An 8-meter wide concrete road pavement s00 meters long is desired to be constructed over a well-compacted gravel road, together wii-h the necessary concrete curbs and gutters on both sides. ln orderio put the subgrade on an even level grade, a 500 cubic meters of sand fiiling is nece"."ry,-or"i *hich the 10-inch concrete pavement will be placed?

Assume the following data: Sand fill, including rolling and watering = p 100 per cubic meter concrete pavement, 10 inches thick (iabor and materials) including currng = P 220 per sq. meter Curbs and gutters = p 12 per linear meter

4 B. C.

The project will cost P 1,207,500.

en etectric utility purchases 2,300,000 kw-hr per month of electric energy from

National Power Cbrporation at P 2.OO per kw-hr and sells all this to consumers per kw-hr should a-fteiJeOucting distribution loss es of 2OYo. At what average rate expenses in monthly other are be-sold to break even if the following tnii

"n"rgy Ite operation:

A. B. c.

2.5Yo of qross revenue P 750,000 P 2.250.000 P 700,000

Taxes Salaries

How much will the project cost allowing 15yo for contingency?

lleoreciation lnterest

P 1,207,000 P 1,207,500 P 1,208,000

P 300,000 P 200.000

Maintenance Miscellaneous

"V.ur.rLet Cr = cost of sand filling, rolling and watering Cz = cost of curbs and gutter (both sides) Co = cost of concrete pavement

A. P 4.90 B. P 5.20 c. P 5.90 D. P 6.30 ,'/,/nt

Let x = rate per kw-hr of energy sold to consumers

",=[#)('oom3) Cr = P50,000

Solving for ExPenses:

.,=[T)(soomtz)

AmountpaidtoNPc=[,

1kw - hr

)1r.ooooo**

-nr;

= 4600000

Cz = P120,000

With 20% losses, only 80% is sold to consumers'

=[TP)(soom[am)

Totalnumber of kw - hr sold = 0.80(2300000) = 1840000 kw - hr

Ca = P880,000

Total cost = C1 + Q2 + Q. Total cost = 50000 + 120000 +

Totalcost=P1,050,000

Contingency = 0. 1 5(1 050000) Contingency = P 157,500

Taxes = 0.025(1 8a0000)(x) BSOOOO

= 46000x + Total expenses = Amt. paid to NPC + Taxes-+ Salary + Depreciation lnterest + Maintenance + Miscellaneous

Qrr.st;iolr

Ilnrrk

l,)rrgirrr.r,rirrg l,)r:.,unrir:s [ry.lairrrt, ll, ,l,iong

Total expenses = 4600000 +-46000x + 750000

300000 + 2OO00O

Present Economy

,ruOOOO + 700000 + Re nta, =

Total expenses = gg00000 + 46000x

tT#)(. -!#)r, weeks)

= 280000

Solving for lncome:

Sarary or 2 nasmen =

lncome = 1,g40,000x

To break-even; income expenses = '1840000x

Total expenses = 12252450 + 280000 + 57600 Total expenses = p 12,590,050

= 8800000 + 46000x

x = 4.90

For Site B:

.'. The average rate the energy be sold is p 4.90 per kw_hr.

An engineer biddino

weers)(z)

= 57600

17940OOx = 88OO0OO

(#{ffi}sz

oozo ms [ f+zk,.liff](2'75kmx7km) )f ),,, " 1;5 -

cost or haurins =

= 13477695

the aspharting of 7 km stretch of road is confronted _on y:'Jr:,ffilJinT orcho-osino sites on w.rt-rr'io-"""i,p *,"

oltd;;'il;

il#"

Cost of installing & dismanfling of machine = 10000

Rentat _ r\errrdr =f

p6,5oo

'"'th

l( l montn ),^^ weeks) 14 *""k.J(32

= 52000

Total expenses = 13477695 + 1OOO0 + 52000 Total expenses = p 13,539,695 The res to is

A. B.

P 949,645 P 962,10.1

c. P 956,807 D. P 974,090

Solving for the difference in expenses; Difference = 1 3539695 _ 12596050 Difference = 949645

.'. Site B is more expensive than site A by p 949,645.

A fixed capital manufacturing

For Site A:

cost of hauting

E retu

depreciation the rate of (z.stm[zr<m)

= 12252450 Cost of installing & dismanfling of machine = 20000

A. 28.33% B. 29.34% c. 30.12% D.30.78%

()trr,sliorr llirrrk

l,)ngrrr,r,r'rtr1i l,l.onorrrr,.., lrt,lrrrrrr,, l( 'l'iorrg

Present Economy 17

Profit Total capitalinvested

Rate of return

'50-

Number of oost

2500000

= 101

iooooooo , 2oooooo _-

5ooo

0 2833

.,=lffi)(rorno,,,)

The rate of return is 28.33 %.

Cr = 505000

A call to bid was advertised in the philippine Daily lnquirer for the construction of a transmission line from a mini-hydroeiectric power piant to the substation

Number of days to dig and erect posts =

TH*

= 33.67days ,say 34 days

day

",=[H)(a+oays[s) Cz = 30600 Number of days to strung wires =

ffi

= 33.57 days , say 34 days

day

..=(H)(s+oays)(+)

A. B

P 745,890.23 P 817,692.00

c. P 789,120.80

Cs = 34000

D. P 829,592.50

..=[H)(s+aavs) Let Cr = cost of posts

C4 = salary of foreman

C2 = salary of 5 laborers C3 = salary of

= 13600

Cs = Cost of wire

4 electricians

Solving for the number of posts:

(pa\. [-J(soss

Cs =20140 Hydro-electric station

Power

. = 505000 + 30600 + 34000 + 1 3600 + 20140

.-,] 50m

50 nt

Total directcost=Cr +C2+ C3 +Ca +C5 = 603340

50m ContingencY = 0. 1 0(603340) 5035 m

Distance between first and last post

= 60334

5035 _ 15 _ 20

:5000

Total cost including contingency = 603340 + 60334 = 663674

Quest,ion Pr

llank

I)rcsont Economy 19

lr)ngirrct,r'irrg l,)<:onornics by Jairut, ll. 'l'iorrg

ofit = 0.25(663674)

Pr ofit

16591 8.50

Total cost of Project = 663674 + 165918.50 Total cost of Project = P 829,592.50

is_3200 The monthly demand for ice cans being manufactured by Mr. Alarde nlcces With a manual ooerated quillotine, the unit cutting cost is P 25'00' An

ne was offered the unit cutting Alarde be able

.'. The estimated cost of the project is P 829,592.50

Upon her retirement after 44 years in government service, Mrs. Safud Araowa lump sum of P 2,300,000. As a hedge against inflation rt of it invested in real estate at pagadian City and the as follows: arao

A. B. C.

30% in T-bills earning 12% interest 35% in money market placement earning 35% in blue chip stock earning 13%

14o/o

lf her annual earnings from the t-bills, money market and stock is p 50,000 how much did she invests in real estate?

A. B. c. D.

A. B. C. D.

10 months 11 months 12 months 13 months

/r,/,/,;.r" For manuallY operated guillotine: Monthty expenses =

[ffi),.r.O

units)

= P 80,000

P 2,091,639.12

For electrically operated hydraulic guillotine:

P 1,916,858.24 P 1,856,120.53 P 1,790,274.78

Cost Per unit = 0.70(25) = 17.50 Monthry expenses =

Let x = amount invested in real estate y = amount invested in T-bills, money market placement and stock Since annual income of y is P 50000, thus 0.30y(0.1 2) + 0.35y(0. 1a) + 0.35y(0.1 3) : 50000

0.1305y = 50000 Y

= 383141.76

x=2300000-y x = 2300000 - 383141 .76 x = 1916858.24

.'. The amount she invested in real estate is P 1,916,858.24.

tTfl)0200

units)

= P 56,000

Amount saved per month = 80000 - 56000

24,OOO

cost of machine Number of months to recover the machine = amount saved Per month

_275000 24000 =11.46 months The machine can be recovered in 12 months'

Quest,io' Ba,k

- l)ngi'e.rirrg

llco,onrics by J:rrr,r,

,l'iong

Present EconomY 21 Net income = 28000 - 400

a new gold mining area in Southern Leyte, the ore contains on the average -ln of.ten ounces of gold per ton. Different methods of processing t.ort"t"o

"r"

follows:

Processinq Method

Cost per ton P 5,500 P 2.500 P 400

A B C

A. B. C. D.

= P 27600

.j. Processing Method A yields the biggest return'

% Recoverv 90 80 70

r\^ ^aaaccanr crarinn of the enoineS. lRf

MotOrS lnC. will pay freight On

Processing method A

Processing method B Processing method C Either of the processing methods B or C

,%zn* For Processing Method A: lncome per ton

A. B.

C. P 650 Per engine D. P 610 Per engine

P 670 Per engine P 630 Per engine

+ooo'][ to ounces)fo.got

- [p \ ounce /(

ton )'

lf shipped in container:

= P 36,000

Net income = 36000 - 5500

cost or rreight

= P 30,500

p +ooo

)[

t"".(!#)

t o ounces

ounce

[#}..

= 1 20000

For Processing Method B: rncome per ton

ton )')ro.aot

= P 32,000 = 20 engines

Net income = 32000 - 2500 = P 29,500

cost per englne = For Processing Method C:

cost of freight

nr-m0", of ensine"

'n20000 1

lncome per ton

- [p

+ooo'1( t o ounces

ounce = P 28000

ton )(o.zot )' ,

= P 6,000

lf shiPPed in crates:

TIP ARLEEUI LEBARY

Quesl,ion Bank -- l,)nginecrirrg F.lconomics by Jaime R. Tiong

Present Economy 23

so szo kg cost of freight per ensine - [eg ][t )

Totat cost =

\ t(g /\ engrne ,

= 5320

[t,'oo l.ks \ru )'

*n)

= 5000

Cost of crate per englne = 50 Cost per engine = cost of freight + cost of crate

Pr oflt = Total income - total cost

= 5320 + 50

= P 5,370

Using ceramics:

Solving for the difference of cost:

14-!l9rsYq}.Y=Sl(s \ hour /1. day /\ month /'

rotar output in 2 morrths time = f

Difference=6000-5370 Difference = 630 .'.

montns)

= 72000liters

Shipping the engine by crates is more economical by P 630 per engine than shipping by container.

rotar income

[*9]t I liter /

rooo liters)

= 1 080000

A paint manufacturing company uses a sand mill for fine grinding of paint with an output of 100 liters per hour using glass beads as grinding media. Media load in the mill is 25 kg costing P 200.00 per kg and is fully replenished in 2 months time at 8 hours per day operation, 25 days a month. A ceramic grinding media is offered to this paint company, costing P 400 per kg and needs 30 kg load in the sand mill, but guarantees an output of 120 liters per hour and full replenishment of media in 3 months. lf profit on paint production is P 15 per liter how much is the difference in profit between the two media?

A. P 436,900 B. P 462,000 c. P 473,000 D.

,9"2',2,*. Using Glass Beads:

rotaroutput

[',090']eo *nl

Iks

= 12000

Profit = Total income - total cost =1080000-12000 = 1 068000

The ceramic is more economical as a grinding media with a difference in profit of P1,068,000 - P 595,000 = P 473,000

P 498,200

rotat cost i

in 2 months time =

[,1qq]lgrsYq$trlf \ hour day ./\ month /'

=+)(z

= 40000liters

rotarincome

([99'ltooo00 liter /'

titers)

= 600000

ii ,ii[t:

;i r,'

t'1

montns)

Question Bank

Engineoring Economics by Jaime R. Tiong

Question Bank

It P1000 accumulates to P15P0 when invested at a simple interest for three years, what is the rate of interest?

A. B. c.

t(r*t

14.12% 15.89% 16.67 % 16.97 %

',,

F = P(1+ in)

15oo=tooo[t+i(s)] i= 0.1667 i

=16 67%

.'. The rate of interest is 16.67 %.

You loan from a loan firm an amount of P 100,000 with a rate of simple interest of 2g7o but the interest was deducted from the loan at the time the money was 5fiowed. lf at the end of one year, you have to pay the full amount oiP 100,000, what is the actual rate of interest?

A. B. c. D.

23.5Yo 24 7 nto

25.0% 25.8%

,'/n7)zrn.' lnterest = 0.20(1 00,000) lnterest = 20,000 Amount received = 100,000 Amount received = 80,000

20,000

Question Bank

Simple Interest & Discount 27

Engineering Economics by Jaime R. Tiong

20,000 = 80,000(i)(1) i = 0.025

J. Reyes borrowed money from the bank. He received from the bank P 1,842 promised to repay P 2,000 at the end of 10 months. Determine the rate of

i=25% .'. The actual rate of interest is 25

12.19%

o/".

12.030h

11.54% 10.29% A loan of P 5,000 is made for a period of 15 months, at a simple interest rate of '1 5%, what future amount is due at the end of the loan period? F = P(1+ in)

A. P 5,937.50 B. P 5,873.20 c. P 5,712.40 D.

2ooo =

P 5,690.12

n+zlt.'l'19)l

\12i1

i= 0.1029 i=10.29%

,9urr*"

The simple interest rate is 10.29 %.

F = P(1+ in) = sooolr.0.',u[E)l 'r --""1' "'"[tz)] F = 5,937.50

.'. The future amount at the end of the loan period is P 5,937.50.

annum, what is the borrowed P 10,000 from a bank with 18% interest per ---amount to be repaid at the end of one r

year?

P 10,900 P 11;200 P 11,800 P 12,000

lf you borrowed money from your friend with simple interest ol 12o/o, find the present worth of P 50,000, which is due at the end of 7 months.

A. B. c. D.

F = P(1+ in)

P 46,728.97 P 47,098.12 P 47,890.12

F=1O,ooo(+0.1S) F = 11,800

P 48,090.21

The total amount to be repaid at the end of one year. is P

,800.

c'*r"rn F=P(1 +in) so,ooo=P[1

..,r(#)]

P = 46,728.97

.'. The present worth is P 46,728.97.

tag of P 1,200 is payable in 60 days but if paid within 30 days it will have a discount. Find the rate of interest.

()uestion

llnnk

l,)ngirrooring lllconorni<:s by Jaime R. Tiong

Simple Interest & Discount 29

h,,,

/,/,,/,i,,,..

0.03(1,200)

Discount = Discount = 36

Solving for rate of discount, d

.80 o=-

2,000 d = 0.04

Amount payable in 30 days = 1,200 - 36 Amount payable in 30 days = 1,164

d = 4o/o

Solving for the rate of interest,

36=1,164(i)[#) i= i

0.3711

= 37.11o/o

1-d 0.04

1- 0.04

The rate of interest is 37.11 %.

= 0.0417

t= 4.17% ,', The rate of interest is 4.1T

o/o.

A man borrowed P 2,000 from a bank and promised to pay the amount for one year. He received only the amount of P 1,920 after the bank collected an advance interest of P 80.00. What was the rate of discount? tIVhrl wlll be the future worth of money after 12 months, if the sum of p 2s,ooo is lltvoatcd today at simple interest rate of 1o/o pet leat?

A. 3.67 % B. 4.OO c. 4.15 D. 4.25 o/o

A, 3, c.

0/o o/o

P 25,168

P 2s,175 P 25,18e P 25,250

Solving for rate of discount, d

.80 d=-

'4ta,, F = P(1+ in)

2,000

d = 0.04

F = 2s,0oo[1* (o.or)!99.1

L '

d = 4o/o

',360]

F =25,250

.'. The rate of discount is 4 %.

;. The future worth is P 25,250. A man borrowed P 2,000 from a bank and promised to pay the amount for one year. He received only the amount of P 1,920 after the bank collected an advance interest of P 80.00. What was the rate of interest that the bank collected in advance? 4.OO

o/o

4.O7

o/o

4.17 4.25

o/o

lltlrt

witt be the future worth of money after 12 months, if the sum of p 25,000 is lnYrcteo today at simple interest rate of 1%o per month?

A. B, c.

o/o

- ,)L

P 28,000

P 28,165 P 28,289

P 28,250

Questiorr

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l,)rrgirrorrringltrconornir:s lrv.Jrrirno

lt.'l'iong

Sirnple lntcrest & Discount 31 Net interest = l- 0.201 Nel interest :0.801

.22,t ,,. F=

P(1+ in)

25,000[1+ o.o1(r z)]

890.36 :0.80t I

Bul

F = 28,000

.'. The future worth is P 28,000.

1,1

12 s5= i=

It is the practice of almost all banks in the Philippines that when they grant a loan, the interest for one year is automatically deducted from the principal amount upon release of money to a borrower. Let us therefore assume that you applied for a loan with a bank and the P 80,000 was approved at an interest rate of 14 o/o of which P 11 ,2AO was deducted and you were given a check of P 68,800. Since you have to pay the amount of P 80,000 one year after, what then will be the effective interest rate?

A. B. c. D.

A. B. c, D,

o/o

(

11.50

o/o

11.75% 11.95% 12.32

o/o

P 5,066.67 P 5,133.33

P 5,050.00 P 5,166.67

F = 5,166.67

,'. The amount due at the end of 75 days is P 5,166.67.

A bueinessman wishes to earn 7% on his capital after payment of taxes. lf the lnoome from an available investment will be tafed at an average rate of 42Yo, what minimum rate of return, before payment of taxes, must the investment offer

lo be justified?

A. B. c,

,%dz,*.

Yo.

r = s,ooolr. (0.16)f=.)l ',(360/l

i= 0.1628 i=16.28%

A. B. c. D.

is 11.75

F = P(1+ in)

11,2oo = 68,800(i)(1)

A deposit of P 110,000 was made for 31 days. The net interest after deducting 20% withholding tax is P 890.36. Find the rate of return annually.

o.1175

rt/,,,,.

.'. The effective rate of interest is 16.28 %.

ooo(i)(#)

lt 0,000 is borrowed for 75 days at 16% per annum simple interest How much wlll be due at the end of 75 days?

o/o

1 1 o,

,'. The rate of return annually

16.32% 16.47

i = 11.75o/o

16.02o/o

16.28

= 1,112.95

12.07 %

12.34% 12.67 12.87

0/o

(lrrcst,irrrr

llrrrrk

Simple Interest & Discount 33

l,)rrgrru,r,t'ittg I!<:ottoutit's lry.lairne It. Tiong

3,200 = (2o,ooo

profft

Rate of return =

invested caPital

i:0.1905

i= 19.05%

Rate of return =

,,, The actual rate

- 3,200)(i)(1)

of interest is 19.05 %

where: p = profit C = invested capital

? .1,000 is borrowed for 75 days at 16 bc due at the end of 75 days?

After taxes are paid, net Profit invested capital Rate of return = 7 oh = O.07 Rate of l9turn -

07: --- 0.42P

0.07

P 4,666.67

'(b^

0'58P

F=P(1 +in)

per annum simple interest. How much

A, P 4,033.33 l, P 4,333.33 C, P 4,133.33

Substituting values: o

o/o

Wlll

F=4ooo[.,.r,.(#)]

= 0.1207

-C =lz.ot% .'. The rate of return before payment of taxes is 12.07

F = 4,133.33

A mgn borrowed P 20,000 from a local commercial bank which has a simple interest of 16% but the interest is to be deducted from the loan at the time that the money was borrowed and the loan is payable at the end of one year. How much is the actual rate of interest.

A. B. c. D.

,,,

o/o

Thoamount due in 75 days is

P 4,133.33.

Agntr Abanilla was granted a loan of P 20,000 by her employer CPM lndustrial lfbrlcator and Construction Corporation with an interest bf 6 % for 180 days on

ho prlncipal collected in advance. The corporation would accept a promissory iOh for P20,000 non-interest for 180 days. lf discounted at once, find the pDoreds of the note.

A, P 18,800 l, P 18,900 C. P 19,ooo

19.05 % 19.34 0/o 19.67 % 19.87 %

P 19,100

,'L.*

"V;/'r.o,,

lnterest = 0.06 (20,000) = 1,200

r= 0.16(20,000) I

Proceeds on the note = 20,000 Proceeds on the note = 18,800

= 3,200

But l= Pin

12OO

r'. The proceeds on the notes is P 18,800.

Substituting the values:

{

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lf you borrow money from your friend with simple interest of 12%, find the present worth of P 20,000, which is due at the end of nine months.

A. B. c. D.

Simple Interest & Discount 35

I,)rtgirtrrrring Econottrics by Jaimtr Il..'lliong

P 18,992.08 P 18,782.18 P 18,348.62 P 18,'t20.45

Mr, Danilo conde borrowed money from a bank. He receives from the bank P1,340.00 and promised to pay P 1,5oo.oo at the end of 9 months. Determine Tala of simple interest

A. B. c. D.

15.92o/o

15.75% 15.45% 15.08 %

(*h,t F = P(1+ in)

F = P(1+ in)

20 ooo = p[r

2f3zq).l -' L'* 0.1 " '-(aooJl

1,500 =

t.scol t.,r?29)l

P = 18,348.62

(360/l

o.11g4 -270

360 i.= 0.1592

The presentwo;th is P 18,348.52.

i =15.92o/o

A man borrowed from a bank under a promissory note that he signed in the aniount of P 25,000.00 for a period of one year. He receives only the amount of P21 ,915.00 after the bank collected the advance interest and an additional of P 85.00 for notarial and inspection fees. What was the rate of interest that the bank collected in advance?

A. B. C. D.

13.05% 13.22%

,', The rate of simple interest is 15.92 %.

Ml, J, Dela Cruz borrowed money P1,340.00 and promised to pay P lh! corresponding discount rate or

A, B, C, D,

13.46 o/o 13.64 %

.7;7)a-,,-.

Amount received = 21 ,915 + 85 Amount received = 22,000

13.15 0/o 13.32 o/o 13.46 o/o 13.73Yo

(tf,.b,. Solving for simple interest rate,

lnterest = 25,000 -22,000 lnterest = 3,000

F = p(1+ in)

1,500=1,340[1+(#)]

3,000 = (22,000xi)(1)

i= 0.1364 i=13.64% .'. The rate of interest the bank collected in advance is 13.64 %.

rec€ive nd of g as the

0.1194

-27o

360 i= 0.1592

i=15.92%

ine nt,,.

9::'tu' lly!

li:l*l'1""1,,*

Ecr)norrrir:s tr.y Jairne R. Tiong

Simple Interest & Discount

Solving for the discount, d

.t d=1- | i+i d=1-

lnterest = 1,500.00 _1,342.00 lnterest = 15g.00

Also, 1

1 + 0.1 592 d = 0.1373 d =13.73%

.'. The discount rate

rsa

r,s+z1nf!)

\4)

i = 0.'1569

1S.gZ%.

i= 15.69%

ffii:ffnil:FJ:',13i'"t ca

h p rice on

,', The simple interest rate is 15.69%.

n"m a merchant who ask p 1,2soat the end o160

TTidll"]l ",.",, -?,

-: l.d

T "l91,r

;rf;' to ;;i

ute

Ddormine the exact simpre interest on p 5,000 invested ror ttre perioo rronof interest is 180/o.

J'nuery 15' 1996 to october rz, rsisi,

P 1,'t24.67 P 1,233.5s P 1,289.08 P 1.302.67

;;ffi:

B, P 664.89 c, P 665.21 0, P 666.3e

F = p(1+ in) 1,250 =

pL.o oa[jq)l (

soo/]

+Eq.1

Solving for the total number of days money was invested, d

P = 1,233.55

.'. The cash price of the television set is p 1,233.55.

A man borrowed money from a roan shark. He rgleryes from the roan shark and

;'rHI[ A. B. C. D.

t",,ti"hfi:'fl?"?:3,1.i",g""0 '" '"p"v

i''r,soo^oo

ir,l'","0'o',", 0,"n",.,

15.47 % 15.69 % 15.80 o/o 15.96 o/o

June July August

September Octoberl-12

= 30 = 31 = 31 = 30

=12 271 days

Substitute values in Eq.

t= s,ooo(o.1s,("1\ . 1366 /

l= 666.39

u^*

Sinrplc lnt,erotrl, & l)iscourlt' 39

(,lut'strorr

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l,)rtgtttocriltg [lconottticr lrv ,lrtirtto lt. Tiong

. The exact slmple lntereet is P 666.39.

,'fi(/,,, F = P(1+

The exact simple interest of P 5,000 invested from June 25, 1995 is P '100. What is the rate of interest?

A. B. c. D.

December ' 1995 to

o/o

3.95 % 3.98 %

.'/./ot;,,

Eq.

Solving for the total number of days, d

June21-30

July

=31

August September October November

Eq'

was invested' d Solving for the total number of days the money

- 30 = $ = 31 May = 30 June = 31 July = 31 August September = 30 October = 31 November = 30 December 1-25 =E-

April 22

3.90 % 3 92

in)

= = = =

247 daYs

Substitute in Eq.

1o,ooo=P[1.(oro)(#)]

P = 8,807.92

Decemberl-25=25 187 dayt

her mother on her birthday is P ,', The amount Nicole received from 8,807.92.

Substitute values: 1oo

: s.ooo(i)f]!Zl ' "t365/

i= 0.0390 i= 3.90% .'. The rate of interest is 3.90 %.

On her recent birthday, April 22,2001, Nicole was given by her mother a certain t on sum of money a Present. an lf-the ac 20% exact simp her w much amount of P 10, birthday?

A. B. c. D.

P 8,807.92 P 8,827.56

P 8,832.17 P 8,845.78

182 days at 5'2%? t rhat it the ordinary interest on P 1,500'50 for P 39.01 P 39.45 P 39.82 P 39.99

,'&,an' l= I

= (1,500,

l= 39.45

50)(o.o5r)[#)

Quest iorr

llrrrrk

lrltrgrrroorirrg lj<:onorrrrr.x lry ,Inirno R. Tiong

Question Bank 3

Nicole has P 20,400 in cash. She invested ll alTo/o from March 1, 2006 to November 1, 2006 at 7% interest. How much is the interest using the Banker's Rule?

A. B. c. D.

P 972.12 P 970.78 P 973.',t2 P 971.83 Tho amount of P 20,000 was deposited in a bank earning an interest of 6.5% per

tnnum. Determine the total amount at the end of 7 years if the principal and

1rrl,,2*.

htlrrcet were not withdrawn during this period. ln Banker's Rule, use exact number of days the money invested and 360 days for the total days in a year. Solving for the exact number of days:

A. t, c, 0,

P 30,890.22

P 30,980.22 P 31,079.73 P 31,179.37

Marchl-31=30 April= 30 MaY = 31

June = 30 July = 31

August = 31

r=e(t+i)n F = 20,OOO(1+

0.065r

F = 31,079.73

September:30 October = 31 November 1 = 1

,'. The total amount at the end of 7 years is P 31,079.73.

Total = 246 A Joan for P 50,000 is to be paid in 3 years.at the amount of P 65,000. What is the dhcrtive rate of money?

l= Pln t= (2o.4oo)(o \ ' /\ I

= 971.83

07)r?€) ,[360/

A. 9.O1% B. 9.14 c. 9.31Yo D. 9.41% o/o

r=e(+i)n 65,000=SO,OOO(1+i)3

t.a=(t+i)3 1.0914 = 1+ i i = 0.0914

Quost,ion

llank

lr)rrginooring Econorrrir.r by.Jaime R. Tiong

(

The present worth is P 40,530.49.

1.0914 = 1+ i i= 0.0914

Since nothing was mentioned about how the interest was compounded, it presumed that the mode of compounding is annually. .'. The effed.t rate is 9.14 %.

$flut

lg the effective rate corresponding to 18% compounded daily? Take 1 year

aqu.l to 360 daYs.

A I

c, D

The amount of P 50,000 was deposited in the bank eaming an interest ol7.So/o annum. Determine the total arnount at the end of S years, if the principal and interest were not withdrawn during the period.

A. B. c.

19,61

0/o

19,440h

1s.31% 19.72%

(**ER=(1+if-t

P 71,78',t.47 P 71,187.47

P71,817.47 P 71,718.47

ER =19.72%

(tzz.

)ortrpottrttl I rrl,t,rtttl nnd Cont.iuuous C

0.18)360 _1

=[r+ 360/ \

,', The effective rate corresponding to 18% compounded daily is 19.72o/o.

F = P(1+ i)n F=

5o,o0o(1+o.o75f

F =71,781.47

.'. The tota! amount at the end of 5 years is P 71,781.47.

Find the present worth of a future payment of P 80,000 to be made in six (6) with an interest of 12Yo compounded annually.

A. B. c.

.7"2*.*

P P P P

40,540.49 40,450.49 40,350.49 40,530.49

yields the same amount as \Mltt nominal rate,.compounded semi-annually, a

Slnpounded

quarterly?

A, 16.09 % l, 16.32 c, 16.45% D, '16.78% 0/o

12,h. ER,

= EP,

[,.T)'-,=(,.0i)'-, o'ro)o l..,*!B)'( 4) ( 2/ [.,* [.,

rB)'

NR 2

='r.16e8556 = 1.0816

16%

()ucstiorr

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Solving for the discount, d

lnlerest

Simple Interest & Discount 37 1,500 00

- .t,342.00

lnterest = 158.00

d=1_1 1+i

Also,

d=1-

1 + 0.1592 d = 0.1373

158 =

d=13.73%

1,342({;)

i= 0.1569 i= 15.69%

,'. The discount rate is 15.92 %.

The simple interest rate is 15.69%. Annie buys a terevision set from a merchant who ask p l,zsoat the end of 60 ty, n J ir, orers to comiute tne 1'\,ll:1vlil-^1" ry^,,lmed.iare cash price on the assumption "'rn ","nani thar mor"y ir *ortn aL ,inr, ,. ,n" cash price?

::f

liil;ft;#.

for the perioo rromorinterest is 18%.

A. P 1,124.67 B. P 1,233.55 c. P 1,289.08 D.

$ermine the exact simpre interest on p 5,000 fnuary 15, 1e96 to october rz,1sigo, ;tiJI]t;invested P 664.89

",B. c. P 665.21 D. P 666.3e

P 1,302.67

F=P(1+in) l= 1,250 =p[r.

o.oa[i9)l (s60i.l

Solving for the total number of days money was invested, d

P = 1,233.55

January

.'. The cash price of the television set is p 1,233.55.

C. D.

.V;z*--

15.47 o/o 15.69 % 15.80 o/o 15.96 %

to repay

1,500.00

1S-31

February March April

man.bollory"-d Iamount rgney from a roan shark. He receives from the roan shark and of p 1,u2.00 and^promised p

What is the simple interest rate?

-+ Eq. 1

tn" Lna

otl

quarters.

May June July August September October 1 - 12

= 16 =

29 (since 19g6 is a leap

=31

=30 = 31 = 30 = 3.1 = 31

=30 =12 271 days

Substitute values in Eq. r

s,ooo(o.18,{!1\ ' '\366'

t= 666.39

year)

(1)ucstrorr

llrrrrlr

Simplc lnt,ereel, & Discoupt 39

lr)rrgrrrrrr.rirrg l,lt:ortorrricx by ,lnime R. Tiong

The exact clmplo lnterest ls P 666.39.

'Aaa

F = P(1+

The exact simple interest of P 5,000 invested from June 21, 1995 to.December 25, 1995 is P 100. What is the rate of interest?

A. B. c. D.

April22-3O

3.95 % 3.98 %

Eq.

=$ = 31 = 30 = 31 = 31 = 30 = 31 = 30

May June July August September October November

3.92%

Eq.

Solving for the total number of days the money was invested, d

3.90 %

in)

Decemberl-25=25

247 days Solving for the total number of days, d

June21-30

JulY August September October November December

1-25

Substitute in Eq.

=g = 31 = 31 = 30 = 31 = 30 =.25

1o,ooo=P[1.rr.,(#)] P = 8,807.92

The amount Nicole received from her mother on her birthday is P

187 day$

8,807.92.

Substitute values: 1oo = i=

s,ooo(i)fEz) "\ 365

What is the ordindry interest on P 1 ,500.50 for 182 days at 5.2 %?

0.0390

A. P 39.01 B. P 39.45 c. P 39.82

i= 3.90% .'. The rate of interest is 3.90 %.

On her recent birthday, April 22,2001, Nicole was given by her mother a certain sum of money as birthday present. She decided to invest the said amount on 20o/o exact simple, interest. lf-the account will mature on Christmas day at an amount of P 10,000.00, how much did Nicole receive from her mother on her birthday?

A. B. c. D.

P 39.99

,ttdlL/,an

P 8,807.92

(1,

5oo, 50)

t= 39.45

P 8,827.56 P 8,832.17 P 8,845.78

--,.It,.

(o.o5r)(ffi)

Quesliorr

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l')rrgrrroorirrg l!)<xrrtorttu x lry JaiIno R. 'liong

Question Bank 3

Nicole has P 20,400 in cash. She invested rl al7% from March 1, 2006 to November 1, 2006 at 7% interest. How much is the interest using the Banker's Rule?

A. B. c. D.

P P P P

972.12 970.78 973.12 971.83

Thc amount of P 20,000 was deposited in a bank earning an interest of 6.5% per annum. Determine the total amount at the end of 7 years if the principal and lntcrest were not withdrawn during this period.

,'/;Zr,ln Banker's Rule, use exact number of days the money invested and 360 days for the total days in a year. Solving for the exact number of days:

A. B. c,

P 30,890.22 P 30,980.22 P 31,079.73

P 31,179.37

Marchl-31=30 April= 30 MaY = 31

r=e(t+if

June = 30

F = 2o,ooo(1+

JulY = 31

o.o65f

F = 31,079.73

August = 31

September:30 October = 31 November 1= 1

Iotal=246 l= Pln t= (2o.4oo)(0.07)f?€) .",/[360J I

= 971.83

,'. The total amount at the end of 7 years is P 31,079.73.

A loan for P 50,000 is to be paid in 3 years.at the amount of P 65,000. What is the

dlbaive rate of money?

A. B.

9.01 9.14

9.31% 9.41%

o/o o/o

r=e(t+i)n 65,000=so,ooo(t+i)3 1.3=(1+i)3 1.0914 = 1+ i i= 0.0914

50,000

......_

61rr0lw

11]lrr_oil1,1,11r,k. _U,,rinooring Economics by Jaime t.s = 1r+i;

R. Tiong

1.0914 = 1+ i i= 0.091a

tfre pres"nt worth is p 40,530.49.

since nothing was mentioned about how the interest was compounded, it is

presumed that the mode of compounding is

annually.

.'. The effect rate is 9.i4 %.

The amount of p 50,000 was depositeo in tne bank eaming an interest of 7.5% annum. Derermine rhe rotar amount at *,e eno oi s i; il;'pri"ipri interest were not withdrawn Ouring ifre p"ii"il -

G;;

A. B. c. D.

/ I

""0

A I c D

19.610/o

19.44% 19.31% 15.72%

(*aEn=(r+if_r

P 71,78't.47 P 71,187.47 P 71,817.47 P 71,718.47

=[r+S)'uo \ 360/ -,

ER =19.72o/o

The effective rate corresponding to 1g% compounded dairy

1g.72%.

F = P(1+ i)n F = so,ooo(1+o.o7s)s

Wtll

F:71,781.47 .'. The total amount at the end of

P 40,530.49

E |

(1+i)n P

80,000 (1+ 0.12)6

P = 40,530.49

r"t",

semi-annually, "o.no1ilffi,a,,..

A, 16.09 % !, 16.32% c, 16.45 % D, 16.78 %

71,781.47.

Find the present worth of a rrtrr" pryrr,"r,t of p g0,000 to be made ln six 1o; with an interest of 12o/o compounded'"nnrrlfy.

A. P 40,540.49 B. P 40,450.49 c. P 40,350.49

nominai

ompounded qr"rt"1y?pounded

yea) ER, =5P,

(,.T)'-,=[,.T)'-,

[,.T)'=(,.T)' [,.T)'=116e8s56

1.T

= 1.0816

._:_,

yields the same amount as 16%

llrrrk

()trcstiorr

(

l,lrrgirrr,r.r'irrg l,lr:rrrrorrrir:s by .Jaime R. Tiong

l!!

o 0816

2 NR = 0.1632

(r , i)o

1,,

But,

-m

8.07

o/o 0/o

NR = 0.'t183 NR = 11.83%

What rate of interest compounded anrtually is the same as the rate of interest of 8% compounded qua(erly? 8.12 8.16

.0.029563

o o2s563 =

A. B. C. D.

t')'

(1ri)a 11236

NR = 16.s2 %

The nominal rate is 16.32o/o.

lorrtltottttrl l rtl.t'r'rl rrtrtl ( lrttl itttttttts ()orrtpotrnding 45

. Ths nominal rate's 11.83%'

o/o

8.24% interest? ffil-O,, of tn".e gives the lowest effective rate of

.V;Z*r-". ER' =EP'

(1+i)1-

t=(t*IB)o -', 4,l

[

(

(1+i)=11+

o.o8

.)4

1+i=1.0824

i= 0.0824 i = 8.24

.'. The rate of interest compounded annually is 8.24%.

A, 12.35% comPounded annuallY a 11.90 % compounded semi-annually C, 12.20% comPounded quarterlY monthll D,

11.60 % comPounded

'(,J*-, Solving for the effective rates:

ER = 12.35 % (if compounded annually, NR = ER

B. ER=(1+i)"-t en=[r.9-119)'-r ER = 0.1225

Flnd the nominal rate, which if converted quarterly could be used instead oI 12oh compounded semi-annually.

A u (: I)

'|.1.93

ER=12.25%

0/o

rn=[r,.9J??)'-r

11.09 % 11,85 % 11.26 0/

ER=0.1277 ER = 12.77o/o

tRr

-ER1

en=(r*9119)"-r

(lrrcslion

llrrnk

l,)rrgirrr,r,r.irrg I,l.orrorrrir.s lry ,lirinrc R. Tiong

(

ER. 0.1224

lrrrrrltrrttntl

l ttl

r,tr"rt ttrttl ()orrt ilrutttts Ooltlpoun dtng 47

But,

.NR l=-

ER =12.24%

.'. Thschoice D provides the lowest rate of interest.

002=!E 6

NR = 0.12

Find the compound amount if p 2,500 is invested at g% compounded quarterry for 5 years and 6 months.

A. B. c. D.

NR = 127o

P 3,864.95 P 3,846.59 P 3,889.95 P 3,844.95

,', The nominal rate is 12"/o.

5 years & 6 months

5.5 years

r=n(t+if .15

2'5oo(1*

!'oe

A, P 6,080.40 !, P 6,020 40 c. P 6,040 20

utol

F = 3,864.95

pi OOo shall accumulate

for 10 years at 87o compounded quarterly, then what lha compound interest at the end of 10 years?

2,s00 i....................

""'> F

P 6,060.20

l*oth',,

r=eft+if

The compound amount of p 2,000 after 5 years and 6 months is

P3,864.95.

F=s,ooo[1*ofu)'o''' F =11,04020

An amount of P 1,000 becomes p 1,60g.44 after 4 years compounded bimonthry Find the nominal interest.

A. B. C. D. ,

lnterest=F-P lnterest = 1 1,040.20 lnterest = 6,040.20

11.89 o/o 12.00 0/o 12.08 o/o 12.32 o/o

5'000

6'040'20' ,'. The compound interest at the end of 10 years is P

l-J2z -/../t//rn

18% compounded semi-quarterly? W-nrt i" the corresponding effective rate of

r=e(+i)n

A. B. c. D.

1,608.44= 1,s66(1 * ;)+to)

1.AOAI+*

=t+i

1.02 = 1+i

i=o.o2

19.24% 19.48

o/o

19.84% 19.92%

f k)*,,.

gp = (r+i)m._1

__l-.

48_ ,,),::Il iorr

llrrrk

()ornllotttrtl I ttlrrrtrrl, nnrl Continuoue Compounding 4!)

-, lr)nginocring li)conornir:x by Jaime R. Tiong t---

rn=lr*o,rtlt ( 8,

F=P(1

+i)2n

, 2ooo + 3ooo

ER = 0.1948 ER = 19.48 %

.'. The corresponding effective rate is

= 2ooo(1*

ol'l'"

[

2.5 = (1.06)2n

. 19.4go/o.

Take log on both sides: log2.5 = tog(t.oo)2^ log2.5 = 2nlog1.06 n = 7.86 years

present worth of a future payment of p 100,000 to be made in 1 0 years interest of 12o/o compounded quarterly.

)

B. P 30,656.86 c. P 30,556.86 D.

,,, Thc money

will increase by P 3,000 at in 7'86 years'

P 30,655.68

ifr-r,mount of P 150,000 was deposited i2the bank earning q1 interqs!9f-]

F (1

+ i)n

P 215,U4.40 P 213,il4.40 P 2Uj53.40 P 255.43.10

100,000

P= ^

f, *

o.rz'14(10)

( l

P = 30,655.68

.'. The present worth is p 30,655.6g.

r=e(t+if F = 1 50,

OOO (1

+ 0.075)s

F =215,344.4O

ln how many years is required for p 2,000 to increase by p 3,000 if interest at 12% compounded semi-annually?

,'. The total amount at the end of 5 years is P 215'344'40'

A. 7.86 years B. 7.65 years C.7.23years

8.12 years

HOn, tong

will it take money to double itself if invested at 5% compounded

f,lnuallY?

A. B.

C. D.

,%*^

__..L

13.7 Years 14.2 years 14.7 years 15.3 years

-S-oZ

r -!tF'

Quesl,iurr

llntrk

I,)rrginooring l,lt:onuurit:x by Jairne Il,. Ti

r=n(t+if

()ontJxrurrtl Ittit'rr,xl rrntl (lont,inuorrs Oompounding 51

,;-

To double the money, F = 2P

ER= (1+i)"-1

2P = P(1+ i)n

z=(r+o.osf

en =

Take log on both sides

o [r* ]o)' -, l. 2)

ER = 0.1449 ER = 14.49 %

log2=nlog1.05 loo2

log1.05

Thc effective rate is 14.49%.

n = 14.2 years

.'. The money if invested at 5% compounded annually will double in 14.2

yeats.

what

Aan lnterest rate of 10% compounded annually, how much will a deposit of ll,tOO be in 15 Years? P 6,265.87 P 6.256.78 P 6,526.87 P 6,652.78

is the corresponding effective interest rate of 1g% compounded semi-

monthly?

,'fu'

A. 19.35% B. 19.84% c. 19.48% D. 19.64%

r=P(1+i)" F = 1,500 (1 + 0.10)15 F = 6,265.87

ER=(1+i)'-r

,.. The amount of P 1,500

will become

P 6,265.87 after 15 years'

en=[r+0.18)24_1

24)

:0.1964

ER = 19.64%

.'. The corresponding effective rate is 19.64 %.

What is the effective rate of 14% compounded semi-annually?

A. B. c. D.

14.49% 14.59% 14.690A 14.79 o/o

P 25,000 in 8 years. How much is that money worth considering interest at 8% compounded quarterly?

fiman expects to receive

i[r

lA. P 13,256.83 'iB. P 13,655.28 ,l c. P 13,625.83 fl o. P 13,205.83

()uort,iorr

llrrrk

l{ltrgirxrorilrg lii:onornir:r bv Jairne lt. Tiong

0r23

'b* ERquarterty = ERsemi -annually

o.oo

( * 2 ))2(8) 1.,

()orrrporurrl Inlorr,nt rrnd (lontinuous Compounding 53

P-n"""""'

[,nl)o -t=(t+o06)2-1 4) 2

P ='13,265.83

)

i= 0.0596. i= 5.96%

.'. The present worth is P 13,265.83.

,,, Thc equivalent rate is 5.96 %.

About how many years will P 100,000 earn a compound interest of p 50,000 if the interest rate is g% compounded quarterly?

A. 4 years B. 5 years C. D.

t{hfl lr the amount

A, t, 0, 0,

6 years 7 years

of P 12,800 in 4 years at 5 % compounded quarterly?

P 15,461.59 P 15,146.e5 P 15,641.59 P'15,614.59

9"2*, r=e(r+if F=P(1 +i)n

loo,ooo + 50,000 = 1oo,ooof1+ o.oe)4N

1.5 = (1.0225)4N

F = 15,614.59

Take log on both sides log1.5 = 4N1o91.0225 N=

,', The amount after 4 years will be P 15,614.59.

-']991'5 4(1o91.0225)

N = 4.56 years

.'. The money will double in 4.56 yearsi or about 5 years.

ly he condition of a will, the sum of P 20,000 is left to a girl to be held in trust fund Oy her guardian until it amounts to P 50,000. When willthe girl receive the lltollcy if the fund is invested at 8 % compounded quarterly?

A. !. compute the equivalent rate of 6% compounded semi-annually to a rate compounded quarterly.

A. B. c. D.

5.12% 5.96 %

5.78% 6.12%

o'05

= 12.Boo(( 't* 4) )4(4)

C. D.

11.23 11.46 11.57 11.87

years years years years

Qucst,ir-rn

lltnk

l,)rrgirxroring Iimnomics by Jaime R. Tiong

(

50,ooo = zo,ooo Ir + 9]?8 2.5 =

Oornglounrl lrrlr,r'oni nnd Continuous Compounding 55

\4n

4-.J

.o2)4"

Take log on both sides:

20,000

...'...-

50,000

1,000

log 2.5 = log(1.02)an log 2.5 = 4n log 1.02 o= loo" 2.5 4log1.O2 n = 11.57 years

. P(1+ l)n

.'. The girl will receive the money in 11.57 years.

1,ooo(1+ o.o5)8 1,477.46

Money left after the principal is withdrawn = 1,477.46 lf P 5,000.00 shall accumulate for 10 years at 4 % compounded quarterly, find the compounded interest at the end of 10 years.

A. B. c.

P P P P

0123891016

1000 = 477.46

2,333.32 2,444.32

2,555.32 2,666.32 Fe = P(1+ i)n

.V.z*r-

F=p(1 +i)n

- 477.46(+

0.05)8

Fz -7o5.42 o'04)4(10)

r = s,ooo -'--- [t( + 4 )

,',

?hr wlthdrawal amount at the end of the 16th year is P 705.42.

F = 7,444.32

lnterest=F-P

lnterest = 7,444.32- 5,000 lnterest = 2,444.32

1,t00.00 was deposited in a bank account, 20 years ago. Today it is worth 10,000,00. lnterest is paid semi-annually. Determine the interest rate paid on this

aount,

.'. The compound interest at the end of 10 years ts P 2,444.32

A sum of P 1,000 is invested now and left for eight years, at which time the principal is withdrawn. The interest has accrued is left for another eight years. lf the effective annual interest rate is 5 %, what will be the withdrawal amount at the end of the 16th year?

A. P 693.12 B. P 700.12 c. P 702.15 D.

F=P(1 +i)n 3,ooo = 1,500

P 705.42

t4-

11 +

ilao

+ i)z(zo1

(|uost,iorr

Ilirrrh

1.0175 = 'l+

lr)ngiruror.rrrg l,lr:orrorrrir.s by Jainre

Il. Tiong

Oorrrporrrrrl lnlr,r'r'sl rrrtrl ( ionl,irruorrs Oompounding 57

l1**

i= 0.0175

ER (t+i)'-1

Solving for the nominal rate:

.NR

0.0175 =

.=l

o'12)4

ER = 0.1255 ER = 12.55 %

!E 2

NR = 0.035 NR = 3.5%

Thc effective rate is 12.55 %.

.'. The interest rate paid on the account is 3.5 %.

l{fndrrln Bank advertises 9.5 % account that yields 9.84 % annually. Find how

fifn

s investm the

A merchant puts in his P 2,000.00 to a With a given interest rate on the annually, how much will he collect at

A. B. c.

r a period of six years. year, compounded

year?

A l, 0,

P 4,626.12 P 4.262.12 P 4,383.12 P 4,444.12

thc interest is compounded. Monthly Bimonthly Quarterly

Annually

(h* en = (1+i)m _t 0.0984 = (',

(

F = P(1+ i)"

1.0984 =

F = 2,000 (1 + 0.1 5)6 F = 4,626.12

o'ogs

"l'

m) -,

[.,* o'ogs')'

(

By trial and error, m = 4.

2,(N0,"""""""""""""""'-. 4

,', Since there are 4 interest periods per year, the interest is compounded quarterly.

.'. The amount the merchant will collect at the end of the 6h year is P4,626.12.

I[tcn A man borrowed P 100,000 at the interest rate of 12Yo per annum, compounded quarterly. What is the effective rate?

A. B. C. D.

',t2.750/o

12.55o/o

12.45

olo

12.35%

witt an amount be tripled with an interest of 11.56%?

A. 9 years B. 1o years C. 11 years D. 12years

Quost,ion llank

(

Cngineering ltrconornir:s by Jaime R. Tiong

F = P(1+ i)n

3P:P(+0.1156)n 3=1.1156n

3,ooo(1 +

= 4,392.30

lotttllottttrl

ttl

lrt,xl rurrl (lorttLinuotrs Oompounding

1o)a

Fs = 5,000(1+ 0.1)1 F3 = 5,500.00

Take log on both sides log3=n1o91.1156

Substitute values in Eq.

loo3 1o91 .1 1 56 n = 10.04 years

P + 5,500 =2,657.34 + 4,392.30 P = 1,549.64

,', The amount of money left in the bank 1 year after withdrawal is

.'. The amount will triple in 10 years.

P1,549.64.

A student plan to deposit P 1,500 in the bank now and another p 3,OOO for the next 2 years. lf he plans to withdraw P 5,000 three years from after his last deposit for the purpose of buying shoes, what will be the amount of money left in the bank after one year of his withdrawal? Effective annual interest rate is 10%.

A. B. c. D.

P 1,549.64 P 1,459.64 P 1,345.98 P 1,945.64

must be invested on January 1 , 1998 in order to accumulate P 2,000 Jrnuary 1,2003? MoneY is worth 6%?

l{CrV much

A, !, c, 0,

P 1,509.34 P 1,249.64 P 1,378.98 P 1,494.52

)

+F3 = F, 1P,

-+ Eq.

5,000---.- Ft P,<.'.."..'-.'.-.-.-.-.--...-....- 2,000

,'. The amount invested on January 1, 1998 is P 1,494.52.

F2 F1

F = P(1+ i)n i

I I

= 1,500(1 + 0.1 0)6

=2,657.34

A nominal interest of 3% compounded continuously is given on the arcount. What is the accumulated amount of P 10,000 after 10 years?

A. P 13,498.59 B. P 13,489.59 c. P 13,789.98 D. P 13,494.52

60 (lrrcsl iorr llirrrk

l,,)

gr rrr,r,

(lorttgrorrnrl Itrlr,roxl rrrrrl Oonlirtrrous Oompoundilrg 61

rin g l,)rrrrronr rr.s by .la rrnc tt. Tiong

F = Pe(NR)N

F = 10,000e0

03)10

F:'10,000e(o

10,000 (1+ 0OElo

P = 4,631.93

F = 13,498.59

.'.

The initial amount that must be deposited is P 4,631.93.

The accumulated amount is P 13,498.59.

ll P 000,000 is deposited at a rate of 11.25 % compounded monthly, determine

A mechani account at what shoul

A. B. c. D.

P P P P

umulate a total of P 1O,OOO in a savings bank pays only 4Yo compounded quarterly,

Ompounded interest after 7 years and 9 months

A, P 690,848.73 !, P 670,65'r.23 c, P 680,649.56

6,176.35

D, P 685,781.25

6,761.35 6,716.53

)

%..'

6.167.35

, I I

F=P(1 +i)n

---'---(,

F = 5oO,OO0[1+

P= (1

+ i)n

7 yr

9 mo.

93 months

0.1125)s3

)

lnterest = 1,190,848.73 Jaterest = 690,848.73

F = 1,190,848.73

10.000

(

lnterest=F-P

o.04 \10(4)

[.'. o

P:6,716.53

500,000

,', The compound interest alf'ar 7 years and 9 months is p 690,848.73.

The initial deposit should be P 6,716.53.

Funds are deposited in a saving account at an interest of 8% per annum. what is the initial amount that must be deposited to yield a total of p 1o,ooo in 10 years?

A. B. c. D.

P 4,196.30 P 4,721.39 P 4,796.03 P 4,631.93

&t lnterest rate is quoted as being 7.5% compounded quarterly. What is the lfrcilve annual interest rate?

A, 7.91% B. 7.51 c. 7.7',1 o/o

o/o

D. 7.31%

'l/.*

,'y'.at*.

fn= P=

+ i)n

L--b-

(1+i)m_r o

[r. Tu)' -,

Question Bank

Engineering Econornit:n lly Jaiuro

'l'rortg

(

ER = 0.0771

ER=7.71

.'.

o/o

3,500.00 of taking pay bY b

3 -_...

oo0 (1 + 0.08)10

F =2.158.92

o/o per from a friend at an interest rate of 1'5 much per year' How 18o/o rate of a at a bank moneY from the bank?

The money in the acdount after 10 years will be P 2,158.92.

A. 7.91% B. 7.51 C. 7.71 D. 7.31%

flllocn years ago P 1,000.00 was deposited in a bank account, and today it is Worth P 2,370.OO. The bank pays interest semi-annually. What was the interest

o/o

tala peid in this account?

A, B, C, D,

9"2*r;.

= P(1+ i)"

The effective interest is7.71 %.

lrrrrpouurl Irrlr,rr,nl rrnrl Oontinuous Cornpounding 63

Borrow money from a friend

5.72%

5.78% 5.84 o/o 5.90 %

F=P(1 +i)n

r=s,boo(i

+0.015)12 F = 4,184.66

F = P(1+ i)"

Borrow money from a bank

F=P(1 +i)n

2,970 =1000

r=9,'5oO(i +0.018)1 F = 4,130.00

.'. You will pay

.02918 = 'l + i i = 0.02918

- 4, 1 30.00 But: i=

P 54'66 less by borrowing the money from the bank'

7,000'"""""""""""'-"">

NR m

Substitute the value of i to get the nominal rate:

o.o2e18 =

IE 2

.NR = 0.05836

A deposit of P 1,000 is made in a bank account that pays 8 7o interest comiounaed annually. Approximately how much money will be in the account after 10 years?

NR = 5.836%

The interest rate paid on the account is 5.84 %.

A. P 2,187.39 B. P 2,145.78 c. P 2,176.45 D.

+ i)2(15)

2.37 =(t+1)3o 1

Solving for the difference: Difference = 4,1 84.66 Difference = 54.66

P 2,158.92

V.e*-.

Eiii

2,370

Questi

.liuginocring

li)<:()nor||r(.3 lrv

,lttttttt,

'l'rorrg

P 200,000 was deposited on January 1 , 1988 at an interest rate of 24 %o 1' 1993? compounded semi-annually. How much would the sum be on January

A. B. c. D.

oolttlxluntl

'-'----=

P 631,627.78

Conrlder a deposit of P 600.00 to be paid back in one year by p 700.00. what is lho rrte of interest, io/o per year compounded annually such that the net present Uorlh of the investment is positive? Assume i > 0.

A, 16.50 % !, ft.75% c, 16.33 %

P 612,890.76 P 621,169.64

P 611,672.18

7"/-s,* 90

'18.67

F = P(1+ i)"

2oo,ooo[,.0:o)"

7oo = 600(1+

20a,000

The amount on January 1, 1993 would be P 621,169.64.

what

,', The rate of interest is 16.67

year and is the present worth of two P 100 payments at the end of the third

fourth year? The annual interest rate is 8%.

A. B. c.

P 150.56 P 152.88 P 153.89 P 151.09

i)1

i= 0.1667 i=16.67%

F = 621,169.64

.'.

0/o

:(,..*

r=e(t+if F=

Irrl,rrroxl, nrrrl Conl,irruous Compounding 65

A flrm borrows P 2,000 for 6 years alg o/o. At the end of 6 years, it renews the lorn for the amount due plus p 2,000 more for 2 years ata %. what isthe lump lUm due?

A. P 3,260.34 B, P 3,280.34 c. P 3,270.34 D,

9"2;:

10.0

t00

P 3,250.34

r=e(t+if Fr = 2,000(1+ o.Oa)B

Fr = 3,701.86

=2,ooo(1+o.o8r

=2,332.80

Present worth = Pr + Pz present worth = 79.38 + 73.50 Present worth = 152.88

The present worth of two P 100 payments is P 1 52'8&'

Total future worth = Ft + Fz Total future worth = 6,034.66

Question llank

Solving for its present worth:

ComJrounrl lrrloroxl rrntl Oontirruous Compounding 67

l,)ngirrccrirtg lt)trrtttttltirs lrv .laime Il. Tiong

With inflation, Capitalto accumulate = 16,000(1 r 0.03)5

Capital to accumulate = 18,548.38

(1 + i)n

6,034.66 _ -=(*oo8)3 P --3,260.34

The capital that must be accumulated is P 18,548.38.

6,034.66

P <'.."""""

Whd

r , i 4-,n

or.,'/lZz,i*.

A, E, c,

2,000

2 000

the effective rate corresponding to 16% compounding daily? Take 1 year

360 days. 17.35 % ',17.4s%

17.55% 17.65%

'lth. en=(t+i)r-t rn=[r+0.16)360_1 \ 360/

Present lumP sum due = 2,000 + P Present lump sum due = 2,000 + 1,260.34 Present lumP sum due = 3,260.34

.'.

ER = 0.1735 ER = 17.35%

The present lump sum due is P 3,260.34

,'. The effective rate is

17.35o/o.

By the condition of a will, the sum of P 25,OOO is left to a girl to Oe nJO in a trust fund by her guardian until it amounts to P 45,000. When will the girl ieceive the money if the fund is invested at 8% compounded quarterly?

A. B.

accumulated?

C. D.

A. P 18,854.38 B. P 18,548.38 c. P 18,458.38 D. P 18,845.38 ,V"zz>-.

7.42years 7.67 years 7.85 years 7.98 years

r=e(t+if he machin ll need the of 2,000, t

cate 5 Years from now' the 8,000' Since there will be a d the amount of P 16'000 only'

45.000 = 25.OOO( 1+ -urvvv-4vrvvv[\r1.8 =1.024^

Without inflation, Capital to accumulate = 16,000

o'08 4

)4n )

Itl!

(luasl.ion

Ilrrnk llngintlcring

lrlconotttit'a by .lairtrc

'l'rrrrrg

Take log on both sides

()orrrgrorurrl IrrLr,roxl, rrlrrl Orlutinuous Compounding 69

It*,-

log1.8 = 4nlog1.02 O= n

loo1.8

41o91.02

P(1+ i)n

F=10,000(1+0.15)s

=7.4?years

F = 20,113.57

The girl will receive the money after 7.42 years. Let P = present worth of the account due to inflation:

P 2O0,OOO was deposited at an interest rate of 24o/o compounded semi-annually

o _ 20,113.57

C. 6 years D. 7 years

("".'...................... 20,1

o= 2oo,ooo[1 -

ry)'^

Take log on both sides 0585 -- Znlog1.12 loo 3.10585

The amount at maturity in terms of year zero pesos is p 15,030.03.

A Empany invests

p 1 o,ooo today to be rgpaid in 5 years in one rump sum no* -rcn protrii'n-pl."nt day pesos is rearized?at

Gmpounded annuary.

A, P 7,563.29 t, P 7,4e8.20

C, D,

P 7,340.12 P 7,623.42

21o91.12 n = 5 years

r=e(r+i)^ .'. The amount wil! become P 621,170 after 5 years'

F=10,000(1+0.12)5 F =17,623.42

ln year zero, you invest P 10,000.00 in a 15Yo security for 5 years. During that time, the average annual inflation is 6 9/o. How much, in terms of year zero pesos will be in the account at maturitY?

A. P 15,030.03 B. P 20,113.57 c. P 18,289.05 D.

13.57

P = 15,030.03

3.10585 = 1 .122n

(r + o.oo)5

r=e(t+i)n

log 3.

+ i)n

where: i = rate of inflation

A. 4 years B. 5 years

621,1 7

After how many years will the sum be P 621 ,170?

0l234

P 16,892.34

Profit=F-P Profit = 17,623.42 -1 Profit = 7,623.42

000

,'. The profit in present day pesos is p 1,623.42.

ll'

Question lJank

Engineering Economicc by Jaime lt. 'l'iong

How long (in nearest years) will it take money to quadruple if it earns 7 compounded semi-annually?

Oo-urllounrl

,-'__

lrrlrrorl tn

A mrn expects to receive P-2-0,000 in 10 years. How much is that money worth tlovv conqidering interest at 6% compounded quarterly?

0zo

A, P',t1,042.89 !, P 11,035.12 c. P 11,025.25

A. 20 years B. 18 years C. 21 years D. 19 years

P 11,012.52

J;**-. F=p(1 +i)2n

E |

+ = p(t* o'07)2n \ 2)

+ i)n

1.,

4 = 11.035)2n

(

Take log on both sides:

o.oo

"1o(10)

P =11025.25

tog4 = tog(1.m5)2n

,', The worth of the money now is p 11,025.25.

log4=2n1og1.035 n = 20.15

.'. The money invested at 7% compounded semi-annually will quadruple about 20 years.

How much should you put into a 10% savings account in order to have P10,000.00 in five years?

A. B. c.

P P P P

tO0,ooo was depositeg

?o 15

years ago at an interest rate of 7% com.pounded

Itltll-annually. How much is the sum now?

A, !, c, D,

P P P P

2,000,033.33 2!000,166.28 2,001 ,450.23

2,002,820.12

6,216.21

6,212.12

20,1 5 years

6,218.21 6,209.21

F=P(1 +i)n

r = soo,ooo(,*o:,)"'

.V"zrz-

F = 2,000,166.28

=---l(1 +

i)n

10,000 (1 + 0.1

0)5

P =6,209.21

.'. You should put into a 10o/o savings account an amount of P 6,209.21.

500,000

The sum of themoney now is p 2,000,166.2g.

Qur:sl,ion

Ilrrrrk

Ir)rrgirror,rirrg Iijrrrrronr ir:r by .laime R Tiong

Solving for interest rate annually: p.ays one percent interest ftheb,* effective anhual interest

rate?

A. B. C. D,

on savings accounts four times

a year.what is

ERannuatty = ERsemiannua,y

(r+i)-r=f,-g)'_,

4.06%

4.12% 4.16

i= o.dsoo

o/o

4.28%

D-

F (1

rn=

(1+i)m_r

sp = (r+o.or4 _1 P,

ER = 0.0406 ER = 4.06 %

+ i)n

25,000

(1+ 0.0506)

:23,795.93

.'. The effective annuat interest rate is 4.06%.

fl,. Alexander Michael owes p 25,000.00 due He agrees to pay p SO,OOO.0O today and t he pay at the end of two years if money is

D-

Ps =

A. P 39,015.23 B. P 39,026.25 c. P 39,056.21 D.

0'906P

Substitute values in Eq. 1:

P 39,089.78

50,000 + O.9O6p 23,795.93 = + 61,561.85 P = 39,026.25

Il;,;ffi;t

Atexander Michaet must pay at the end of two years is

FlndthedifferenceoP H il; ;. i;"i.,;Hf "i

depos it

A. P 1,510 B. P 1,530 c. P 1,550

25,000

75,000

P 1,570

Solving simple interest:

'3-A+OoSO6p

JT1"",.i

il:

Question Bank

Engineoring Economics by Jaime R. Tiong

Compound Intereet and Continuous Compounding 75

l= Pin r= 50,000(0.10x3) I

='15,000

SolVing for compounded interest: F = P(1+ i)n

Ps=

F = 5O,OO0(1 + 0.1 0)3 F = 66,550

150,000 + 280,000 1,,

lnterest=F-P lnlerest = 66,550 lnterest = 16,550

^'s = -

o.os'14(n)

430,000

1.or2s4n

50,000 Substitute values in Eq.

Difference = 16,550- 15,000 Difference = 1,550

142,728.64

.'. The difference between simple interest and compound interest is

25g,51 1 sO = -1SO'OO9

1.0125*"

396,240.2-

P1,550.

430',000^

1.01254n 1.01254n = 1.0852

Take log on both sides lf money is worth 5% compounded quarterly, find the equated time for paying a loan of P 150,000 due in 1 year and P 280,000 due in 2 years.

A. B.

C. D.

4nlog1 .0125 -- logl.0852

1.52 years 1.64 years 1.69 years

:1.64 years

1o91.0852

41o91.0125

1.72years The equated time for paying is 1.64 yeans.

.V;z*-..

P1+Pr=P,

For a loan acquired six years ago, a man paid out the amount of P 75'000.00. The interest was computed al18% compounded annually. How much was the

D_ (1

_ P-= 'r

-+Eq.

+ i)n

A. P 27,367.28 B. P 27,278.36 c. P 27,782.36

150.000 ', o.o5\4(1)

['* ,

P 27,872.63

Pt =142,728.64

280,000 Solving for the borrowed amount, P

76 ('lrrcst iorr llrrrrk l')ttgtrtclt'ittg lt)txrltotltl"t lry'lrrrrrro ll 'liong

('r rrn lrrrrr rrrl

tLet P

PF (1

_ 9-

Inl

lrl'sl

nrl ( brr( irrtrorrs ()ornpoun

tling 77

prcscnl verlue of inherrtance when boy is 6 years old

012...6789

i)n

...

2021

75.000 (1+ 0.18)o

P =27,782.36

.'. The borrowed money was

P 27,782-36.

be born to them theY will Place a birthday, the child will receive the an interest of 18% comPounded' e to make on the birth of a child to

li x

i)n

10.000 (r +

o.o+)5

P = 5,552.64

them?

A. B. c. D.

r,Il

The value of the inheritance as of the boy,s 6th birthday is p 5,552.64.

P 15,367.18 P 15,249.13

lLr

P 15,722.16

P 15,482.64

A man who won P 300,000 in a lottery decided to place so% of his winning in a trust fund for the college education of his son. lf the money will earn M% per year compounded quarterly, how much will the man have at the end of 10 years when his son will be starting his college education?

.72r,*Solving for the deposit on the child's birth, P

D_ (1

A. B. c. D.

+ i)n

300.000

9:_L

P 593.12012 P 593,452.12

.'. The couple will deposit the amount ol P 15,249'13'

(

A. P 5,552.64 B. P 5,549.10 c. P 5,522.12 D. P 5,582.63

inheritance will be paid in a the present value of the st is compounded annually?

F:593,888

::.t ,,\

P(1+ i)n

F = 1so.ooof1-

Lr<

I l,z,zn

P =15,249.13

P 592,739.96 P s93,888.96

+ 0.1 8)18

On his 6,h birthday a boy is lump sum of p t O,OOO on n inheritance as of the boy's Assume i = 4oh.

lriill

('l+

o'14l4(10)

The man will have P 593,888.96 at the end of 10 years.

78 (lrrt'sri,. Il:r.k

l,)rrgrrrr,r'r'r.g ri<:...lrrir:s rry Jairne R.

Tio.g

(

1l r l)rl

lf the sum of P 15,000 is deposited in an account earning 4o/o plt annum compounded quarterly, what will be the deposited amount at the end of 5 years./

A. B. c. D.

,,(x),000

P 18,302.85 P 18,450.89 P 18,512.83 P 18,638.29

Itrrrnrrl wrlrth = 300,000 + g0,375.51 rtornrrl worth = 3g0,375.51

trrlvtng lor the difference of present worth of proposals A and B:

r=e(r+i)"

10,000

=6'074'70

D_ F ,r-{r*,1n E

o-'

[(t * i)" (1+i)n

680, OOo

2.500.000

[tr*o.rz)o -r] ^'t = -2,ooo (l*olz)\olz)

Cost of dam = 2,500,000 = P

nLl

--+ Eq.

t"l

ftr * o.oal" - r]

''-

(r+0.12)5

= 0'567427 F

Substitute values in Eq. 1:

(1+ 0.08)"(0.08)

(1.08)" -1 = 0.294(1.08)" 1 = (1.08)n(0.706) 1.4164 = (.1:08),

P|+P2 = 10,000 6,07

P Equivalent cash flow

4.70 + 0.567427 F = 1 0,000 F -- 6,917.72

The amount the businessman will pay at the end of the 5th year is P6,917.72.

165

166

Annuity Question Bank - Dngineering Econo,mics by Jaime R. 'l'iorrg

Engr Richard Flores agreed to pay the loan he is borrowing from a developmbnt bank in six annual end-of-the-year payments of P 71 ,477 .7O. lnterest is 18% per annum compounded annually and is included in the yearly amount he will be paying the bank. How much money Engr. Flores is borrowing from the bank?

A. B. c.

P P P P

_/ 18000

nlrt*i)" p= L' '- -t'lI i(1 +

--+Eq.

i)"

Solving for equivalent monthly interest:

250,000 ERmonmty =

260,000

270,000 280,000

AAAAA

ERannuatly

;

1+ i = 1.00949

i= 0.00949 Substitute the value of P and i in Eq' 1:

P = loan

P= 7

1,477 .7ol(+ o. r a)6

- r]

P = 250,000

n[(r*o.oos+uF'-1] olfur4r(1. ooo%nr A:701.82

ooo =

AAAAAA

0.1 8(1 + 0.1 8)6

Solving for the monthly financing charge, B:

^'1215,000(1.12) H=

.'. The amount of money Engr. Flores is borrowing from the bank is P250,000.00

B = 150.00

Total monthly installment = 7O1.82 + 150 Total monthly installment = 851 .82 Barbette wishes to purchase a 29-inch flat-screened colored W at Gillamac Appliance Center an amount of P 20,000.00. She made a lownpayment of P5,000.00 and the balance payable in 24 equal monthly installments. lf financing charge is 12% for each year computed on the total balance to be paid by installment and interest rale 12Yo, how much would Barbette pay every month for the colored TV? What will be the actual cost of the money?

A. B. c.

Solving for actual cost of money:

851.82[(1+i)'?4

36.71% 36.21%

15,000 =

i(1

35.89 %

D.35.23o/o

Zz"a)

By trial and error:

i=0.0264 Solving for the monthly installment without taking into account the financing:

But

.NR l=m

P = 15,000

167

+ i)24

-1]

Annuity

Question lJank - Engineering liconornics by Jairnc ll,. 'l'iorrs

169

Let SV = salvage value of the machine 5 years from now:

0.0264 = IE 12

F=P(1+i)"

NR = 0.3168 NR = 31.68%

1,683,062.077

SV :80,oo0(1+ 0.07)5

sv=112,204.138

lf converted to annual effective rate; ER"nnu"tty = ERmontny (1

(t * o.ozu)l2 - t :0.3671 i

+ i)i -.1

AAAA i....................

i:36.710/o

+112,204.138 = 1,683,062.077 F = 1,570,857.939

Engr. Omayan of Main Engineering decided to purofrase a machine which is to be used foi their refrigeration and airconditionlng wonfts at an amount of P '1 ,200,000. The useful life of the machine is estimated b be 5 years with a salvage value of P 80,000 as based on current prices. The average annual rate of inflation during the next 5 years will be 7%. The machine will be replaced with a duplicate and the firm will accumulate the necessary capital by making equal end-of-year deposits in a reserve fund that earns 6% per annum. Determine the annual deposit.

P 278,664.U

I !12244p8

But F

nftr*o.oo)'-r]

1,570,857.939

o06

A = 278,664.54

P 277,189.56 P 279,180.00

Referring to the constructed cash flow:

.'. The actual cost (effective rate) of the money is 36.71 %.

A. B. c. D.

.'. The annualdeposit is P 278,664.54.

P 280.673.12

Cost of the machine at present = 1 ,200,000 Let C = Cost of machine 5,years from now: F = P(1+ i)"

C = 1,200,000(1+ o.o7)s

= 1,683,062.077

Note that the interest used was due to inflation only since nothing was mentioned about its interest. The interest of 6% stated in the problbm was implied to be that of the annuity (fund) only. Salvage value at present = 80,000

Engr. Omayan of Main Engineering decided to purchase a machine which is to be used for their refrigeration and airconditioning works at an amount of P 1,200,000. The useful life of the machine is estimated to be 5 years with a salvage value of P 80,000 as based on current prices. The average annual rate of inflation during the next 5 years will be 7%. The machine will be replaced with a duplicate and the firm will accumulate the necessary capital by making equal end-of-year deposits in a reserve fund. lf money is worth 6% per annum, determine the annual deposit.

A. B. c. D.

P 367,890.12 P 366,062.33 P 365,089.34 P 364,890.43

170

Quest,ion

llank

Annuity

lt)ngincering !)conornics by Jairrru It. 'l'iorrg

This problem is similar to the previous problem only that all amounts this time is to earn an interest of 6%. Cost of the machine at present = 1 ,200,000 Let C = Cost of machine 5 years from now:

C = 1,200,000

, a car costing P 150,000 is to be purchased ad of paying cash. lf the buyer is required to pay 4,000 each month for 4 years, what is the effective interest rate on the diminishing balance?

A. B. c.

F = P(1+ i)n + o. 1 3)s

C =2,210,922.215

35.28% 35.82o/o

34.89% 34.29o/o

Note that the interest 13% used was actually the sum of the interest rate and the inflation rate.

Balance = 1 50,000 Balance = 1 10,000

Salvage value at present = 80,000

40,000

Let SV = salvage value of the machine 5 years from now: F = P(1+ i)"

SV = 80,000(1+ 0.13)5 SV =

147,394.81

. ira A.r.rE

Referring to the constructed cash flow: i

F + 1 47,394.81

2,210,922.21 5

F = 2,063,527.40

A AA,4AAAAAAAA A

:..,........................,.,....-

n[(r*i)" P=

I 2,063,527.40 =

-r]

0.06

147,394.81

4,ooo[(1+

110,000=

iffi

By trial and error:

A = 366,062.33

i= 0.0255 .'. The annua! deposit is P 356,062.33.

-r]

+ i)"

i(1

n[{r*o.oo)u

AA,UAAAAAAA

AAAAA

But,

171

.NR m

0.0255 =

!B 12

NR = 0.306 NR = 30.6%

i)4(12)

- 1]

AAA,+4AAAAAAA

172

Question llank -. llngineering Economics by Jaime R. ,l,iong

Arrnrrity 173

Converting into effective rate:

ER=(1*!)'-1

Present worth = Pr + Pz Present worth = 1,277,O34.56 + '148,643.63 Present worth = 1,425,678.19

En=1,r+0.306)12_1

12)

Leasing the building:

ER = 0.3528 ER = 35.28%

AAA

Effective rate is 35,28% or 30.6% compounded monthly

.'. The effective interest rate on the diminishing balance is 35.2g %.

Solving for the difference in present worth, D Engr. Clyde Vasquez plans to purchase a new office building costing P1,000,000. He can raise the building by issuing 10%,2}-year bond that would

D = 1,425,678.1 D = 233,779.27

I-

1,1

91,898.92

... The difference in amount between buying the building and leasing the

building is P 233,779.27.

A. B. c.

P P P P

233,779.27 233,070.12

JJ Construction Firm had put up for sale of some of their heavy equipments for construction works. There were two interested buyers submitting their respective 'bids for the heavy equipments. The bids are as follows:

234,070.U 234,560.12

.V.l,*rn

P 10,000,000 PaY nually for 8 years.

the balance

Buying the building:

PaYa

the balance PaYable 00 payable P2,000,000 allY for 7 Years.

How much is the difference between the two bids if money is worth 10% effective?

A. P 346,520.05 B. P 346,980.12 c. P 347,019.45 D. P 347,733.29

------------:-

(t + i)"

C = 1.000.000 To compare the two bid, compare/their present worths:

1,000,000 -Gor,

For Buyer A's Bid:

= 148,643.63

Downpayrnent = 0.20(1 0,000,000) Downpayment = 2,000,000

174

(lrrcstiou

llunk

l,)nginr-.ering lJconornir:s by.lairnc

Anrruity

,l,rorrg

175

Solving for the semi-annual interest rate:

A's Bld

ER""r;.nnr"1 = 0 10

(1+i)2

-1=0.10 1+i =1.04881

i= 0.04881

hL)

2,000,000

n[1r*i;"-rl i(1

+ i)n

0.04881(1+ 0.04881)14 Present worth of A's Bid = 2,000,000 + p

-+ Eq.

P = 4,987,192.91

Substitute the value of P in Eq. 2 P

D-

present worth of B,s bid = 2,000,000 + 4,997,192.91 Present worth of A's bid = 6,987,1 92.91

r] ' - l

1,ooo,ooo[(1+ o.ro)B

Difference between two bids = 7 ,334,926.20 Difference between two bids = 347,733.29

0.10(1+ 0.'10)B P = 5,334,926.20

Substitute the value of P in Eq.

6,987,192.91

.'. The difference in amount between the two bids is P 347,733.29. 1

Present worth of A's bid = 2,000,000 + 5,334,926.20 Present worth of A's bid = 7 ,334,926.20

Excel First Review & Training Center, lnc. is expanding its school facilities starting 2001. The program requires the following estimated expenditures:

For Buyer B's Bid:

P 1,000,000 at the end of 2001 P 1,200,000 at the end of 2002 P 1,500,000 at the end of 2003

B's Bid

To accumulate the required funds, it establish a sinking fund constituting of 15 uniform annual deposits, the first deposit has been made at the end of 1992. The interest rate of the fund is 2o/o per annum. Calculate the annual deposit.

A. B. c. D.

2,000,000

<............. .

Present worth of B's Bid = 2,000,000 +

t1 2

/).t

P 346,520.05

P 346,980.12 P 347,019.45 P 347,733.29

176

Question Bank

Engineering Economics by Jaime R. Tions

Annuity Substitute values in Eq.

177

\21062A + A = 788,493.j2 + 927639.03 + 1,136,8j2.54 1

062A = 2,852,944.69 A= 217,679.01

Solving for the balance on January 1 , 2002 (i.e. end of 2001)

The annual deposit is P 217,679.01

xcel First Review & Training Center, lnc. is expanding its school facilities starting 2001. The program requires the following estimated expenditures: P 1,000,000 at the end of 2001 P '1,200,000 at the end of 2002 P 1,500,000 at the end of 2003 Solving for annual deposit, A:

+A =P, +P,

F (1

+P,

+ i)"

1000,000

'1- (+o.W P1

= 788,493.18

1,200,000

Pz =

927,639.03

Eq.

To accumulate the required funds, it establish a sinking fund constituting of 15 uniform annual deposits, the fiist deposit has been made at the end of 1992. The interest rate of the fund is 2o/o pet annum. Calculate the balance in the fund on

January 1,2OO2.

A. B. c. D.

P P P P

2,185,902.11

2,195,600.03 2,165,399.54 2,175,380.00

.%2r.,.

'2-(1+orop'f

D _ 1'500'000 '3 - (1+ooa14 Ps = 1,136,812.54

o_ t4-

n[1r.i)'-r] i(1

+ i)n

jPr,"i<"":.'""

p. al

t.2M

<.............

"""""', l,M

il,l

,t ,89

'91

'93

'9s

',97

'99

'0t

178 fluestion llank -

l!ngineering li]conornics bv Jairnc ll.'l'iorrg

Annuity

Solving for annual deposit, A

Po+A=P,,+Pr+P.

End of 2001, or 1

Jan. 1, 2002

F (1

l-1i=-

Eq.

+ i)n

1,000,000 (1

+ O.02)''

= 788,493.18

1,200,000

A A A A A A A A A

(1+ 0.02)'"

Pz = 927,639.03

Balance=F-1,000,000

1,500,000

(1+ 0.02)1a P3

= 1,136,812.54 0.o2

e[1r*i1'-rl

nL)

r, =-

F = 3,195,600.03

i11+i1"

n[1r*o.oz;'o P-LJ

=12.10624

Balance = 3,195,600.03 - 1,000,000 Ealancs.- 2,1 95, 600. 03

-rl

o.o2(1+0.02)la

Substitute values in Eq.

.'. The balance in the fund on January 1 ,

2OO2

is 2,195,500.03.

12.1062A + A = 788,493.12 + 927639.03 + 1 3. 1 062A = 2,852,944.69 A= 217,679.01

1, 1 36,81

2.54

J & N Pawnshop sells jewelries on either of the following arrangements:

Cash

price:

Discount of

1Oo/o

of the marked price

lnstallment: Downpaymenl of 20o/o of the marked price and the'balance payable in equal annual installments for the next 4 years.

Solving for the balance on January 1 ,2002 (i.e. end of 200'1):

lf you are buying a necklace with a marked price of P 5,000, How much is the difference between buying in cash and buying in installment? Assume that money is worth 5%. I

A. B. c. D.

P 40.76 P 41.90

P4354 P 45.95

,%l*on Compare their present worths:

179

180

Queslion

llank

Arrnui(,y

llngineoring Economics by Jaime R. 'l'iorrg

Cash basis:

The present worth of his earnings must be equal to the present price of the property.

P = 0.90(5,000)

Prfue

P = 4,500

Present worth = 4,500.00

lnstallment basis:

I,A00A

P -"""""""""""""""""'..

Downpayment = 0.20(5,000) Downpayment = 1,000

r=-

n[1r*i;"

hLJ

AAAAA

-rl

i(1 + i)n

r,ooo[(r+o.o.ef P=

-r]

0.05(l + 0.05)a

P = 3,545.95

Cash price = Pt +

Present worth = 3,545.95 + 1,000 Present worth = 4,545.95

Solving for annual net income, A:

Difference = 4,545.95-- 4,500 Difference = 45.95

Eq.

A=350,000-135,000 A = 215,000

.'. Buying in cash is economical by P 45.95.

Bon Fing Y Hermanos, lnc. has offered for sale its two-storey building in the commercial district of Cebu City. The building contains two stores on the ground floor and a number of offices on the second floor.

A prospective buyer estimates thai if he buys this property, he will hold it for about 10 years. He estimates that the average receipts from the rental during this period to be P 350,000.00 and the average expenses for all purpose in connection with its ownership and operation (maintenance and repairs, janitorial services, insuFance, etc.) to be P 135,000.00 He believes that the property can be soid for a net of P 2,000,000 at the end of the 10'n year. lf the rate of return on this type of investment is 7oh, determine the cash price of this property for the buyer to recover his investment with a 7o/o return before income taxes.

A. B. c. D.

P P P P

2,526,768.61

2,490,156.34 2,390,189.00

2,230J20.56

181

5,ooo[(1 + O.oz)10

- r]

0.07(1+ 0.07)10 P1

= 1,510,070.03

'z-,r*y D _ 2'000'000 '2 - (+o.W P2 = 1,016,698.58

Substitute values in Eq.

Cash price = 1,5'10,070.03 + 1,016,698.58 Cash price = 2,526,768.61 .'. The cash price of the property is P 2,526,768.61.

182

Question Bank

Engineering Econornics by Jairne R. 'l'iorrg

Annuity

E ,

^r--

183

+ i)n

36.000 (1+ 0.10)

P =32,727.27

net present value of the low price strategy.

A. B. c. D.

.'. The net present value of the high price strategy is P 32,721.27.

P 34,389.12

P 34,490.10 P 34,518.89 P 34,710.74

i(1

+ i)n

slq

20;

P= D_

Jan Michael plans to purchase a new house costing P 1,000,000. He can raise the building by issuing a 1Oo/o,20 year old bond that would pay P '150,000 interest per year and repay the face amount at maturity. lnstead of buying the new house, Jan Michael has an option of leasing it for P 140,000 per year, the first payment due one year from now. The building has an expected life of 20 years. lf interest charge for leasing is 12o/o, which of the following is true?

+ 0. 1oF -11 ---"-=

20,000

0.10(1 + 0. 10)'

A. B. C. D.

Lease is morei economical than borrow and buy. Lease has same result with borrow and buy. Borrow and buy is half the value than lease. Borrow and buy is economical by almost a hundred thousand than lease.

34,710.74

,9;2,r.,. .'. The net present value of the low price strategy is P 34'710'74'

Compare present worth of borrow and buy and lease..The Smaller the present worth the better for Jan Michael. For borrow and buy: (A = 150,000, i = '10%)

net present value of the high price strategy.

A. B. c.

P 32,727.27 P 33,737.34 P 33,747.20 P 33,757.89

'/-/,2--

150,ooo[(1 + o.r o)'zo 0.10(1+ 0.10)20

P1=1,277,034.56

- r]

184

Qucst,ion 13rrrrk I!)rrginotrrirrg llt:onornics by Jaime It. 'l'iorrg

Annuity

For lease: (A = 140,000, i = 12oh)

185

Gross income per year = 150,000 Expenses per year = 40,000 + Expenses per year = 140,000 Net income per year = 150,000 Net income per year = 10,000

70,OOO

1O,OOO

+ 0.1 O(1 SO,OOO) + 5,000

14O,OOO

olz Pz -'

AAAAAAA ;

P2 = 1,374,540.64

.'. The borrow and buy option is economical by 1,374,540.64 = P 97,506.08 than lease.

1,277,034.56 1

=' = Pt

Engr. Harold Landicho is considering establishing his own business. An investment of P100,000 will be required which must be recovered in 15 years. lt is estimated that sales will be P 150,000 per year and that annual operating expenses will be as follows: Materials

P 40,000

Labor

P 70.000 P 10,000 + 10% of sales P 5,000

Overhead Sellino exDenses

Engr. Landicho will give up his regular job paying P 15,000 per year and devote full time to the operation of the business. This will result in decreasing labor cost by P 10,000 per year, material cost by P 7,000 per year and overhead cost by P 8,000 per year. lf he expects to earn at least 2Oo/o of his capital, is investing in this business a sound idea?

A. Yes, it is a sound idea. B. No, it is not a sound idea. C. Neither yes nor no because it simply breakeven. D. lt depends on the current demand of the market. .VZ.*r," CASE 1: He does not resign from his job.

o,ooo[(1 + o.zo)15 /

- r'l'l

0.20(1+ 0.20)15

= 46,754.73

CASE 2: He resigned his job.

Gross income per year = 150,000 The new annual operating expenses are now as follows:

Material

=40,000-7,000

33,000

Labor = 70,000 - 10,000== 60,000 Overhead = 10,000-8,000 + 15,OOO = 17,000 + 10% of sates Selling = 5,000 Expenses per year = 33,000 + 60,000 + 17,000 + O.1O(150,000) + Expenses peryear = 130,000

1 2

AAAAA

S,OOO

186

Question Bank

Annuity

lt)ngineering Economics by Jaime R. '['iorrg

187

A=150,000-130,000 A = 20,000

A Civil Engineering student borrowed P 2,000.00 to meet college expenses during his senior year. He promised to repay the loan with interest al 4.5 % in 10 equal semi-annual installments, the first payment to be made 3 years after the date of the loan. How much will this payment be?

20,ooo[(1+ o.zo)15

Pz= P2 =

A. .P 252.12 B. P 261.89 c. P 273.90 D. P 280.94

- r]

0.20(1+ 0.20)15

93,509'45

.'. lt is not a sound idea since the present worth of the income for both case 1 and case 2 are less than the investment which is P 100'000.00.

A housewife bo\ght a brand new washing machine costing P 12'000 if paid in cash. However, she can purchase it on installment basis to be paid within 5 years. lf money is worth 8% compounded annually, what is her yearly imortization if all payments are to be made at the beginning of each year?

A. B. c. D.

P P P P

2,617.65 2,782.93 2,890.13 2,589.90

P=F

--) Eq.

n[(1. P=

')" (1

- 1]

+ i)"

A AA

8.866A

I I

Vi/,,2,-"

Eq.

F = P(1+ i)n

F = 2.ooo( 1+

(

0'045

2) )5

=2,235.355

Substitute values in Eq.

n[1r*o.oalo -r.l J L O.OS(1+ 0.08)a

P=F

AAAA

P = 3.312A

Substitute values in Eq. 1:

P ...'..".."""""""""""""i

12,000=A+3.3124 12,000 = 4.312A.

A --2,782.93 .'. The yearly amortization of the housewife is P 2'782.93.

8.866A =2,235.355 A = 252.12 The paymentis P 252.12.

188

Question Bank

Engineering Economics by Jaime R- '['iong

Annuity

A father wishes to provide P 4O,OO0 for his son on his son'sZ1"t birthday. How much should he deposit every 6 months in a savings account which pays 3% compounded semi-annually, if the first deposit was made when the son was 3 % years old?

A. B. c. D.

The investment that gives a bigger rate of return is the better investment Compare the rate of return of the two situations:

Domestic operation:

P 829.68 P 815.80 P 830.12 P 846.10

AAAAAAAAAA 0 1 2

40,000 =

4 '..

^[(,.T)"-,i --j3-5

Pt + Pz =

8,000,000 +

Eq.

A = 846.10 .'. The deposit every six months in a saving account is P 846.10.

A group of Filipino-[4echanical Engineers formed a corporation and the opportunity to invest P 8,000,000 in either of the two situations. The first is to expand a domestic operation. lt is estimated that this net year end cash flow of P 2,000,000 each year for 1 at that time, the physieal assets, which would no longer a operating and involve building would opportunity The alternative P 5,000,000.

P-=

E t

' ('t+i)" ^ 5,000,000 -'= *,)* Substitute values in Eq.

2,ooo,ooo[(1+ i)'o

is true?

A. B. c. D.

Foreign operation yields bigger rate of return. Domestic and foreign operations yield the same rate of return. Domestic operation has double the rate of return of the foreign operation. Foreign operation yields approximately 3% less rate of return than

- r]

-----* Dt.-. By trial and error:

i= 0.2381 i

= 23.81o/o

Tffi

= 8,ooo,ooo

189

190

Question Bank

Engineering Economics by Jaime It. 'l'iorrg

Annuity

Foreign operation:

A th at at

8,000;000

I pay P 90,OOO cash, P

60,000 at nnual payments starting with one es as to the principal and interest ?yment which must be made for 6

years.

;

Pa .""1

A. B. c. D.

A AAAAAA P' 1."

P 66,204.14 P 65,701.67 P 67,901.34 P 68,900.12

-+ Eq.2

P+ = 8,000,000

Solving for P1:

4,ooo,ooo[(1+ i)? -

i(1

+ i)7

P,+P.+90,000=400,000 P1

4,ooo,ooo[(1+i)? i(1

-r]

+ i)7 (1 + i)3

+ P. =

310,000

= 8,000,000

By trial and error:

i= 0.2080 i=20.80% The foreign operation has a rate of return whlch ls 3% less than the

domestic operation.

191

o[t,+o

oz)6

-r]

0.07(1+ 0.07)6 P2

4.7665A

--r Eq.

192

Quostion Bank

o- '3

[trnginoering Economics by Jaime

ll. 'l'iorrg

Annuity

F (1

+Dn

15,ooo[(1+ o.oa)s

4,7665A

0.08(1+ 0.08)5

Ps =

- r]

't - {t*0.97;' I

193

3'89094

Substitute values in Eq.

'z-*y

52,4o6.32 + 3. 8ebeA

= 59,890.65

o _ 59,890.65 'z-,r*ooty

:::::.,,.

Pz = 51,346.58

.'. The annual payment is P 66,204.14.

Fair prepayment = 51,346.58 Two years ago, the rental for the use of equipment and facilities as paid 5 years in advance, with option to renew the rent for another years by payment of P 15,000 annually at the start of each year for the renewal period. Now, the lessor asks the lessee if it could be possible to prepay the rental that wilt be paid annually in the renewed 5 years period. lf the lessee will consider the request, whatwould be the fair prepayment to be made to the lessor if interest is now

figured alSo/o?

A. B. c.

P 51,84330 P 51,346.58 P 52,OO2.45 P 52,740.20

.'. The fair prepayment to be made to the lessor is P 51,346.58.

NCT Bui an estim .equal ye

n additional building at the end of 10 years for 00. To accumulate the amount, it will have eaming 13%. However, at the end of the Sth year, it was decided to have a larger building that originally intended to an estimated cost of P8,000,000.00. what should be the annual deposit for the last 5 years?

A. B. c. D.

,7"22*..

P 734,391.48

P 742,890jt0 P 738,900.45 P 740,010.86

Let P2 = fair prepayment to the lessor

n[{r*i)" P= i(1

-r]

F = 5,000,000

+ i)"

Solving for annual deposit to accumulate 5,000,000 in 10 years, A:

AAAAA n[{r*o.rs)'o 5,000,000 =

0.13

A =271,447.78

-r]

194

Question Bank

- Engineering Economics

by Jaime R' 'I'iong

Let B = last 5 annual deposits -+ Eq.

Fz+Fs=9000,000

n[{r*i)" F,r:

-r]

Question Bank

8,000,000 1

1 10

zt t,++t .tel(r F,r=

+ o. r s)5

- r]

0.13

An investment, consisting.of deposits of P 1,000, P 1,500 and p2,000 are made the end of the 2nd year, 3-'d year and 4th year, respectively. lf money is worth 10% what is the equivalent present wo(h of the investment?

BBBBB

Fr = 1;759'055'07

F. =F,r(1+i)' Fa

= 1,759,055.07(1+ 0.13)5

= 3,240,944.94

Ft t"'- """"n i"""""""""'-

Fz l7.r

A. B. c. D.

P 3,129.89 P 3,319.45 P 3,372.12 P 3,490.09

e[1r*i;"-rl J L

E-

a[{r*o.ra)'-r] Fz= Fz

0.13

=6480278

Substitute in Eq.

6.480178 + 3,240,944.94 = 8,000,000 B = 734,391.48

.'. The annual deposit for the last 5 years is P 734'391.48.

Let P = Equivalent present worth P =Pt

+Pz

where:

P1 P2

., _ c [1-

Eq.

= present worth of uniform arithmetic gradient = present worth of annuity

(1+

i)-" n

"-tL--(1-if] D _ soo [r-(r*0.10)-4 4 '1- oroL olo -1r;oro)il

196

Question Bank Pr =

Ilnilbrnr Arithrnetic and Geometric Gradients 197

Engineering Economics by Jaime R. Tiong

2'189'06

^ "

500

1,500 (1+ 0.10)'

=1,126'97

2,000 D_ '6 -

500

o+0.'to)a

;

P6 = 1,366.03

Substitute values in Eq. 2 P = 826.45 + 1,126.97 +

366.03

P = 3,319.45 it checks! 5oo P3

[(1

+ o. r o)3

- r]

.'. The equivalent present worth is P 3,319.45.

0.10(1+ 0.10)3 P3

=1,243.42

o 'z -

Miss Calledo deposited P 1,000, P 1 3'd year and 4th year, respectively in annum. How much is in the account

D t3

11+i) 1,243.42 r^=' (1+ 0.10)

A. B. c. D.

P2 = 1,130.39

Substitute values in Eq.

P 4,880.00 P 4,820.00 P 4,860.00 P 4,840.00

,177o/,.n.

P=2,189.06+1,130.39 P = 3,319.45

Let F = equivalent future worth

F=Fl+F2

-ct*,no* S/Zor*.

-+Eq.3

Using compound interest formula:

where: Fr = future worth of uniform arithmetic gradient

P=Pa+P.+Pu -+ Eq.2

clrr+i)n -1 I -i =rl-i-"1

Fz = future

.'' _- soo [(t+o.to)4 -t _ol

1,000

=-(1 + 0.10)'

=826.45

worth of annuity

o.1o

I,500

Pt P6

I o.to

Fl = 3'205

of the 2nd year, .

rned 10% per

198

Question Bank

Engineering Economics by Jaime R. Tiong

Llnifonrr Arithmetic and Geometric Gradients 199

Miss Calledo deposited p

500

sa0

>F2 Solving for F2

.t2---_ o[(l+i)" -r] i

1,000, respectivel u

A. B. c. D.

only?

P 670.81 P 690.58 P 660.53 P 680.12

5oo[(1+ o.ro)q - r] _ Fz=

2nd year, O% per

3'o year and 4h year, annum. What is the equivalent

Let A = equivalent periodic payment for the uniform gradient

o.10

n n=cilI

Fz = 1,655

(1+ i)n _

Substitute valUes in Eq. 3

n=sool1

F=3,205+1,655

I 1.1

10.10 (1+0.r0)4 _1_l

F = 4;860

A = 690.s8 The equivalent uniform deposit is p 690.5g.

-M,& 9"2*r^ F=2,000+Fa+Fr -+ Eq.4 Fe

= P(1+ i)n

An amortization of a debt is in a years, p 4,500 on masecono fourth year. What is the

equivale

500(l + 0.10)1

A. P 15,093.28 B. P 15,178.34 c. P 15,890.12

Fs = 1'6so

Fg = 1,000(1 + 0.1 0)1

=1'210

Substitute values in Eq. 4 F

:2,000

+ 1,650.+ 1,210

F = 4,860 it checks

.'. The equivalent future worth is P 4,860.00.

P 15,389.82

ffil; the

J3""11T"

Sy".

200

Question Bank

Uniform Arithmetic and G€ometric Gradients 201

Engineering Economics by Jaime R. Tiong

The cash flow in the previous page is equivalent to the figure below:

n.=c[1" ' (1+i)" Li

,t.

ll,,

I.500

I -1.1

1-l

10.05

(1+ 0.05)"

- 1.1

Ar = 719'53 Let A = equivalent uniform payment of the. amortization:

A:5,000-A1 A=5,000-719.53 A = 4,280.47

s,000 5,000 s,000

s,000 :

=Pt-Pz

Eq.

P,r=

4280.474280.474,280.47 4,28.0.47 :

P;:,<"""""'

=17,729.75

n-.| i -fl.'r]

o_c[t-1t*i;" ''-TL

D_soo[t-1t*o.os)4 ''-

oosL oJ5

Pz =

P = 17,729.7 5

(1+oGf]

2'551'41

4,280.47l

P = 1 5,1

Substitute values in Eq.

2,55',t.4',1

15,178.34

.92-24,. ,%:;*,.". Let 41 = equivalent uniform payment due to uniform arithmetic series only

(t+o.os) -t.1

O.O5(1+ 0.05)a

78.3a it checks

The equivalent present worth of the debt is P tl5'178.34.

"""""""j

Unilbrm Arit,hnrot,ic and Geometric Gradients 203

2OZ Question Bank - Engineering Econnmics by Jaime R. 'l'iong

An amortization of a debt is in a form of a gradient series of P 5,000 on the first year, P 4,500 on the second year, P 4,000 on the third year, P 3,500 on the fourth year. Determine future amount of the amortization if interest is 5%.

A. B. c. D. .t'l"Zz,i,u.

P P P P

18,030.56 18,290.12 18,621.89 18,449.37

n[1r*i;'-rl J L

oI1-

i(1+i)n

_ t1-

s, ooo

+ o.os)a

[(1

- r]

o.o5(1+0.05)a

P1=17'729.75 F, =P,(1+i)" Ft = 17,729.75(1+ 0.05)a

1=21,550.62

_c[t-(t+i)*o ''-iL

-l

" * o.o5)-4 4 I -' - 5oo [t - (t o.os (1+ o.o5)'] ' o.o5 [ i

(1+i)".1

=2'551'41

Fz = Pz(1+ i)n

Fz=2,551.41(1+0.05)a Fz = 3'1O1'25

Solving for future worth, F: F

=1-Fz

-+ Eq. 1

Substituting values to Eq.

F =21,550.62-3,101.25 F = 18,449.37

.'. The future amount of the amortization is P 18'449'37'

ffiofadebtisinaformofagradientseriesofP5,oooonthefirst g:. f :1 ?^what s"Lo no vqa r,. P G;; ilEoil; ",h "tn" 1;999 :i -11i, pavment l;lll"is ?l"tn the " equiialent u1it9r1 periodic i;dh'G;;. wn"ri. is 5%' equirai"nt uniform periodic payment if interest

A. P 4,280.47 B. P 4,378.17 c. P 4,259.68 D.

P 4,325.12

204

Question Bank

tlnif

Engineering Economics by Jaime R. Tiong

Comparf the equivalent annual salary of the two ball clubs:

7"2r.,.'-

For LoqiAngeles Lakers, annual salary = $ 5,000,000 Solvir/q for the equivalent annual salary of Chicago Bulls:

n o.=c[1^t-"Li (r*D"-rl] Ar=6oo,ooot#

f;jful]

=2,235,276.31

Let 41 = equivalent uniform payment due to uniform arithmetic series only

Total annual salary of the NBA superstar with the Chicago Bulls, A:

n n.=c[]' (1+i)"

A=3,000,000+Ar

-1.1

' -

A = 3,000,000 + 2,235,276.31

A= 5,235,276.31

o.=uoo[ ' 10.05 (1+ 0.05)4 _ 1_l

... The offer of Chicago Buits is $235,276.31 more annually'than that

of LA

Lakers'

Ar = 719'53 Let A = equivalent uniform payment of the amortization:

A=5,000-41 A=5,000-719.53

John Grisham, author of the best selling novel "The Chamber" sold its copyright to warner Bros. for the rights to make it into a motion picture. Mr. Cristram's nas options between the following Warner Bros' proposals: An immediate lump sum payment of $ 5'000'000. B. An initial payment of $ 2,500,000 plus 4% of the movie's gross receipts for the next 5 years which is forecasted as follows:

A = 4,280.47 .'. The equivalent uniform payment is p 4,280.47.

End of year 1

2 3 4 5

A. Chicago Bulls offer is smaller than that of l-A Lakers B. Chicago Bulls offer is exactly the same as l-A Lakers, chicago Bulls offer is just few dollars more p6r year than that of l-A Lakers, ID. Chicago Bulls offer is per over $150,000

year than that of LA Lakers,

Gross Receipt $ 10,000,000.00

4% of Gross ReceiPt $ 400,000.00

$ 8,000,000.00 $ 6,000,000.00 $ 4,000,000.00 $ 2,000,000.00

$ 320,000.00 $ 240,000.00 $ 160,000.00

80,000.00

lf money is worth 10% and Mr. Grisham will not receive any royalty after the fifth year of Lxhibition of the movie, by how much is proposal A bigger than proposal B?

A. B. 9"2r2.,,

P 1,532,630 P 1,390.090

c. P 1,478,100 D. P 1,289.450

206

Question Bank

Llnifurrn Alit,lrrnotic antl Geometric Gradients 207

Engineering Economics by Jaime lt. 'l'iorrg

Proposal A will provide the author a present worth of g 5M. Solving for the present worth of proposal B: Let 41 = equivalent uniform payment of the uniform arithmetic gradient

n I n.=c[1(1+i)" -1l Li ,1-

A.=8o,ooo[ 10.10

What is thq pregent worth of the given cash flow diagram if interest rate is i10%? i

(1+0.10)5 _1j

A. P 2,200.34 B. P 2,089.37 c. P 2,191.49 D.

P 2,119.49

Al :144,810.08 Let A = equivalent uniform payment of proposal B:

A = 400,000 -'144,81 0.08 A = 255,189.92

680.24448

This is an example of uniform geometric gradient since the next term in the series is a fixed percentage of the previous term. Solving for the fixed percentage, 1+ r:

1+l=-

540 500

'l+r=1.08 Solving for the present worth of the royalty

Thus, r = 0.08 Since r is not equal to i, then x + 1 . Use the following formula:

zss,r as.sz[(r + o.ro)5

G f1-x") o_ '

1+i[ 1-x

- r]

0.01(1 + 0.1 0)5

1+r

P =967,370.57

1+i 1+ 0.08

Total present worth of proposal B = 2,500,000 + gOT,37O.S7

1+0.10

= 3,467,370.57 Solving for the difference between the two proposals: Difference = Difference --

5, 000, 1,

000

- 3,467,37

0.57

532, 629.4

Proposal A is P 1,532,630 bigger than propoeal B.

0.9818

-r =

5oo

Ir - o.osreu I u*s!10)L 1-or818 ]

P =2,191.49

.'. The presentworth of the given cash flow is P 2,191.49'

208

Question Bank

Engineering Economics by Jaime R. Tiong

[lrrrlbrrrr Arit,ltrno[it: ttnd (]oomettic Gradients 209

Solving fQr its future worth, F: \

What isithe equfualent future worth of'the-gi'ven cash flow diagram if interest rate is

F = P(1+ i)\

10o/o?

F = 3,529.54"

A. P 3,529.54 B. P 3,259.34 c. P 3,925.64 D.

=2,19L+qt*O.tO;u

.'. The equivalent future worth of the given cash flow is P 3'529'54'

P 3,295.54

7;zt-,,

the other sister? This is an example of uniform geometric gradient since the next term in the series is a fixed percentage of the previous term.

A. B.

Jocelyn, P 671.'18 Jocelyn, P 763.27 Joan, P 671 .18

Solving for the fixed percentage, 1+ r:

1+r=540

,9oz*r.,.

500

'l+r=1.08

To compare, solve the future amount of the two investments:

Thus, r = 0.08

For Joan:

Since r is not equal to i, then x + 1. Use the following formula:

G (t-r"') D_ ' - t+rIr-r.1

x=-1+r 1+i x=

1+ 0.08 1+ 0.10

x = 0.9818

'_ D

5oo

[r - o.sar au I 0*olo)L 1-oss1s l

P =2,191.49

2,357.9s Solving for the fixed percentage, 1+ r:

1+r -

1.100 --:-1,000

1+r=1.10

210

Question Bank

L.lnilbrnr Arithmotic and Goometric

Engineering Economics by Jaime R. Tiong

Gradients

211

Thus, r = 0.10 Since r = 0.10 is not equal to I = 0.14, then x+ 1. Use the following formula:

- c (t-r") H=-t 1+i[ 1-x

-l 1+r x=

1+O.14

1t{-e!!!!!

A. B. c.

1+i 1+0'10

x = 0.9&t9

Engr. Herqro, be\eving that life begins at 40, decided to retire a1{ life-at ase +0. neVisnis to have upon his retirement the sum of P 5,000,000. On his 21"t-birthday, he deposited a certain amount and increased his deposit by 15% each yeai until he will be 40. lf the money is deposited in a super savings account which earns 15% interest compounded annually, how much was his initial deposit?

1,000 [t - o.so+s,o I (1.oJOL 1-0s64, ]

P 17,253.18 P 17,566.33 P 17,672.77 P 17,490.21

P =7,508.45

since the first deposit was made when Engr. Hermo was 21, then the value of n is 20.

F = P(1+ i)n

r --15%

F = 7,508.45(1+ 0.14)10 F=

i=15%

27,835.48 Since, r = i, then

For Jocelyn:

= 1. Use the following formula:

+Eq.1

1+r

Solving for the present worth of P 5,000,000' E

' ('1

+ i)"

_ 5.000.000 P== (1 + 0.1

5)"

P = 305,501.39

F-

Substitute values in Eq.

305,501

3e=dtrb

0.14 F =27,072.21

Joan has P 763.27 more than Jocelyn.

G:17,566.33 .'. Engr. Hermo's initial depositwas P 17'566.33.

212

Question Bank

Engineering Economics by Jaime R. Tiong

Question Bank 5

A newly-agq.uired equipment requires an annuar maintenance costs of p 1or0o0. lf the annubt rndintenance cost is increased by 2oo/o each v"r, v"ar for 10 years, what is the estimated present worth of ihe maintenince costs it money is

"*w

worlh 15Yo?

A. B. c. D.

P',t0,s,712.33

P 106,101.37

P 107,490.12 P 108,890.11

2OVo

15o/o

A. B. c.

x=1+r

A company issued 50 bonds of P 1,000.00 face value each, redeemable at par at the end of '15 years. To accumulate the funds required for redemption the firm established a sinking fund consisting of annual deposits, the interest rate of the fund being 4%. What was the principal in the fund at the end ol ,l2h year? P 37.002.54 P 37,520.34 P 38,010.23

P 38,782.34

9;z*r...

1+i _1+ O.20

Amount needed by the company after fifteen years = 50(1,000) = 50,000

1+ 0.15

x =1.04348

LetA=annualdeposit

c D_ ' - 1-l[t-r") i-r.,J 1o,ooo

-t.o+s+a,o') ' _- 1*ols[-1-l.rrn4l ) D

P = 106,101.37

AAAAAAAAAAAAA

.'. The estimated presentworth of the maintenance cost is p l06,1ol .37.

n[{r*i)'-r] ,F= But F = 50,000 Substituting values:

n[1r*o.o+';'u 50,000

-r'l

A=2,497.06

2'14 Question Bank - Engineering Economics

by Jaime R. Tiong

Bonds

Solving for the principal at the end of 12th year.

1011

A P 1,000,000 issue of 3%, 1S-year bond was sold at 95%. What is the rate of

C. D.

3.O

o/o

3.4% 3.7

o/o

4.O

o/o

n[tr*i)"-r] F'rz

= I

24e7.06[(1+ o.o+)1'? Ftz =

,, _rrl(+i)"

-r]-(1+ir c

+Eq.1

"n-- ;6*y

- r]

0.04

Fe = 37,520'34 .'. The principal in the fund at the end of the 12th year is P

,520-34.

A local firm is establishing a sinking fund for the purpose of accumulating a sufficient capital to retire its outstanding bonds at maturity. The bonds are redeemable in 10 years and their maturity value is P 150,000.00. How much should be deposited each year if the fund pays interest at the rate of 3%?

A. B. c. D.

P P P P

13,084.58 13,048.85 13,408.58 13,480.58

Zr Zr Zr Zr Zr Zr Zr Zr Zr Zr Zr Zr

Vn = Vn

0.95(1,000,000)

= 95,000

C = 1,000,000

.V"z*...

Zr = 1,000,000(0.03) Zr = 30,000 Substituting values to Eq. 150,000

0.03 A = 13,084.58

.'. The annual deposit is P 13,084.58.

zr[1r * i)"] ,"" =-lnY -

95,000

30, ooo

[(1

* iI

c (1+i)n u

------+ i(1+ i)15

,,,

ooo, ooo

-11;

Dl5-

21 5

216

Question Bank

Bonds 217

Engineering Economics by Jaime R. Tiong

By trial and error:

i:0.03 i

=3Yo

.'. The rate of interest of the investment is 3%.

vn = 90,614.93'. .'. The man must pay the ceilificate the amount of P 90,614.93.

A man was offered a Land Bank c,ertificate with a face value of P 100,000 which is bearing interest of 6% per year payable semiannually and due in 6 years. lf he wants to earn 8o/o semiannually, how much must he pay the certificate?

P 1,000 bond which will mature in 10 years and with a bond rate of 8% payable annually is to be redeemed at par at the end of this period. lt is sold at p1,030. Determine the yield at this price.

A. P 90,123.09 B. P 90,614.93 c. P 90,590.12 D. P 90,333.25

A. 7.56%, B. 7.65 c. 7.75% o/o

9"2a.".

,,v"=-

zrf(r*i)" -1]

;1'1*;y -(1+ir

-+ Eq.

7.86%

Zr ZrZr Zr Zr Zr Zr Zr Zr Zr Zr ZrZr Zr Zr Zr Zr Zr Zr Zr Zr Zr

,, _rrl(*i)'-1] , c -ll;;n--,r*;y,

-+Eq

Zr = 1,000(0.08)

Zr=80 Zr = 3,000

C = 1,000 Substitute values in Eq.

i;^l

zrf1r * c :vn-______!__________ '(1+i)" i(1+i)"

Substitute the values in Eq.

., _zr[{r*i)"] , c [emailprotected]!n--(t*n "n

2'l8

Question Bank

Bonds 219

Engineering Economics by Jaime R. Tiong i

oooto.oe)

r..dao =

kr+

i)10

i(1+i)"

- r]

- 0.0385

i= 3.85%

1009_

(1+i)''

of interest is 3.85 %.

By trial and error:

i= 0.0756 i--7.56%

A P1,000, 6% Uond pays dividend semiannually and will be redeemed al 11Oo/o on June 21, 2004. ll is bought on June 21,2OO1 to yield 4% interest. Find the price of the bond.

The yield is 7.56 %.

You purchase a bond at P 5,100. The bond pays P 200 per year. lt is redeemable for P 5,050 at the end of 10 years. What is the net rate of interest on your

A. B. c. D.

P P P P

1,122.70 1,144.81 1,133.78 1,155.06

investment?

A. B. c. D.

3.56% 3.85

o/o

Fr=1,ooo[ry)

3.75%

Bought on

Will be redeemed

lune 21, 2001

June 21, 2004

3.68 %

Fr=30

,V/or,r-

Vn: 5,100

30 3A

30 I I

200 200

C = 1,100

200 I

V" = 1,144'81

C = 5,050

.'. The price of the bond is P 1 ,144.81.

220

Question Bank

Engineering Economics by Jaime R. 'l'iorrg

Question Bank

,-1;;

A voM has a sellingprice of P 400. lf its selling price is expected to decline at a rate of 10o/o per 4nrfum due to obsolescence, what will be its selling price after 5

years?

A. P 222.67 B. P 212.90 c. P 236.20 D.

P 231.s6

Using Matheson Formula:

k=',-,8 \lco

o.1o=1-FE v

400

cu :(o.gou 4o0/ *T Cs =236'20

.'. The selling price at the end of the

Sth

year is p 236.20.

A machine costs P 8,000 and an estimated life of 10 years with a salvage value of P 500. What is its book value after 8 years using straight line method?

A. B. c. D.

P P P P

2,000.00 2,100.00 2,200.00 2.300.00

,9zo*Solving for annual depreciation charge, d

222

Question Bank

l)trprcr:rnl iorr arrd Depletion 223

tr)ngineering Economics by Jaime R. 'l'iorrg

. Co-Cn d=n

ABC Corporation makes it a policy that for any new equipment purchased, the annual depreciation cost should not exceed 20% of the first cost at any time with no salvage value. Determine the length of service life necessary if the depreciati used is the SYD method.

8.000-500 10

d=750 Solving for book value after 8 years, C6

C. =Co-Dm c8 =8,000-750(8) Ce = 2,000

A. 7 years ,/ B. 8 years C. '9 years ./ D. 10 years ./

The book value after 8 years of using is P 2,000.00. ln SYD method, the largest depreciation is on the first year. Analyze first year depreciation, d1 A telephone company purchased a microwave radio equipment for P 6 million, freight and installation charges amounted to 4% of the purchased price. lf the equipment will be depreciated over a period of 10 years with a salvage value of 8%, determine the depreciation cost during the 5"'year using SYD.

A. B. c. D.

d, =(co

-.")r,:"*

But:rvear=n(n*1) 2

P P P P

626,269.09

d, =(co

623,209.09 625,129.09 624,069.89

d" =

o 2oc^ = c"

Solving for d6preciation during the 5th year, ds

years

The length of service life necessary is 9 years.

But

Years =

r-r)

"[n+1]

n+1=10

-..)rEk

c^r-?-) "In+1]

But: d1= 0.20C0

6,000,000 + 0.04(6,000,000) = 6,240,000 Cn= 0.08(6,240,000) = 499,200 Co =

=(co

../;/;r*,,

-o)--!

n(n+1I

-:=-

company purchases an asset for P 10,000.00 and plans to keep it for 20 years. the salvige value is zero althe end of 20th year, what is the depreciation in the third year? Use SYD method.

2 Substituting, d5 = (6, 24o,ooo

- 499'2OO)!#

d5 = 626,269.09

The depreciation cost during the

Sth

year is P 626,269.09'

A. B. c. D.

P 857.14 P 862.19 P 871.11 P 880.00

224

l)oplor:urliorr rrrtt.l l)cplution 225

Engineering Economics by Jaime R. 'l'iong

Question Bank

The total depreciation in the firCt three years is P 48,000.00.

'-/,./u1,;',,

Solving for depreciation at the end of 3'd year, d3

-+ Eq.

A machine has an initial cost of P 50,000 and a salvage value of P10,000 after 10 years. What is the book value after 5 years using straight line depreciation?

A. B. c. r

vears -2o(2o

+1)

- r,o

Substitute values in Eq.

P 30,000.00 P 31,000.00

P 30,500.00 P 31,500.00

Solving for the annual depreciation charge, d

=(1o,ooo

-o)4-3

ds =857'14

Co-Cn n

.50.000-10.000 .'. The length of service life necessary is 9 years.

'10

d = 4,000 Solving for the book value at the end of 5th year, Cs

An asset is purchased for P 500,000.00. The salvage value in 25 years is P100,000.00. What is the total depreciation in the first three years using straightline method?

A. P 45,000.00 B. P 46,000.00 c. P 47,000.00 D.

Cs=Co-Ds cs =50,000-4,000(5) c5 = 3o,ooo The book value after 5 years is P 30,000.00

P 48,000.00

?).,r^n. Solving for the annual depreciation charge, d

-c_ n

ct =

500.000

-=-

- 100.000 ' 25

d = 16,000 Solving for the total depreciation in the first three Years, Dg Ds = 16,000(3) Ds = 48'000

An asset is purchased for P 9,000.00. lts estimated life is 10 years after which it will be sold for P 1,000.00. Find the book value during the first year if sum-ofyears' digit (SYD) depreciation is used.

A. B. c. D.

P 7,545,45 P 7,320.11

P 7,490.00 P 7,690.12

V/,,t;"n.. Solving for depreciation on the first year,

d,:(co-c")=+.years z

226

Question Bank

Engineering Economics by Jaime R. Tiong

Deprocintion and Depletion 227

But

)

vears

n(n +

Cr=Co-D, cr =530,000-96,000(4)

2 1o(10 + 1)

Ca = 146'000

.'. The book value at the end of the 4th year is P 146,000.00

Substituting,

=(9,000-1,000)#

dt =1,454'54 Solving for the book value after first year,

C' = Co -d1 cr = 9,00o- 1,454.54

--7,545.45

P 1,344.21 P 1,245.45

.V;z;r,,,,

The book value during the first year is P 7,545.45.

F Let dr = first year depreciation

d, = (co The cost of equipment is P 500,000 and the cost of installation is P 30,000. lf the salvage value is 10o/o ol the cost of equipment at the end of 5 years, determine the book value at the end of the fourth year. Use straight-line method.

A. B. c.

P P P P

146,320.50 146,000.00 146,230.50 146,023.50

-.")r#"*

vear

)

vear =

)year

n(n +

2 10(1'1)

------:-----!'

= 55

Substituting;

Vt,,,,,,

Co = 500,000 + 30,000 = 530,000 Cn = 0.10(500,000) = 50,000

Solving for the annual depreciation charge, d:

= (2o,ooo

- 12'ooo)r+l 't55/

dt = 1'454'54 .'. The depreciation for the first year is P 1,454,54.

c^-c_ n

_ _ 530,000-50,000

A machine has an initial cost of P 50,000 and a salvage value of P 10,000 after 10 years. Find the book value after 5 years using straight-line depreciation.

d = 96,000 Solving for the book value at the end of the fourth !ear, Ca

A. P 31,000.00 B. P 31,500.00 c. P 30,000.00 D.

P 30,500.00

228

Question Bank

l)ngineering Econornics by Jaime R. 'l'iorrg

l)o;rrocint,iort and

Depletion 229

.'ll,/,r., Yo=

Let Cs = book value after 5 years

o/o

Solving for annual depreciation charge, d:

xtoo

= 8ok

.'. The straight line method depreciation rate is 8%.

r-"0

4'ooo 50,000

vn n

- 10,000

50,000

A machine costing P 45,000 is estimated to have a book value of P 4,350 when

d = 4,000 Solving for Cs

Cs=Co-Ds cs =50,000-4,000(5)

B. 32.25o/o 32.00% D. 32.75o/o

C5 = 30'000

.'. The book value after 5 years of using is P

3O,OOO.OO

k=1-"8 !co

A machine has an initial cost of P 50,000.00 and a salvage value of P 10,000.00 after 10 years. What.is the straight line method depreciation rate as a percentage of the initial cost?

A. B. c.

\145,000 k = 0.3225 k = 32.25o/o

o/o

8% 7%

D. I

[emailprotected]

.'. The annual rate of depreciation is 32.25

o/o

o/o.

V;1itr.. Let d = annual depreciation

d-co-cn n

,{ _ 50,000

10,000

10 d = 4,000

Solving for the percentage of depreciation with respect to the initial cost: %=

4x100 co

An asset is purchased for P 120,000.00. lts estimated life is 1 0 years, after which it will be sold for P 12,000.00. Find the depreciation for the second year using the sum-of-years' digit method.

A. B. c. D.

P 17,578.13 P 17,412.43 P 17,344.67 P 17,672.73

V;z*,,, d, =(co

-c,).4 jyears

230

Question Bank

n(n +

)

vears

)

vears

)

years = 55

Depreciatiort and Depletion 231

Engineering Economics by Jaime R. Tiong 1)

2 10(1

An asset is purchased for P 9,000.00. lts estimated economic life is 10 years after which it will be sold for P 1,000.00. Find the depreciation in the first three years using straight line method.

/o\

+ '\55/

d2 = (120,000-12,000)l

dz =17,672'73

A. P 2,400.00 B. P 2,412.34 c. P 2,250.OO D.

P 2,450.00

.%7i*--.

.'. The depreciation for the second year is P 17,612.73.

A machine costing P 720,000 is estimated to have a book value of P 40,545.73 when retired at the end of 10 years. Depreciation cost is computed using a constant percentage of the declining book value. What is the annual rate of depreciation in %?

A. B. C. D.

25 26 27

d=800

o/o

Solving for total depreciation in the first three years, D:

o/o o/o

D = d(3)

28%

D = 800(3) D

.V;/;r,,Using Matheson Formula

=2,400

.'. The total depreciation in the firct three years is P 2'400.00.

k=1-"8 lco

[emailprotected]'il\n i/

720,000

k = 0.25

k=25% .'. The annual rate of depreciation is 25%.

An engineer bought an equipment for P 500,000. He spent an additional amount of P 3O,0OO for installation and other expenses. The estimated useful life of the equipment is 10 years. The salvage value is x % of the first cost. Using the stiaight line method of depreciation, the book value at the end of 5 years will be P 291,500. What is the value of x?

A. B.

O.2

0.4

c. 0.3 D.

0.1

Let Cs = book value at the end of 5 years

Cs=Co-Ds -r

Eq'1

232

Question Bank

Depreciation and DePletion 233

Engineering Economics by Jaime R. Tiong

Co :500,000 + 30,000

co = 530,000

n d

Solving for annual depreciation, d:

800.000

---ro

- 35,000

d = 76,500

,..-Co-Cn

The annual depreciation charge is P 76'500'00'

530,000-x(530,000) 10

d:53,000-53000x Total depreciation for 5 years, Ds:

""l"rri" iS

Ds = d(5) D5

The The initial cost of a paint sand mill, including its installation, is P 800,000. Tle years for 10 is depreciatio.l.' Bli approved tife oi this machine "tl]'?lg9; ill is P 50,000 and the cost of dismantling is estimated to of the value book is the what p aight-line depreciation' O" of six end machine at the Years?

=(53,000-53,000x)5

Book value at the end of 5 years, Cs

Cs=Co-Ds 291,500 = 530,00

A. P 341,000.00 B. P 343,Q00.00 c. P 340,000.00

P 342,000.00

5(s3,000 _ 53,000x)

- 53,000x) = 238,500 53,000 - 53,000x = 47,700

5(53,000

53,000x = 5,300

x=0.1

Co = 800,000 Cn = 50,000- 15,000 = 35,000

Solving for annual depreciation charge, d:

.'. The value of x is 0.1. n

The initial cost of a paint sand mill, including its installation, is P 800,000. The BIR approved life of this machine is 10 years for depreciation. The estimated salvage value of the mill is P 50,000 and the cqst of dismanfling is estimated to be P 15,000. Using straight-line depreciation, what is the annual depreciation charge?

A. B. c. D.

P 75,500.00 P 76,000.00 P 76,500.00 P 77,000.00

.V.z*-. Co = 800,000 Cn = 50,000- 15,000 = 35,000

Solving for annual depreciation charge, d:

. 800,000 - 35,000 o= io d = 76,500 Solving for book value at the end of 6 years' C6 Co = Co

-Do

co =800,000-76,500(6) Co = 341'000

The book value at the end of the 6th year is P 341,000'00'

234

Question Bank

Deprocial,ion and DePletion 235

Engineering Economics by Jaime R. Tiong

6=Co-Cn n

A unit of welding machine cost P 45,000 with an estimated life of 5 years. lts salvage value is P 2,500. Find its depreciation rate by straight-line method.

A. B. c. D.

. 410,000 - 60,000 -20

18.89 %

d = 17,500

19.21% 19.58%

years, cro Solving for the book value (appraisal value) at the end of 10

19.89 %

Cls=Co-D16 cro = 41o,0oo - 10(17,500) Cro'= 235' o0o

6=Co-Cn

...Theappraisalvalueofthemillattheendofl0yearsisP235,000.00.

,_45,000-2,500 .J

d = 8,500

10 An equipment costs P 1O,OOO with a salvage value of P 500 at the end of years. catcutate the annual depreciation cost by sinking fund method at 4 %

Depreciation rate = -9

interest.

Depreciation

,"t" = 9:590 x 45,000

loOo/o

Depreciation rate = 18.89 %

A. P 234,000.00 B. P 235,000.00 c. P 234,500.00 D.

P 235,500.00

.'. The depreciation rate is 18.89 %.

.V;z,t*,,

d = 791.26

A. P 234,000.00 B. P 235,000.00 c. P 234,500.00 D.

P 235,500.00

Co = 360,000 + 5,000 + 25,000 + 20,000

C" = 410,000 Solving for annual depreciation charge, d

.'. The annual depreciation cost is P 791'26'

year by 20% of its A machine initially worth P 50,000 depreciates in value each it is 9 years old. when its book-value year. Find of that value at the beginning

A. P 6,710.89 B. P 6,400.89 c. P 6,666.89 D. P 6,512.78

236

Question Bank

L)oprecial,ion and Depletion 237

Engineering Economics by Jaime R. Tiong

.V,z"arsince the percentage of depreciation is constant every year to be 20% of the book value, use the Matheson Formula

*=1_.8 !Co

o.2o=1-'E 50,000

life is 10 years An asset is purchased for P 9,000.00. lts estimated economic in the first three depreciation the Find 1,000.00. P aftei wnicn it will be sotd for method' digit sum-of-years' years using

A. P 3,279.27 B. P 3,927.27 c. P 3,729.27

Vffi=ouo

722*-

,tn

50,000

P 3.792.72

Solving for depreciation in the first three years, Da

= 0.80s

Cg = 6,710.89

.'. The book value when it is 9 years old is p 6,7i0.g9.

A consortium of internationaltelecommunication companies contrActed for the purchase and installation of fiber optic cable linking [ianila citv anaCeou city at a total cost of P 960 million. This amount includesireigtrt ana insiaiiaili:nehaiges estimated at 10% of the above total contract'price. lf fie cable shal be depreciated over a period of 15 years withzero sarvage varue anJ money is worth 6% per annum, what isthe annual depreciation-charge?

A. P 41,04p'903.40 B. P 41,211,158.40 c. P 41,254,000.40

+ '"t':''l=55 - 10(10 1)

(e,000-1,ooo)*

= 1,454'54

-"")thk

(e,ooo-1,ooo)*

= 13oe.oe

-.,,)k

= @,ooo

dl = (co

-.^)t**"

d, = (co

d. = (co

Substitute in Eq.

P 41,24.253.40

)years=ry lyears

-+ Eq.

Dg=dr+dr+d.

- 1,000)*

= 1163.64

Ds =1,454.54+ 1'309.09 + 1,163'64 Ds =3,927 '27

since a rate of interest is given in the probrem, use qinking fund method.

.'. The depreciation in the first three years is P 3,927 '27 '

Solving for annual d€preciation charge, d

,n_(Co-Cn)i (1+i)" -1 d

_ (s60,000,009 - 0x0.06) (1+ 0.06)15

d= 41,244,253.40 .'. The annual depreciation charge is p 41,241,253.40.

resale value at the end of A radio service panel truck initially cost P 560,000. lts ui"irriir" is esiimated.at P 150,000 ?J T:iT,"l.declining

il?'-G;;fli"

oatancemethod,determinethedepreciationchargeforthesecondyear.

A. P 99,658.41 B. P 99,128.45 c. P 99,290.00 D. P 99,378.45

238

Question Bank

l)r:prot:iat,iort anrl Depletion 239

Engineering Economics by Jaime R. Tiong

9Z** efinery developed an oil well which is estimated to s of oil at an initial cost of $ 50,000,000. What is the the year where it produces half million barrels of oil? Use computing dePletion.

Using the Matheson formula: k=

1-"8 !Co

A. $ 500,000.00 B. $ 510,000.00 c. $ 525,000.00

k = 1- uE!g,ogq \l 560,000

k = 0.2316

$ 550,000.00

Tabulating the depreciation charges: Year 0 1

Depreciation 0

Book Value 560.000

6(560,000) =1 29,696 0.2316(430,304) = 99.658.41

330,645.59

0.231

430,304

Unit depletion 121"

$50'ooo'ooo

5,000,000 Unit depletion 131s = $10.00 per barrel

Totaldepletion during the year = (500'000)(10)

.'. The depreciation on the second year is P 99,658.41.

Total depletion during the year = $ 5,000,000 .'. The total depletion charge during the year is $5,000,000'00' Shell Philippines, particular y6ar of except for depleti for that particular

A. B. c. D.

P 9,358.41 P 9,228.45

P 9,250.00 P 9,308.45

9,Zz*," iJsing Depletion Allowance Method: Depletion charge = O.22(5O,000,000) = '1 1,000,000 Depletion charge = 0.50(18,500,000) = 9,250,000 Use the smaller value, P 9,250,000

,'. The allowable depletion allowance is P 9,250.00.

ome for a all deductions etion allowance as 22o/o.

240

Question Bank

Engineering Economics by Jaime R. Tiong

Question Bank I

J,;*

The first cost of a certain equipment is P 324,000 and a salvage value of P 50'000 at the end of its life of 4 years. lf money is worth 6% compounded annually, find the capitalized cost.

A. P 540,090.34 B. P 541,033.66 c. P 540,589.12 D. P 541,320.99 .7;za-. Capitalized cost= Co+

P+

Eq.

The maintenance cost is due to the depreciation of the equipment only Total depreciation = Co Co Co

-Cn =32a,000-50,000 - Cn =274,OOO

Since, there is a given interest rate, use sinking fund method in solving for annual depreciation charge. i

'-[tr*ir-r] ^_(co-c")

\ ' .{- 274.000(0.06) [t',.

o.oo)' - r]

d--62,634.07 Solving for the present worth of all depreciation charges, P

242

Question Bank

(lrrpit,rrliz,rrrl

Engineering Economics by Jaime R. Tiong

62,634.07[(1+0.06)4

(bst & Anntral Cost 243

Capitalized cost = 325,000

-1]

.'. The capitalized cost is

0.06(1+ 0.06)a

,/ /

P 325,000.00.

P = 217,033.66

Another way to solve for the value of P

o[(r*i)'-r] i(1

+ i)"

A piece of equipment was purchased for P 4,000,000. The useful life is estimated at 5 years with no salvage value. The equipment is estimated to be used for 1,000 hours per year. Operating costs are as follows:

Fuel Oil Grease

,_ffi[t'.'1"-'] i(1

: l0liters per hour, P 10 per liter : 1 liter Per hour, P 20 Per liter : 2 liters per hour, P 5 per liter

ti insurance, is Problem

+ i)n

nt uses 6

P 60,000 each. Cost of money is 15% per age, etc. is 15% per year. The interest tabie the figure.

Determine the cost of using the equipment and tires, in pesos per hour.

^r--

A. P 2,202.90 B. P 2,101.80 c. P 2,000.90

274.OOO

(1+ 0.06)" P = 217,033.66

Substitute in Eq.

P 2,303.70

,7"2r*-

Capitalized cost = 324,000 + 217,033.66 Capitalized cost = 541,033.66

P = 4,000,000

.'. The capitalized cost is P 541,033.66.

An item is purchased for P 100,000. Annual cost is p 1g,ooo. Using interest rate 8%, what is the capitalized cost of perpetual service?

4.000.000 =

0.15(1+ 0.15)5

A = 1,193,262.21

P 325,000

ln terms of pesos Per hour, Hourly cost of equiPment =

Capitalized cost = Co +

AAAAA

n[{r*o.rs)u-r]

A. P 310,000 B. P 315,000 c. P 320,000 D.

Annual cost of equiPment onlY

1,193,262.21

year

Yeat -----.-.:-= 1.1 93.26 1,000hours

Capitalized cost = 100,000 +

8,000 0.08

Annual cost of insurance, taxes' storage, etc':

Annual cost of insurance, etc = 1,193,262.21(0.15) = 178,989'33

244

(lrrgrit,ulizorl Oont & Anrrual Cosl 245

Question Bank

Engineering Economics by Jaime R. Tiong Total annual cost = 1 1193.26 + 178.99 + 130 + 593.26 + 107.39 Total hourlY cost = 2,202.90

ln terms of pesos per hour, Hourly cost of insurance, etc =

178.989.33 -'-----x

The total cost of using the equipment and tires is P 2,202.90 per hour.

vear

'--' =178.99 1,000hours

yeil

Hourly fuel, oil and grease costs:

A corporation uses a type of motor which costs P 5,000 with life 2 years and final value of P 800. How much could the corporation 3fford-to pay.for another ""tr"b" typ" Jt motor of the same purpose whose life is 3 years with a final salvage value of P 1,000. MoneY is worth 4%.

Fuel cost = 10('!0) = 100 per hour Oil cost = 1(20) = 20 per hour Grease cost = 2(5) = 10 per hour Hourly cost of fuel, oil and grease = 100 + 20 + tO = D

A. P 7,892.13 B. P7,157.40 c. P7,489.21 D. P 7,300j2

130

hour

Annual cost of depreciation: Using sinking fund method since the interest is given

SlnZr,r."

, (co-c")i

The two types of motors must have the same annual cost'

ftt*i)"-t]

For Motor 1: Using sinking fund method in computing depreciation:

d= 593,262.21 Annuar ln term of pesos per hour, Hourly cost of depreciation =

592.262.21

year

""o

ffi

vear

1,000hours

Annuat cost

+ co(i) +

(5'-000

- 800)(0

o.c.

94) + (5,oooXo.oa)

Itt*o.o+)'-t.1

Annual cost of tires (in pesos per hour):

Annualcost = 2,258.82

P = 60,000(6) P = 360,000

For Motor 2: Using sinking fund method in computing depreciation:

360,000-

n[(r*o.rs)'-_r]

AAA

'ffi.

0.15(1+ 0.15)5

ln terms of pesos per hour,

107'393'60x---yegl--

year

Annuat

A = 107,393.60

Hourty cost of tires

Annuar

1,000hours

= 107.39

"or,

*"1 =

j"' - 9") I + co(i) + o.c. [tt*i)"

-'11

(?o -

000X0

0j)

[(t+o'o+)" -t]

Annual cost = 0.32035C0 Annual cost = 0.36035C0

+ (co )(0.04)

- 320.35 + 0.04C0 - 320.35

246

Question Bank

()rrlritalizod Cost & Annual Cost 247

Engineering Economics by Jaime R. Tiong

Equating annual costs:

2,258.82 =.0.36035C0

- 320.35

Co =7'157 '4O

,'. The corporation could afford to pay P 7,157.40 for anot[er motor.

At 6%, find the capitalized cost of a bridge whose cost is P 250M and life is 20 years, if the bridge drust be partially rebuilt at a cost of P 100M at the end of each 20 years.

A. B. c.

A motorcycle costs P 50,000 and has an expected life of 10 years. The salvage value is estimated to be P 2,000 and annual operating cost is estimated at P 1,000. What is the appropriate rate of return on the investment if the annual revenue is P 10,000?

A. B. C. D.

P 297,308,323.10 P 298,308,323.10

P 296,308,323.10 P 295,308,323.10

,%h*.

12.12% 12.54

o/o

12.72o/o 12.ggo/o

,:%bTo get the rate of return, i, equate annual cost and annual revenue: Using sinking fund method in computing depreciation:

Annuat

*r,

- c") I + c^(i) + o.c. = Ico

l{t*i)"-tl r50. 000 2. 000) (i) 10,000=. - - ,=' -," +(50,000Xi)+1,000 l(1+i)'"-t] e,

ooo =

jg-qqq!= + 50, oooi i[ - t]

[(t.

By trial and error: i=

o.1272

i=12.72% The appropriate rate of return of the investment is 12.72 %.

Capitalized cost = Co +

Solving for the equivalent interest rate in 20 years,

i = (1+ NR)20

i = (1+

-1 0.06)20 - 1

i--2.2071

i=220.71% Capitalized cost = 250,000,000

109:qqq900 2.2071

Capitalized cost = 295,308,323.10 .'. The capitalized cost of the bridge is P 295,308,323.10.

248

Question Bank

Cnpitalized Coet & Annual Cost 249

Engineering Economics by Jaime R. 'liorrg Total annual cost = 319,833.49(2) Total annual cost = 639,666.98

A newly hired Mechanical Engineer of CPPI was made to evaluate on the following condition.

FOR SECOND EQUIPMENT:

A packing equipment that cost P 500,000 will last for 20 years and have a scrap value at that time of P 50,000. Repair will average P 30,000 per year in which the e P 20,000 per month. The as many units per year. lt costs 25 years at a cost of P 1,OOO,OOO. Repairs for this machine will average P 25,000 per year and operating expenses will be P 30,000 a month.

Annuat

B.'

c. D.

= Ico 9") i

o.c. -r Eq. 2

[tt*i)'-t.]

Substitute values in Eq.2

P120,978.20 P120,962.20 P120,988.20 P120,999.20

Annual cost

518,678.78

Difference = 639,666.98

Compare the annual cost of the two packaging equipments:

+ co(i) +

O.C. = 25,000 + 30,000(12) O.C. = 385,000

lf money is worlh 8% effective, How much will be saved each year if the more economical equipment is used?

"o",

518,678.68 = 120,988.20

.'. The amount that will be saved each year if the more economical

equipment is used is P 120,988.20.

FOR FIRST EQUIPMENT: Annual cost = depreciation + interest of first cost + operating cost Using sinking fund method in computing depreciation:

Annuar

"o",

= Ico 9") i

[(t*i)"

+ co(i)+ o.c.

Eq.

For its proposed expansion, an ice plant company is selectlng from two offers of ice cans. The data on the offers are as follows:

Total cost Annual maintenance No. of vears. Life

-t.1

Solving for total operating cost, O.C. = 30,000 + 20,000(12) O.C. = 270,000

Annuat

"o.t

(509

P 640.000.00

00, - 50'9-00)(9' 08) (r+o.oa)'?o

-r

+ (500, 000)(0. 08) + 270, 000

Annual cost 1'tg,gsz.qg // ,/

Offer B 1.6 mm thick 90.000.00

For their replacements, the company is putting up a sinking fund to eam 16% interest compounded annually. lf the money to purchase the ice cans is to be borrowed at 20% annual interest and the tax on the first cost is 2%, what is the difference in the annual cost of the two offers?

o.c.

Substituting values in Eq.

Offer A 1.8 mm thick P 720,000.00 P 60.000.00

Since the,(econd equipment can produce twice as many units than the first equipmCnt, we will consider acquiring 2 units for the first equipment to balance the production, thus

A. P 34,489.01 B. P 34,OM.92 c. P 34,320.12 D.

P 34,617.22

,%)r,*. For Offer A: Using sinking fund method in computing depreciation

250

Qtrestion llank

Annuatcost =

jt'-:')

(lrst

251

r,,r,, ,.

For alternative A:

(220'000 -0)-(0' 1 9) (t *

lrrpilrrlizorl ()ost & Attttuitl

l+co(i)+o.c.

f(t+r) -r]

Annual cost = Annual cost

(

Engineeling Dcortrlrnics by Jaime Il. 'l'iorrg

+ (720, 000)(0 .2o) + o.o2(7 20, 000) + 60, 000

Using sinking fund method in computing depreciation

o.to)'' - t.]

Let x = number of days the equipment is used in one year period

241,738.61

Annuat cost

For Offer B:

-9") +co(i)+o.c. = it'(1+i) I I L") -1 |

Using sinking fund method in computing depreciation

(c^ Annual 6ss1 =

-c_)

l(t*i) Annual cost

Annual cost = i

+ Cn(i) + O.C.

-1

Annual cost = 151,508.78

-t.1

(6-40'000 -0X0' 1-6)

(700'000

+ (6a0, 000)(0.20) + 0.02(6a0,000) + 90, 000

00'000X0' 1 8)

+ (700,000)(0.1 8) + 500x

500x

For alternative B

ftt*o.to)'-t.]

Annual rental = 1,500x

Annual cost = 275,743.53

Equating: Solving for the difference between the two offers: Difference = 27 5,7 43.53 Difference =34,004.92

x = 151.5 days

The difference between offer A and offer B is P 34,004.92.

A company is considering two alternatives with regards to an equipment which it needs. The alternatives are as follows: Alternative A: PURCHASE Cost of equipment Economic life ............ .... Salvage value ... Daily operating cost ........

P 700,000.00 10 years P 100,000 P 500.00

Alternative B: RENTAL at P 1,500 per day

At 18% interest, how many days per year must the equipment be in use if alternative A is chosen?

A. B. C.

152 days 154 days 156 days

151,508.78 + 500x = 1,500x 1,000x = 151,508.78

241,7 38.61

.'. The equipment must be used lor 152 days per year.

A new broiler was installed by a textile plant at a total cost of P 300,000 and projected to have a useful life of 15 years. At the end of its useful life, it is estimated to have a salvage value of P 30,000. Determine its capitalized cost if interest is 18% compounded annually.

A. P 323,500.33 B. P 322,549.33 c. P 332,509.33 D.

P 341,240.33

.Vn/oln.'

Capitalized cost = Co + present worth of all depreciation charges Let P = present worth of all depreciation charges using sinking fund method

252

Question Bank D

Engineering Economics by Jaime R. Tiong

Cepitalized Coet & Annual Coet 253

_ 300'000-30'000

Annuatcost

(1+Ol8f

9-:") [(t+i)"

P =22,549.33

Annual cost Capitalized cost = 300,000 + 22,549.33 Capitalized cost = 322,549.33

l+co(r)+o.c.

-t]

ffiffi

({200)(0'12) + 1'200(0'01)

Annualcost = 224.38

.'. The capitalized cost is P 322,549.33.

FOR TREATED POLE: Using sinking fund method in computing depreciation

A man planned of building a house. The cost of construction is P 500,000 while annual maintenance cost is estimated at P 10,000. lf the interest rate is 6%, what is the capitalized cost of the house?

A, B. c. D.

P 666,000.00 P 666,666.67

I?'

Annuatcost

Annuat cost

,!")l+co(i)+o.c.

f(t.r[ -t]

(co -0)(9--12)

- +(co)(0.12)+co(0.01)

[(t+o.tz)'" -t.]

P 633,333.33 P 650,000.00

Annual cost

0.14388 Co

Equating annual costs:

Capitalized cost = Co +

0.14388C0 =224.38 Co = 1'559'50

Capitalized cost = 5OO,0oO +

]lPqq 0.06

Capitalized cost = 666,666.67

The maximum justifiable amount that can be paid'for the treated pole is P 1,559.50.

.'. The capitalized cost is P 666,666.67.

An untreated electrical wooden pole that will last 10 years under a ce(ain soil condition, costs P 1,200.00. lf a treated pole will last for 20 years, what is the maximum justifiable amount that can be paid for the treated pole, if the maximum return on investment is 12o/o. Consider annual taxes and insurance amount to be 1 % of the first cost.

A. B. c. D.

P 1,612.01 P 1,559.50 P 1,789.23

A granite quarry purchased for P 1,600,000 is expected to be exhausted at the end of 4 years. lf the resale value of the land is P 100,000, what annual income is required to yield an investment rate of 12%o? Use a sinking fund rate of 3%.

A. P 550,540.57 B. P 550,540.57 c. P 550,540.57 D.

P 550,540.57

92b.".

P 1,409.38 Using sinking fund method in computing depreciation

,V./,ir,rn FOR UNTREATED POLE: Using sinking fund method in computing depreciation

Annuat

"o.1

$5f") ftt.i[

-t1

+ co(i)+

o.c.

254

Questiou Bank ,Annual

cost

Engineering Economics by Jaime R. Tiorrg (1,600,ooo

-1

00,000)(0.03)

()npil,alizod Oost & Annual Cost 255

where:

+ (1,600,000X0.12)

Q = discharge in cublc meter / second or

[{r*o.os)'-r] Annual cost

550,540.57

= density of water = 9810

E = head in meters

For an assufance of rate of return of 12o/o,lhe one used in computing annual cost, the annual income must be equal to annual cost.

For full load operation, efficiency =

7Oolo

(e'810x43) P_ 4s.5

Annual income = Annual cost Annual income = 550,540.57

3,600 0.70 7,616.375 watts

.'. The annual income required is P 550,540.57.

7.6'16375 kw

cost of power = (7.616375 rw1[2a hours][tso An ice plant owner decided to install a water pump from a nearby clean fresh water source to supply its water requirements. The capacity of the pump is 45.5 cubic meters per hour against a head of 43 m with an efficiency of 70% at full load an 40% al half load. The total cost of pumping unit including the piping system to the plant is P 700,000 with zero salvage value at the end of its estimated life of 20 years. The prime mover of the pump is an electric motor and cost of power is P 2.10 per kw-hr. Taxes, insurance and maintenance cost is 1.5% of the first cost per year. The pump will operate 200 days continuous operation per year at half load and 150 days at full load. lf the cost of money is 12o/o, how much will it cost the owner to operate the pump per cubic meter?

Cost.of.power = P 57,579.80 per year For half load operation, efficiency = 40yo

45.5 (e,810X43)-

' _- 3,600 o4o D

P = 6,664.328 watts

P = 6.664328

cost of power = (6,664.328 my[24 hours )[ zoo oavs )f - P2' 10 '(

A. P 0.84 B. P 0.20 c. P 1.15 D.

oavs)rE!)

'( day /1. yeat./l.tw-nr/

day /( year /(kw-hr/)

Cost of pow€r = P 67,'1 70.60 per year

P 1.35 Total cost of power = 57,579.80 + 67,170.60 Total cost of power = 124,75O.4O

9"2a... Using sinking fund method in computing depreciation

Annuar

"or1

-9") I + co(i)+ o.c f(t+i)" -t.l

= Ico

Eq.1

Analyzing operating cost, O.C.

Solving for operating cost; O.C. = 1 24,7 50.40 + 0.0 1 5(7OO,O0O)

o.c. = 135,250.40 Substituting values in Eq.

Operating cost = cost of power + cost of taxes, insurance and maintenance Solving for cost of power:

P=QorE,

Annuar cost

Annual cost

tffiffi

+ (700,000)(0 .12) + 13s,2s0.40

228,965.546

Solving for the cost per cubic meter:

256

Question Bank

Engineering Economics by Jaime R. Tiong

Oepitalizod Cost & Annual Coet 257

Let V = total volume pumped per year

Annualcost = 2,716.67

V = volume during full load + volume during half load (

For Machine B:

+s.sm')(z+nours)(rSodavs\ / t \(+s.sm3'\(z+nours\t. 200davs'l

'=[--.](. ., _ 273,000

d,y

]t v..-1.[zJ[ n' ]t *, ll.-y;)

Annual cost = depreciation + interest of first cost + operating cost

year Therefore,

Annual cost

a^ - vn vo

+ c^i + o.c.

(20.000 - 6.000)

---G-+(20,000)(0.08)+350

Annualcost = 4,283.33

228.965.546 P0.839

273,000

Annual cost

Difference = 4,283.33 -2,716.67 Difference = 1,566.66

.'. The cost to operate the pump is 84 centavos per cubic meter. .'. The machine to choose is machine A and the difference of annual costs

is P 1,566.66. Make an economic analysis to determine which of the following two machines capable of performing the same task in a given amount of time, should be purchased. The minimum attractive return is 8%.

First cost

Estimated life Salvaqe value Annual maintenance

Machine A P 10.000 6 years 0 P 250

Machine B P 20.000 14 vears P 6,000

P 300

Basing on annual cost, using straight line method of computing depreciation, determine which machine you would choose and how much is the difference in its annual cost?

A. B. C. D.

Machine B, P Machine B, P Machine A, P Machine A, P

1,205.06 1,566.67 1,205.06 1566.67

,'%h-.

A decision by the supervising mechanical engineer must be made whether to replace a certain engine with a new one or to rebore the cylinders of the old engine and thoroughly recondition it. The original cost of the engine 10 years ago was P70,000, to rebore and recondition it now will cost P 28,000 but would extend its useful life for 5 years. A new engine will have a first cost of P 62,000 and will have an estimated life of 10 years. lt is expected that the annual costs of fuel and lubricants with the reconditioned engine will be about P 20,000 and this cost will be 15% less with the new engine. lt is also believed that repairs will be P 2,500 a year less with the new engine than with the reconditioned one. Assume that neither engine has any salvage value when retired. lf money is worth 10%, how much money is saved each year if the old engine will be replaced?

A. B. c. D.

P 2,796.12

P 2,123.45 P 2,340.12 P 2,555.30

Compare annual cost of either engine. For Machine A:

For reconditioned engine:

cost: depreciation + interest of first cost + operating cost to tn Annual cost = * coi + o.c. -n Annual

Annuat cost

(10.000 - 0)

= #+(10,000)(0.08)+2S0 6

Using sinking fund method in computing depreciation

Annual cost

(c^

-c-)

[(t.

i)"

- t]

-+C^i

+O.C.

258

Question Bank

Engineering Economics by Jaime R. Tiong

Annuat cost

Annual cost

(?8'000

0)-(0'1-0)

ftr.o rof -rl

Capitalized Cost & Annual Cost 259 + (28,000)(0.10)+ (20,000 + 2,500)

29,886.33

Using sinking fund method in computing depreciation

Annual cost ^rrrru'rr uosl

Using sinking fund method in computing depreciation

Annuat rr ruar ^r

"o.,

jco - c") I =

= """' "o.,

[{t.i)" -t

+ c^i + o.c.

+o'c'

0)fi-T

Annuat cost

Annuar cosr

$2,ooo

,2oo)(0.2o)

+ (52'000)(0 20) + (8'300 + 27

.rl;',o"T-

'600)

For diesel driven unit:

(9''ooo - o)(o'10)

I;;r

-C ) i

FE+coi

For new engine:

Annuar

'(C"

+ (62' 000)(0' 1 0) + (20' 000)(0'85)

Annual cost = 27,090.21 Difference = 29,886.33 -22,090.2i Difference = 2,796.12

.'. The amount that witl be saved each year is the old engine is replaced with a new one is P 2,796.12.

Using sinking fund method in computing depreciation

Annual wo'( cost

lco

-C,)

fGf-f+coi+oc' Annuat cost

(98,_250 _ 1i,9oo)(0.20)

ffr. ozol'o_l-

+ (e8'250)(0'20) + (5'800 + 18,800)

Annualcost = 47,576.44 Solving for the difference:

Excel Review Center is considering the purchase of a generating set to the electrical needs during brown outs.

The first offer he received is for a gasoline driven unit with an estimated life span of 5 years and would cost the review center p 52,000. The annual maintenance cost is estimated to be P 8,300, annual operating cost would be p 27,600 and salvage value would be P 7,200. The second offer the review center received is for a diesel engine driven unit with an estimated life span of l0 years. Annual maintenance cost is P 5,800, annual operating cost is p 1g,goo and with salvage value of P 1 1,900. The price of the unit is 98,250.

Difference = 52,320.21 _ 47,S7 6.44

Difference = 4,243.22

]ff ,1i:::llffi,U:5;"" ffiJ#;,,"f ", p 4,7 43 7 t c hea pe r tha n the

sewa ge havin g .or" crete, cast iron and steel. Because of -r 'J^^^ ^!,,

"orro"r]h

How much cheaper is the annual cost of the diesel driven unit than the gasoline driven unit?

A. B. c. D.

P 4,612.04 P 4,743.77 P 4,490.12 P 4,500j2

For gasoline driven unit:

____---.ljv

q,,v vqDr rrun ptpe nave the same annual costs. I

260

Question Bank

-Ungineering Econotttit:s bv Jairne

ll. 'liong

Question Bank 9

,7Lr,"Compare annual cost of the three bids For reinforced concrete PiPe: Using sinking fund method in computing depreciation

Annuat cost

= Jt' -:") I +coi+o.c. ftr*i)'-t1

Annuatcost

Annual cost

+(770,000)(0.05)+900

A manufacturer produces certain items at a labor cost per unit of p 315, material cost per unit is P100, variable cost of P 3.00 each. lf the item has.a selling price of P 995, how many units must be manufactured each month for the rnanufacturer to breakeven if the monthly overhead is P 461 ,600?

A. 782 B. 800 c. 806

40,417.64

For cast iron piPe:

,7"2a,

Using sinking fund method in computing depreciation

Annuat cost

Annuar cosr

Annual e,ost

: Itt -:") I + coi+ o.c

Let x = number of units to be manufactured per month to break-even

ftt*i)" -t1

= =

(190

000

Material cost = 100x

0)-(0'0-5)

Labor + (790,000)(0.05) +

600

[(t*o.os)" -t.]

cost

= 315x

Variable cost = 3x Total expensss = (100x + 315x + 3x) + 461,500 Total expenses = 418x + 461,600

40,402.68

For steel piPe: Using sinking fund method in computing depreciation Annuar cost

f+"!

+ coi+ o.c.

[(t*i)'-t1 Annuatcost Annual cost

= =

577x = 461,600

+(580,000)(0,05)+3,600 35,370.51

The steel pipe has the least annual cost.

x=800 .'. The number of units to be manufactured each month to breakeven is 800 units.

262

Question Bank

Engineering Economics by Jaime R. Tiong

Break-evenAnalysis 263 Expenses = 1 1Sx + 76x + 2.32x +42g,OOO Expenses = .1g3.32x + 42g,O0O

The annual maintenance cost of a machine shop is p 69,994:rf the cost of making a forging is p 56 per unit and its selling price is n f SS pei unit, flnd the number of units to be forged to break_evei.'

fpffi

To breakeven: lncome = Expenses

A. 892 B. 870 c. 886 D.

600x=j93.32x+42g,000 406.68x = 428,000 x = 1,052.42 r 1,053 units

862

,tEJXXlfr

of units to be manufactured each month to breakeven !s

Let: x = number of units to be forged to break even

lncome = 135x

Expenses=69,994+S6x To breakeven; lncome = Expenses

135x=69,994+56x 79x = 69,994 x = 886 units

Steel

yearly fixed operating cost of $ 200,000. drum - Each produce and sells $ 200. What is the manu manU.--.-,-, e vreo^-Eve, sales volume i" dr.;;;::;; A. 5,031 B. 5,0,17 c. 5,043

5,000

.'. The number of units to be forged to breakeven is gg6 units. Let: x = number of drums to be sold out per year to breakeven lncome = Expenses

A manufacturer produces certain items at a labor cost of p 1 15 each, material cost of P 76 each and variabre cost of p2.32 each. rf the item has a unit price of P600, how many units must be manufactured each month tor tn" ,anuticiiiel to break even if the monthly overhead is p 42g,000.

A. 1,033 B. 1,037 c. 1,043 D.

200x=200,000+160x 40x = 200,000 x = 5,000 drums

The number of drums to be sord each year to breakeven is 5,000 drums.

1,053 kcases that sell for p 65.00 each. lt costs to operate its plant. This sum inclules rent, Let: x = number of units to be manufactured per month to breakeven

11_oe"**,;;fii;o:#,x,;"JlUi"i;,;y.ffi Ttoid taking a loss?

,#il"ii,"^11[?,1ffi ;

lncome = 600x Expenses:

Labor

Materials

= 115x

76x

Variable = 2.32x

Overhead = 428,000

9;2r.,..

264

Question Bank

Engineering Economics by Jaime R. f iong

Broak'cven Analysis 265

To avoid taking loss means that the minimum production must be enough to breakeven.

Let: x = number of units to be forged to breakeven

lncome = Expenses

Let; x = number of cases to be sold each year to breakeven lncome = Expenses

125x=56x+70,000 69x = 70,000 x = 1,014.49 units

65x=50x+35,000

15x = 35,000 x = 2,333.33 cases

2,334 cases

1,015 units

The number of units to be forged to breakeven is 1,015.

.'. The number of cases to be sold each year to breakeven is 2,334.

A company which manufactures electric motors has a production capacity of 200 motors a month. The variable costs are p 150.00 per motor. The average selling price of the motors is P 275.00. Fixed costs of the company amount to p zo,ooo per month which includes taxes. Find the number of motors that must be sold each month to breakeven.

A. B. c. D.

A manufacturing firm maintains one product assembly line to produce signal generators. Weekly demand for the generators is 35 units. The line oper-ates for 7 hours per day, 5 days per week. What is the maximum production time per unit in hours required of the line to meet the demand?

A. B.

C. D.

160 157 153

1.0 1.2 1.4 1.6

hour per unit hours per unit hours per unit hours per unit

,(y'.Z,,/*,,

163

Let x = number of hours to produce 1 unit v-l-ll-

Let: x = number of motors to be sold each month to breakeven

+ 20,000

)/ Sdavs\(Thours) ll-l

\35units/\ week x = t hour per unit

lncome = Expenses

275x= 150x

/ lweek

day )

.'. The maximum production time is

hour per unit.

125x = 20,000 x = 160 motors

.'. The number of motors to be sold each month to breakeven is 160.

The annual maintenance cost of a machine is p 70,000. lf the cost of making a forging is P 56 and its selling price is p 125 per forged unit. Find the numbeiof units to be forged to break even.

A. B. c. D. .72,*,,

1,000 1,O12 1,015 1,018

Compute for the number of blocks that an ice plant must be able to sell per month to break even based on the following data; Cost of electricity per block Tax to be paid per block Real Estate Tax Salaries and Wages Others Selling price of ice

A. B. c. D.

1.220 1,224

1,228 1,302

- P 20.00 - P 2.00 - P 3,500.00 per month - P 25,000.00 per month - P 12,000.00 per month - P 55.00 per block

266

Question Bank

l}roak'even Analysis 267

Engineering Economics by Jaime R. Tiong

.7'/u,,,,,

... The number of units to be producod and sold per month

to breakeven is

120. Let: x = number of blocks to be sold per month to breakeven

lncome = 55x Expenses:

ElectricitY Tax

= 20x

=2x

Fixed charges = 3,500 + 25,000

tZ,OOO =

40,500

To breakeven: lncome = Expenses

55x=20x+2x+40,500

33x = 40,500 x = 1,227 .3 blocks

,228 blocks

,'. The number of blocks to sel! per month to breafieven is 1,22g.

100

104

110

112

,92d,., Let x = number of motors to produce per month to break-even

Expenses=4000+78,000

JRT lndustries manufactures automatic voltage regulators at a labor cost of P85.00 per unit and fixed charges on the business are P 15,000 e p 20.00 [er unit. lf g0.00 each, how many the automatic voltage units must be produced an

Expenses = 82,000

materi per regu

A. B. c. D.

lncome = 750x To break-even, lncome = Expenses

120 124 128

750x = 82,000

x=109.33*1'10 units

130

... The number of motors

.VZ,./,r,,. Let: x = number of units to be produced per month to breakeven

lncome = 580x Expenses:

Labor = 85x Materials = 350x Variable = 20x

Fixed charges = 15,000

To breakeven: lncome = Expenses 580x = 85x + 350x + 20x+ 15,000

A. B.

C. D.

125x= 15,000 x = 120 units

.V"Z*,*.

1,000 1,040 1,100 1,120

feet feet feet feet

to produce each month to breakeven is 110.

268

Question Bank

Engineering gconomics by Jaime R. Tiong

Break-evenAnalysis 269

For enameled wire:

To breakeven, lncome = Expenses

Let x = length of wire

1800x = 1,000x + 38,333.33 800x = 38,333.33 x = 47.92: 4Sunits

Cost of wire = 0.55x Cost of soldering = '1.65(a00) = ggg Total cost = 0.5Sx + 660 For tinned wire:

lf 200 units are produced per month, there wiil be profit since it needs onry 48 units to break-even.

Let x = length of wire

Solving for profit:

Cost of wire = 0.75x Cost of soldering = 1.15(400) = 460 Total cost = 0.75x + 460

lncome = 200(1,800) = 360,000 Expenses = 1,000(200) + 3g,333.33 Expenses = 238,333.33

ln order to get the rength of wire that wiil give the same cost, equate totar cost of enameled wire and total cost of tirined wire.

0.55x+660=9.75r*460 0.2x=200

Profit = lncome

- Expenses Profit = 360,000 - 238,333.33 Profit = 121,666.61 .'. There is a profit of p 121,666.67..

x = 1,000

The length of cable to make cost of installation the same is i,000 feet.

A local factory assembring carcurators produces 400 units per month and sefls them at P 1,800 each. Dividends are 8% on the 8,000 snaiei wrn paivarue p ot 250 each. !1eO operating cost per month is p 2O,OOO. Other 6sts are p '1 per unit. lf 200 units weie produced per month, oetermine tne piont ,000 or toss.

A. B. C. D.

Profit of P 121,666.67 Profit of P 21,666.67 Loss of P 121,666.67 Loss of P 21,666.67

Nike shoes manufacturer produces a pair of Air Jordan shoes at a labor cost of p 900.00 a pair and a material cost of p B0O.O0 a pair. The fixeri charoes.an the nth and the variaLle cos pair. 1,000 per pair of shoes sell at must be produced each

A. B. c. D.

2,590

2,632 2,712 2,890

VZaLet x = number ofcalculators produced per month to break_even Expenses = 1,000x + 2S,OOO + g,000(250) Expenses = 1,000x + 38,333.33

lncome = 1,800x

(H)

Let x = number of pairs of shoes produced each month to break_even Expenses = 900x + 800x + 400x + 1,000x + 5,OO0,O0O Expenses = 3,1 00x + 5,0OO,OOO

lncome = 5,000x To break-even,

lncome = Expenses

,iI

270

Question Bank

Engineering Economics by Jaime R. Tiong

Break'evenAnalyeis

5,000x = 3,100x + 5,000,000 1,900x = 5,000,000 x = 2,631 .6 x 2,632 pairs of Air Jordan Shoes

271

0.75n = 0.0417n + 0.50n + 2000

0.75n=0.5417n+2000 0.2083n = 2000 n = 9601.54 holes

.'. The number of pairs of Air Jordan shoes to produce each month to

breakeven is 2,632.

Sayn=9601 holes The number of rivet holes for the multiple-punch machine to pay for

itself is 9,601.

minute. This method requires P0.50 per holes to set multiple punch machine and an installation cost of P 2,000.00. lf all other costs are the same, for what number of rivet holes will the multiple punch machine pay for itself?

A. B. c. D.

' A new Civil Engineer produces a certain construclion material at a labor cost of P per 7.4O P cost of per piece and variable per piece, P 38.50 cost of material 16.20 piece. The fixed charge on the business is P 100,000.00 per month. lf he sells the hnished product at P 95.00 each, how many pieces must be manufactured in eacl" month to breakeven?

A. 3,045 B. 3,035 c. 3,030

9,601

9,592 9,581

9,566

3,040

9"2*""

,V"z:ro,-.

Let x = number of pieces to be manufactured per month to breakeven

Using first method, the Drill Press:

Cost of machinist per hole

lncome = 95.00x

20:25

27 Cost of machinist per hole = 0.75

Expenses:

Labor Materials

Using the second method, the Multiple punch Machine:

Variable Fixed charges

Cost of machinist per hole

= 8(60) 1-0-=

= 16.20x = 38.50x

7.40x = 100,000

To breakeven:

Cost of machinist per hole = 0.0417

Additional cost for setting up th€ machine per hole = 0.50 lnstallation cost = 2,000.00

lncome = Expenses 95.00x = 16.20x + 38.50x + 7.40x+ 100,000 32.90x = 100,000 x = 3039.5 say 3040 Pieces

Let n = number of rivets holes that will make the multiple punch machine pay for itself

Total cost using drill pr€SS = total cost using multiple puhph machine

The number of pieces to be manufactured each month to breakeven is 3,040.

272

Question Bank

Engineering Economics by Jaime R. Tiong

Break'evenAnalyeie 273 To breakeven,

Co uni

n Transmission

ume is 500,000 Fixed expenses tot a year?

A. B. c.

lncome = Expenses

50)

O.Sx = 80,000

x = 160,000

The breakeven point is 160,000 units.

P 168,000 P 170,000 P 172,000 P 174,000

Pro&ction

pacity due to portation of their Solving for the total profit, P:

'l9O,0OO.O0 and

loss?

500,000 units P0.50 tncome x -year

A. P 87,450 B. P 88,960 c. P 88,450

lncome = P250,000

Expenses = P80,000 Profit = lncome

costs are P variable costs are P 0.348 per unit. What is the current profit or

Expenses

P 87,960

,7o/,sru.,..

Profit=250,000-80,000 Profit = 170,000

Solving for the current profit or loss:

.'. The present total profit is P 170,000.

Pr oduction

0.62(700,000)

Production = €a,000

nsm s 50 exp nt in

A. B. c.

160,000 162,000 165,000 170,000

.V./,:,r,r,,

sent O.SO)

rotat expens"r

(ffi)(+s+,ooo

Total expenses = 341,032 Total income = 430,000 Pr qfit = total income - total expense

Profit= 430,000 -341032 Profit = 88,960 .'. The current profit is P 88,960.

Let x = number of units produced to breakeven

lncome = 0.5x Expenses = 80,000

units)+ 1eo,ooo

274

Question Bank

Engineering Economics by Jaime R. Tiong

ation of an automobile part with a production ear is only operating at62% of capacity due to foreign currency to finance the importation of their me is P 430,000.00. Annual fixed costs are P190,000.00 and variable costs are P 0.348 per unit. what is the breakeven point?

A. 294,560 B. 291,000 c. 290,780 D.

295,490

It'r'it k-rrvt'tt nnrilrll\ -275

.,af

At A certain firm has the capacity to produce 650,000 units of product perl'' present, it is operating al62% capacity. The firm's annual income is. . ,sts"r" P4,160,000.00. Annual fixed costs are P 1 920,000.00 and the variableni equal to P 3.56 per unit of product. What is the firm's annual profit or lo5)

A. B. c. D.

P 814,320 P 815,230 P 816,567 P 817.239

,%,n -.

.7;/)zr-.

Pr oduction = 0.62(650,000)

Let x = number of units to breakeven

Production = 403,000

Solving for the expenses:

Total expenses = 1,920,000 + 3.56(403,000) Total expenses = 3,354,680

Variable cost = 0.348x Fixed cost = 190,000

Total Income = 4, 1 69,000

Solving for the income:

Pr ofit = total income - total expenses

Pr oduction = O.OZIzOO,OOO;

Profit = 4,169,000 - 3,354,680 Profit = 814,320

Production = 434,000 Totat expens""

s+a)1oao,ooo = [0unrt \ ,\

Tqtal expenses

units)+190,000 /

341,032

Totalincome = 430,000 settins prics

.'. The firm's annual profit is P 814,320.

= i:9'999 434,}Uo

0.00. 3.56

= o.es1 per unit

p r,gzo,oo'o.oo

in" ""0does ""ti"5;ec0sls the firdl at volume of sales

breakeven?

Total income = 0.99x

A. B. c. D.

To breakeven:

lncome: Expenses

,re?r. At A certain firm has the capacity to produce 650,000 units of product peil . The flrm's annual income is c0! is op

P 3,354,680 P 3,534,880 P 3,155,690

P 3,254,680

__-

0.991x = 0.348x + 1 90,000 0.643x = 190,000

Solving for the flrm's annual profit or loss:

x = 295,490 Pr oduction = 0.62(650, 000)

The current breakeven point is 295,490.

Pr oduction = 403,000

are

276

Question Bank

Brenk'ovon Analyeis 277

Engineering Economics by Jaime R. Tiong

Total expenses = 1,920,0Q Total expenses : 3,354,680

+ 3.56(403,000)

\ \ '\ Total income = 4,169,000 ln as much as the total expenses is P3,354,680, then the volume of sales must be equal to this value in order breakeven. Thus, volume of sales = P3,354,680 .'. The firm's volume of sales to breakevpn is P 3,354,680.

.'. The number of thresherg to be produced and sold annually to

breakeven is 25.

The direct labor cost and material cost of a certain product are P 300 and P 400 per unit, respectively. Fixed charges are P 100,000 per month and other variable qosts are P 100 per unit. lf the product is sold at P 1,200 per unit, howmany units must be produced and sold to breakeven?

A. 280 B. 250 c. 260 D.

A small shop in Bulacan fabricates threshers for palay producers in the locality. The shop can produce each thresher at a labor cost of P1,800.00. The cost of materials for each unit is P 2,500.00. The variable costs amounts to P 650.00 per unit while fixed charges incurred per annum totals P 69,000.00. lf the portable threshers are sold at P 7,800.00 per unit, how many units must be produced and sold per annum to breakeven?

A. B. c.

25 26 27

270

Let x = number of units produced per month to breakeven Expenses: Direct labor = 300x Direct material cost = 400x Variable = 100x Fixed = 100,000

cost cost charges

Total expenses = 300x + 400x + 100x + 100,000 Total expenses = 800x + 100,000

,7"22*^ Let x = number of threshers produced and sold in 1 year Expenses: Labdr

Total income =1,200x

To breakeven:

cost

= 1,800x Materials cost = 2,500x Variable cost'= 650x Fixed charge = 69,000

Total expenses = 1,800x + 2,500x + 650x + 69,000 Total expenses = 4,950x + 69,000 Total income = 7,800x To breakeven: Total income =Total expenses

7,800x=4950x+69,000 2,850x = 69,000 x =24.2 say 25 threshers

Total income = total expenses

1,200x=800x+100,000 400x = 100,000

x=25O .'. The number of units to be produced and sold monthly to breakeven is 250.

278

Question Bank

Engincering Ilcouourir:s by Jairnc

llrr'nk-cvon Analvsis 279

'l'iorrg

'-/nl,t;,, The following data for year 2000 are drqllable for Cagayan Automotive Company which manufactures and sells a single arlto-motive product line: Unit selling price Unit variable cost Unit contribution margin Total fixed costs

..... ..... ..... .....

P P P P

Let x = number of units produced per month to breakeven Expenses: Labor

40.00 20.00 20.00 200,000.00

What is the breakeven point in units for the currentyear?

A. B. c.

cost

= 230x

Material cost = 370x Variable cost = 10x Fixed charges = 1,000,000 Total expenses = 230x + 370x + 10 x + 1,000,000 Total expenses = 61 0x + 1,000,000

10,000 10,100 10,050 10,200

Total income = 750x

To breakeven: Total income = total expenses Let x = number of units produced per month to breakeven Expenses: Variable cost Fixed

charges

750x=610x+1,000,000 140x = 1,000,000

= 2Ox = 200,000

x=7,142.8 say 7,143 ynits .'. The number of sets to produce per month to breakeven is 7,143.

Total expenses = 20x + 200,000 Total income = 40x

To breakeven: Total income = totalexpenses

4Ox=2Ox+ 200,000 20x = 200,000 x = 10,000

.'. The breakeven point for the current year is 10,000.

The cost of producing a small transistor radio set consists of P 230.00 for labor and P 370.00 for material. The fixed charges in operating the plant is P 1 ,000,000.00 per month The variable cost is P 10.00 per set. The radio set can be sold for P750.00 each. Determine how many sets must be produced per month to break even.

An item which can be sold for P 63.00 per unit wholesale is being produced with the following coSt data: Labor cost Material cost Fixed charges Variable cost

= P 10.00 per unit = P 15.00 per unit = P 10,000.00 = P 8.00 per unit

What is the breakeven point sales volume if one out of every 10 units produced is defective and is rejected with only full recovery on materials?

A. B. c. D.

P 25,011 P 25,111 P 25,121 P 25,132

Let x = number of units produced per month to breakeven

280

Question Bank

Engineering Economics by Jaime R. Tiong

Question Bank

Expenses:

Laborcost Material

= 10x

cost = 15x

= 8x

Variable cost ,/ Fixed charges = 10,990 Total expenses = 10x + 15x + 8 x + 10,000 Total expenses = 33x + 10,000 Total income = 63x

A. T-bills B. Bank note C. Check D. Coupon

To breakeven: Total income = total expenses

63x=33x+10,000 30x = 10,000 x = 333.33

say 334 units

[Pffi)(ss+ rnit") \ unit /' Breakeven sales volume --P4,042 Breakeven sates votume =

lf 1 out of 10 (10%) is defective and rejected: Number of units sold per month = 0.90x Total expenses = 10x + 15x +8 x + 10,000 - 15(0.10x) Total expenses = 31.5x + 10,000

A. B. C D.

Devaluation Deflation lnflation Depreciation

Posl ME Boord Exom It is a series of equal payments occurring at equal interval of time.

= 56.7x

To breakeven: Posl illE Boord Exom The place where buyers and sellers come together.

Total income = total expenses 56.7x = 31 .5x + 10,000

A. B. C D.

25.2x= 10,000 x = 396.8 say 397

Breakeven sates votume =

(ffi)t

r"*1

Breakeven sales volume =P25,011

The breakeven sales volume is P 25,011.00.

Posr

Market Business Recreation center Buy and sell section ECE

Boqrd Exom

A market whereby there is only one buyer of an item for which there are no goods substitute.

Monopsony Oligopoly Monopoly Oligopsony

6. Post E(E Boord Exom It is a series of equal payments

occurring at equal interval of time where the first payment is made after several periods, afterthe beginning of the payment.

Post ECE Boord Exsm Reduction in the level of national income and output usually accompanied by the fall in the general price level.

A. Annuity B. Debt C. Amortization D. Deposit

Total income = 63(0.90)x

A. B C D

Post E(E Boord Exom The paper currency issued by the Central Bank which forms part of the country's money supply.

A. Perpetuity B. Ordinary annuity C. Annuity due D. Deferred annuity t.

Posl lhE Boord Exom The total income'equals the total operating cost.

A. B. C. D.

Balanced Sheet ln-place value Check and balance Break even - no gain no loss

8. Post IllE Boord Exom Kind of obligation which has no condition attached.

A. Analytic B Pure C. Gratuitous D. Private 9. Posl ftlE Boord Exonr Direct labor costs incurred in the factory and direct material costs are the costs of all materials that go into production. The sum of these two direct costs is known as

A. B.

GS and A expenses Operating and maintenance

costs

282

Question Bank

C. D.

Engineering Economics by Jaime R. Tiong

Prime cost

OandMcosts

Posl tE Boord Exomm An index of short tem paying ability is called

10.

lf.

receivable turn-over profit margin ratio current ratio acid-test ratio

Post

ilE

Boord lxom

infinite period of time.

A B

I7.

Demand

Supply

c. D.

Stocks Goods

Posl

A. B.

President Board of Directors Chairman of the Board Stockholders

Posl

A. B.

19.

A. B

Posl f,lE Boord Exom Additional information of prospective bidders on contract documents issued prior to bidding date.

c D.

Equitable Public Private Pure

20.

Posl illE Boord Exom Decrease in the value of a physical property due to the passage of time.

Technological assessment Bid bulletin

A. B.

lnflation Depletion Recession

Posl

ECE

26. Porr E(E Boord

Post E(E Boord Exom It is defined to be the capacity of a commodity to satisfy human want.

A. B. G. D.

Discount Luxury Necessity Utility

27.

A. B. C. D. 25.

Fair value Market value Book value Salvage value

A. Perfect competition B. Oligopoly C. Monopoly D. Elastic demand

Post

ECE

Boord Exom

somewhat the same quantity even though the price varies considerably.

28.

A. B C. D.

Utilities Necessities Luxuries Product goods and services

Post

ECE

Boord Erom

A condition where only few individuals produce a certain product and that any action of one will lead to almost the same action of the others.

A. B. C. D.

Oligopoly Semi-monopoly Monopoly Perfect competition

29.

Post ME Boord Exon Grand total of the assets and operational capability of a corporation.

A. Authorized capital B. lnvestment C. Subscribed capital D. Money market

Posl E(E Boord Exom

This occurs in a situation where a commodity or service is supplied by a number of vendors and there is nothing to prevent additional vendors entering the market.

Exom

These are products or services that are required to support human life and activities, that will be purchased in

24.

Post ECE Boord Exom It is the amount which a willing buyer will pay to a willing seller for a property whbre each has equal advantage and is under no compulsion to buy or sell.

283

A. Utilities B . Necessities C Luxuries D. Product goods and services

Company Partnership Corporation Boord Exom

Test

These are products or services that are desired by human and will be purchased if money is available after the required necessities have been obtained.

23.

Boord Exom Type of ownership in business where individuals exercise and enjoy the right in their own interest.

14.

Sole proprietorship

A. nominal rate B. rate of return C. exact interest rate D. effective rate

Boord Exom

an asset a liability an expenses an owner's equity

A. B. C. D.

We may classify an interest rate, which specifies the actual rate of interest on the principal for one year as

Post IllE Boord Exom What is the highest position in the corporation?

Post tlE Boord April Exom Consists of the actual counting or determination of the actual quantity of the materials on hand as of a given date.

Delict Escalatory

22.

18.

13.

A. B. C. D.

A. Physical inventory B. Material update C. Technological assessment D. Material count

Posl lhE Bocrd Exom An association of two or more individuals for the purpose of operating a business as co-owners for profit.

Work-in process is classified as

Posl tlE Boord Exom Estimated value at the end of the useful life. Market value Fair value Salvage value Book value

Depreciation Annuity Perpetuity lnflation

Posl tE Boord Exom The quantity of a certain commodity that is offered for sale at a certain price at a given place and time.

12.

A. B. C. D.

Depreciation

21.

I6.

An artificial expenses that spreads the purchase price of an asset or another property over a number of years.

A. Depreciation B. Sinking fund C. Amnesty D. Bond

Fost IilE Boord Exom

A series of uniform accounts over an

A B. C. D.

Mull;ipla Ohoico

30.

Posl IIIE Boord Exom

The worth of the property equals to the original cost less depreciation.

A B C. D.

Scrap value Face value Market value Book value

284 31.

Question Bank

Posl

Engineering Economice by Jaime R. Tiong

Eoord Exom

- A---- profit margin

B. gross margin C. net income D. rate of return

Money paid for the use of borrowed capital.

A. B. C. D.

Discount Credit lnterest Profit

Post tE Boord Exom Liquid assets such as cash and other assets that can be converted quickly into cash, such as accounts receivable and merchandise are called

37.

Posr

A. lnitial investment B. Current accounts C. Woking capital D. Subscribed capital

A. B. C. D.

39.

Queslion A market situation where there is only one seller with many buyer.

A Monopoly B. Monopsony C. Oligopoly D. Oligopsony

Boord Exom The provision in the contract that indicates the possible adjustment of material cost and labor cost. Secondary clause Escalatory clause Cohtingency clause Main clause

tl0.

A. B. C. D. 36.

book value capital recovery depreciation recovery sinking fund

Post tlE Boord Exsm

Gross profrt, sales less cost of goods sold, as a percentage of sales is called

Queslion

A market situation where there are few sellers and few buyers.

A. Oligopoly B. Oligopsony C. Bilateral oligopoly D. Bilateraloligopsony

35.

Posr tlE Boord Exom The present worth of all depreciation over the economic life of the item is called

Fair value Market value Salvage value Book value

Posi tlE Boord [xom Those funds that are required to make the enterprise or project a going [emailprotected]

Posl tE Boord Exorn The length of time which the property may be operated at a profit.

3f.

A. B. C. D. 38.

33.

A. Physical life B. Economic life C. Operating life D. All of the above

42. Qucrllon A market situation where there are only two buyers with many sellers.

A. B. C. D.

Boord Exom Worth of the property as shown in the accounting records of an enterprise.

32.

A. total assets B. fixed'assets C. current assets D. none ofthe above

Posl

Mult,iplo Choice

{1.

Queslion A market situation where there is one seller and one buyer.

A. B C. D.

Monopoly Monopsony Bilateral monopoly Bilateral monopsony

t[3.

Queslion

Qucrtlon The payment for the use of borrowed money is called

A. B. C. D. 48.

ft[.

Queslion Defined as the future value minus the present value.

A. lnterest B. Rate of return C- Discount D. Capital 15.

loan

maturity value interest principal

Oueslion

The interest rate at which the present work of the cash flow on a project is zero ofthe interest earned by an investment.

earn.

A. Present worth factor B. lnterest rate C. Time value of money D. Yield

285

f7.

Duopoly Oligopoly Duopsony Oligopsony

The cumulative effect of elapsed time on the money value of an event, based on the earning power of equivalent invested funds capital should or will

Test

A. Effective rate B. Nominal rate C. Rate of return D. Yield 19.

Question

The ratio of the interest payment to the principal for a given unit of time and usually expressed as a percentage of

the principal.

A. lnterest B. lnterest rate C. lnvestment D. All of the above

Question

The flow back of profit plus depreciation from a given project is called

A. B. C. D.

50.

Oueslion The true value of interest rate computed by equations for compound interest for a 1 year period is known as

capital recovery cash flow economic return earning value

f5.

0ueslion The profit derived from a project or business enterprisg without consideration of obligations to financial contributors or claims of other based on profit.

A. Economic return B. Yield C. Earning value D. Expected yield

A. B. C. D. 51.

expected return interest norninal interest

effective interest

Question

The intangible item of value from the exclusive right of a company to provide a specific product or service in a stated region of the country.

A. B. C. D.

Market value Book value Goodwill value Franchise value

286 52.

Question Bank

Oueslion

The recorded current value of an asset is known as

A. B. C. D. 53.

c. D.

scrap value salvage value book value present worth

Queslion

Scrap value of an asset is sometimes

Ordinary annuity Annuity due Deferred annuity Perpetuity

value?

55.

c. D.

Deferred annuity Delayed annuity Progressive annuity Simple annuity

Book value Going value

Queslion

B. D.

60.

63.

57.

Market valu Goodwill value Fair value Franchise vblue

Question

are made at the end ofeach payment period starting from the first period.

Declining balance method SYD method

67. 0ueslion

Ordinary annuity Aniruity due Deferred annuity Perpetuity

6f.

c. D.

A. B. C. D.

annuity

capital recovery annuity factor

amortization

68.

The reduction ofthe value of an asset due to crnstant use and passage of time

A. B. C. D. 65.

Scrap value Depletion Depreciation Book value

Question

an amount equal to the total depreciation of an asset at the end of the asset's estimated life.

A. B. C. D 69.

Straight line method Sinking fund rnethod Declining balance method SYD method

Question

The function of.interest rate and time that determines the cumulative amount of a sinking fund resulting from specific periodic deposits.

A. B. C. D.

Queslion

A method of computing depreciation in which the annual charge is a fixed percentage ofthe depreciated book value at the beginning of the year to which the depreciation applies.

Declining balance method Sinking fund method Straight line method SYD method

A method of depreciation where a fixed sum of money is regularly deposited at compound interest in a real or imaginary fund in order to accumulate

Queslion

The amounts of all payments are equal. The payments are made at equal interval of time The first payment is made at the beginning of each period. Compound interest is paid on all amounts in the annuity.

methods cannot have a salvage value o'f zero?

Queslion

Oueslion

Book value Fair value Goodwill value Going value

Question The value which a disinterested third party, different from the buyer and seller, will determine in order to establish a price acceptable to both parties.

demand factor sinking fund factor present worth factor

As applied to a capitalized asset, the distribution of the initial cost by a periodic changes to operation as in depreciation or the reduction of a debt by either periodic or irregular prearranged program is called

Which is NOT an essential element of an ordinary annuity?

56.

B. D.

operation.

load factor

Question A type of annuity where the payments are made at the start of each period, beginning from the first period.

Scrap value Salvage value

An intangible value which is actually operating concern has due to its

A. B. C. D.

59.

Ouestion What is sometimes called second hand

C D

Questlon

A mathematical expression also known as the present value of an annuity of one is called

Which of the following depreciation Ouestion It is a series of equal payments occurring at equal intervals of time where the first payment is made after several periods, after the beginning of the payment.

book value salvage value replacement value future value

51.

A. B. C. D.

62.

58.

known as

A. B. C. D.

Multiple Choice Test 287

Engineering Economics by Jaime R. Tiong

70.

Sinking fund factor Present worth factor Capacity factor Demand factor

Queslion

The first cost of any property includes

61.

Queslion

Sfaight line method

Sinking fund method SYD method Declining balance method

A is a periodic payment and I is the interest rate, then present worth of a

perpetuity =

A. Ai B. AiN c. An/ D. Ni

65.

Queslion A method of depreciation whereby the amount to recover is spread uniformly over the estimated life of the asset in terms of the periods or units of output.

A B

Straight line method Sinking fund method

A. B. C. D.

the original purchase price and freight and transportation charges

installationexpenses initial taxes and permits fee all of the above

Tl.Oueslion ln SYD method, the sum of years digit is calculated using which formula with n = number of useful years ofthe equipment.

288

Question Bank

n. "

n(n

o C. D.

n(n+1)

o.t

72.

76.

Queslion A distinct legal entity which can practically transact any business transaction which a real person could

-1)

2 2 n(n +1) n(n

do.

A. B. C. D.

-1)

Queslion

Capitalized cost of any property is equal to the

A. B. C. D. 73.

Mult,iplo Choico

Engineering Economics by Jaime R. Tiong Which is TRUE about PartnershiP?

Sole proprietorship

Enterprise Partnership Corporation

77. annual cost first cost + interest of the first cost first cost + cost ofperpetual maintenance first cost + salvage value

Question Double taxation is a disadvantage of which business organization?

A. B. C. D.

A. B. C.

78.

Sole proprietorship Partnership Corporation Enterprise

A. B

79.

Sole proprietorship Corporation Enterprise Partnership

Queslion

What is the minimum number of incorporators in order that a

75.

Question

An association of two or more persons for a purpose ofengaging in a profitable business.

A. B. C. D.

Sole proprietorship

4.3 B.5 c. 10

D.7

80.

Ouestion

I ln case of bankruptcy of a partnership, A B

Enterprise Partnership Corporation

The minimum number of incorPorators to start a

Ouestion Represent ownershiP, and enjoYs certain Preferences than ordinary stock. A. B.

C. D.

Authorized capital stock Preferred stock Common stock lncorporator's stock

partnership

the partners may sell stock to generate additional capital.

A. Bond B. T-bill C. Preferred stock D. Common stock 87.

Queslion

A form of fixed-interest security issued by central or local governments, companies, banks or other institutions. They are usually a form of long-term security, buY maY be irredeemable, secured or unsecured

A. B. C. D

Queslion Represent the ownershiP of stockholders who have a residual claim on the assets ofthe corporation after all other claims have been settled.

A. Authorized caPital stock B. Preferred stock C. lncorPorators stock D. Common stock 85.

Question

The amount of comPanY's Profits that the board of directors of the corporation decides to distribute to ordinary shareholders.

Bonds

T-bills Certificate of dePosit

All of these

88.

Queslion A type of bond where the corporation pledges securities which it owns (i.e. stocks, bonds of its subsidiaries).

A B. "C D.

84.

liabilities.

mortgage on certain assets of the corporation.

83.

the partners are not liable for the liabilities of the partnership. the partnership assets (excluding the partners personal assets) only will be used to pay the the partners personal assets are attached to the debt of the

A certificate of indebtness of a corporation usually for a period not less than '10 years and guaranteed bY a

corPoration are onlY liable to the extent of their investments.

corporation be organized?

B. Partnership C. Enterprise D. Corporation

lt is the not best form of business organization.

C. lts life is dependent on the lives of the incorPorators. D. The stockholders of the

organization?

A. B. C. D.

lts capitalization must be equal for each Partner.

corPoration is three.

Question

Which is NOT a type of business

Depreciation Depletion lnflation

another.

86. 0uestion

82.-Queslion Which is TRUE about corporation?

Ouestion

The lessening ofthe value ofan asset due to the decrease in the quantity available (referring to the natural resources, coal, oil, etc).

lt has a PerPetual life. lt will be desolved if one of the partners ceases to be connected with the PartnershiP. lt can be handed down from one generation of Partners to

289

A. Dividend B. Return C. Share stock D. Par value

81. Oucrtion

A. B.

Test

Mortgage bond Registered bond CouPon bond Collateral trust bond

89.

Question A type of bond which does not have security except a promise to pay by the issuing corPoration.

A. Mortgage bond B. Register bond C. Collateral trust bond D. Debenture bond

290

Question Bank

90. 0uestion A type of bond issued jointly by two or more corporations.

A. B. C. D. 91.

5 years 6 years 7 years

B. lnvestment in subsidiarY companles C. Furnitures D. Patents l0I.

96.'Question

Debenture bond Registered bond Collateral trust bond

fhe 72 rule of thumb

Queslion Lands, buildings, plant and machinery are examples of

is used to

determine

A. B. C. D.

Question

Equipment obligations bond Debenture bond Registered bond lnfrastructure bond

how many hew many dquble how many million how many money

years money wlll triple years money will years to amass

years to quadruple the

97.

Question To triple the principal, one musl,rse

Question lf the security of the bond is a mortgage on certain specified asset of a corporation, this bond is classified as

A. integration B. derivatives C. logarithms D. implicit functions

A. registered bond B. mortgage bond C. coupon bond D. joint bond

98.

Queslion A currency traded in a foreign exchange market for which the demand is consistently high in relation to its supply

93.

Queslion A type of bond where the corporation's owners name are recorded and the interest is paid periodically to the owners with their asking for it.

A. Money mprket B. Hard currency C. Treasury bill D. Certificate of deposit 99. Queslion

A Registered bond B. Preferred bond C. lncorporators bond D. All of these l

9tl.

Question Bond to which are attached coupons indicating the interest due and the date when such interest is to be paid.

A. B. C. D. .95.

4 years

B C D

92.

Joint bond

A type of bond whose guaranty is in lien on railroad equipments.

A. B. C. O.

Multiple ()hoice'l'est

Engineering Econornics by Jaime R. Tiong

Registered bond

Coupon bond Mortgage bond Gollateral trust bond

Everything a company owns and which has a money value is classified as an asset. Which of the following is" classified as an asset?

A. lntangible assets B. Fixed assets C. Trade investments D. All of these 100. Queslion Which an example of an intangible

Question

An amount of money invested at interest per annum will double in approximately

ASSCT? 12o/o

Cash

A. current assets B. trade investments C. fixed assets D. intangible assets 102. Oueslion An increase in the value of a capital asset is called.

A. Profit B. caPital gain C. caPital exPenditure D. capital stock 103. 0ueslion The reduction in the money value of a capital asset is called

A. capital exPenditure B. caPital loss C. loss D. deficit 104. 0uestion It is a negotiable claim issued bY a bank in lieu of a term dePosit.

A. Time dePosit B. Bond C. CaPital gain D. Certificate of dePosit

106. Oucrtion It denotes the fall in the exchange rate of one currency in terms of others The term usuallY aPPlies to floating exchange rates.

A. CurrencYaPPreciation B. CurrencYdevaluation C. CurrencY float D. CurrencYdePreciation 107. 0ueslion

The deliberate lowering of the price of a nation's currency in terms of the accepted standard (Gold, American dollar or the British Pound).

A. CurrencY aPPreciation B. CurrencY float C. CurrencY devaluation D. CurrencYdePreciation 108. Queslion The residual value of a comPany's assets after all outside liabilities (shareholders excluded) have been allowed for.

A. Dividend B. Equity C. Return D. Par value 109. Question A saving which takes Place because goods are not available for consumption rather than the consumer really want to save.

A. B. C. D.

105. Queslion

Any particular raw material or primary product (e.9. cloth, wool, flour, coffee...) is called

A. utility B. necessitY C. commoditY D. stock

291

ComPulsory saving Consumer saving Forced saving All of these

ll0.

Question A document that shows proof of legal ownershiP of a financial securitY.

A. B. C. D.

Bond Bank note CouPon Check

292

Question Bank

Question Defined as the capacity of commodity to satisfy human want.

A. B. C. D.

13.

increase, increasing the other factors of production will result in a less than proportionate increase in output',. A. B

c. D.

Debenture Goodwill Capital gain lnternal rate of return

Demand Suppty Utility Market

B. D.

I 15. Question " When free competition exists, the price of a product will be that value where supply is equal to the demand,,.

A. B. C. D.

Lawofdemand Law of supply and demand

Sunk cost

Opportunity cost Replacement cost lnitial cost

supply demand supply and demand

Cost of goods sold Variance Overhead Payback

Ouostion The simplest economic order quantity (EOQ) model is based on which of the following assumptions? A. B. C.

Shortages are not allowed. Demand is constant with respect to time. Reordering is instantaneous. The time between order placement and receipt is zero. All of the choices

ll9.

Question ln economics, a "short-term,, transaction usually has a lifetime of

122.

Quesrion

ln replacement studies, the existing process or Piece of equiPment is known as

challenger defender

liability

asset

l23.0uestion ln replacement studies, the new process or piece of equipment being considered for purchase is known as

A. challenger B. defender C. asset D. liabilitY 124. 0uestion mgans that the cost of the isset is divided into equal or unequal parts, and olrly one of these parts is taken as an bxPense each Year. A. B.

A. B.

C. D.

3 months or less 1 year or less 5 years or less 10 years or less

l20.Ouestion Law of diminishing return Law ofsupply

18.

Question

Demand Supply Market Utitity

Oucclion An imaginary cost representing what will not be received if a particular strategy is rejected

A. A.

The quantity of a certain commodity that is bought at a certain price at j given time and place.

A. B. C. D.

I2l.

diminishing return

Qucstion An accounting term that represents an inventory account adj ustment.

Question

A. B. C. D.

Law of Law of Law of Law of

The quantity of a certain commoditv that is offered for sale at a certEin price at a given time and place.

llf.

I 16. Question "When one of the factors of production is fixed in quantity or is difficult to

Discount Necessity Luxuries Utility

I 12. Question It is the proflt obtained by selling stocks at a higher price than its original purchase price.

A. B. C. D.

Multiple Choice Teet 293

Engineering Economics by Jaime R. Tiong

ln the cash flow, expenses incurred before time = 0 is called A.

receipts

disbursements

sunk costs first costs

c. D.

Capitalizing the asset Expensing the asset Depreciating the asset Artificial expense

125. Question lndicate the CORRECT statemenl about dePreciation. A. B.

The depreciation is not the same each year in straight line method. The declining balance method can be used even ifthe salvage value is zero.

The sum-of-years' digit rnethod (SYD), the digits 1 to (n + 1) is summed.

Double declining balance dePreciation is indePendent of the salvage value.

126. Quegtion An artificial deductible operating expense designed to compensate mining organizations for decreasing mineral reserves.

A. B. C. D.

Deflation Reflation DePletion lnflation

127. Queslion The change in cost per unit variable change is known as A.

sunk cost

incrementa! cost fixed cost semi-variable cost

c. D.

128. Question What type Qf cost increases step-wise?

A. SuPervision cost B. Direct labor cost C. Semi-vaiiable cost D. Operating and maintenance

cost

129. Oueslion Which of the following is NOT a variable cost?

A. Cost of miscellaneous B. lncome taxes C. Payroll benefit costs D. lnsurance costs

suPPlies

130. Oucstion Which of the following is NOT a fixed cost?

A. Rent B. Janitorial service exPenses C. Supervision costs D. DePreciationexPenses

294

Question Bank

l3l.

Question

The annual costs that are incurred due to the functioning of a piece of equipment is known as A

General, selling and administrative expenses Prime cost Operating and maintenance

Total cost

Multiplo Choicc'l'est 295

Engineering Economics by Jaime R. 't'iong

COSTS

I36.

Ouestion Which of the following is NOT a direct labor expense?

A. B. C. D.

c. D.

lnspection Testing Supervision

The sum ofthe direct labor cost and the direct material cost is known as A. B.

prime cost total cost indirect manufacturing expenses

total cost

\ I33. Oueslion

Resparch and development costs and adm\nistrative expenses are added to the fEctory cost to give the of the product

A Asset accounts B. Bank accounts C. Liability accounts D. Owner's equitY accounts

All are administrative expenses

A. total cost B. marketing cost C. manufacturing cost D. prime cost l3t[.

0uesrion The sum of the prime cost and the indirect manufacturing cost is known as A. B.

c. D.

I35.

factory cost research and development cost manufqcturing cost total cobt

Question

The manufacturing cost plus selling expenses or marketing expenses equals A.

total cost

indirect production cost

administrative cost miscellaneous cost

Ouestion The journal and the ledger together are of the known simply as company.

A. accounting system B. the books C bookkeePing system D. balance sheet I44.

A. Testing B. Drafting C. Prototype D. Laboratory Which is not a factory overhead expense?

A. Pension, medical, vacation benefits B. Expediting C. Quality control and inspection D. Testing l4L Queslion Bookkeeping consists of two steps, namely recording the transactions and categorization of transaction6 Where are the transactions (receipts and disbursements) recorded?

149. Oueslion The ratio of the net income to the owner's equity is known as

A. price-earning ratio B. profit margin ratio C. return of investment D. gross margin

Question

A. Assets = Liability + Owner's equitY B. Liability = Assets + Owner's equitY C. Owner's equity = Assets + LiabilitY D. Owner's equitY = LiabilitY -

Research and development expenses includes all EXCEPT one. Which one?

140. Queslion

A gross profit to net sales B. net income before taxes to net sales C. quick assets to current liabilities D. net income to owner's equitY

The basic accounting equation is

139. Question

I50.

tl5.

0uestion

Payback period is the ratio of

A. initial investment to net annual profit B cost ofgoods sold to average cost of inventory on hand C. gross profit to net sales D. net income before taxes to net

Assets

Oueslion

An acid test ratio is a ratio of

One of the following is NOT a selling or marketing expense. Which one? Advertising Commission lnsurance Transportation

I48.

lf3.

138. Queslion

A. B. C D.

A Current ratio B. Acid test ratio C Gross margin D. Return of investment

EXCEPT:

137. Question

A. Marketing B. Accounting C. Data processing D. Office supplies

short{erm paying abilitY?

Columnar Statement of account

The following are ledger accounts

Assembly

Qucction

What is considered as an index of

142. Queslion

EXCEPT:

132. Queslion

I{7.

Joqrnal Lpdger

sales

Queslion

The ability to convert assets to cash quickly is known as

A. B. C. D I46.

solvency liquidity leverage insolvencY

Queslion

The ability to meet debts as theY become due is known as

A B. C. D

solvencY leverage insolvencY liquidity

Ouestion

A secondary book of accounts, the information of which is obtained from the journal.

A. B. C. D.

Balance sheet Ledger

Worksheet Trial balance

152. Queslion The present worth of cost associated with an asset for an infinite period of time is referred to as

296

Question Bank

A. B. C. D.

annual cost capitalized cost increment cost operating cost

153. Oueslion A stock of a product which is held by a trade body or government as a means of regulating the price of that product.

A. B. C. D.

158. Quesrion Refers to the order quantity that minimizes the inventory cost per unit time.

A. B. C. D.

Economic order quantity Social order quantity Public order quantity Private order quantity

159. Ouestion

Stock pile Hoard stock Buffer stock

Withheld stock

What is referred to as an individual who organizes factors of production to undertake a venture with a view to proflt?

154. Ouestion A negotiable claim issued by a bank in lieu of a term deposit is called A.

Cheque

T-bills Currency Certificate of deposit

c. D.

155. Question firm which is owned ' of individuals for

B. C. D.

Corporation Enterprise Partnership

156. Queslion A document which shows the legal ownership of financial security and entitled to payments thereon.

A. B. C. D.

Coupon

Contract Bond Consol

A. Agent B. Enterpreneur C. Salesman D. Commissioner l60.Ousstion The money that is inactive and doeS not contribute to productive effort in an economy is known as

A. B. C. D.

idle money hard money soft currency frozen asset

16l.

Qucstion ln counting the number of days when computing simple interest,

A. the first day is included B. the last day is excluded C. the first day is included and the last day is excluded D. the first day is excluded and the last day is included

162. Queslion ln the so-called "Banker's Rule",

157. Queslion A government bond which have an indefinite life rather than a specific maturity.

A. B. C. D

Coupon

T-bi[ Debenture Consol

Mtrlt,iplo (lhoice'l'est 297

Engineering Econornics by Jaime R. Tiong

A. the number of days in 1 year is 360 days B. the number of days in 1 year is 365 days C. the number of days in each month is 30 days D. the number of days in 'l year is 366 days

l67.0uortlon

163. 0ucstlon To discount an amount F for n conversion periods means

A. B.

to find the present which is n periods to find the present which is n periods

to find the present value on a day which is (n-1) periods before F is

value on a day after F is due value on a day before F is

due

due to find the present value on a day which is (n+'l) periods before F is due

164. Question ln the formula for compound interest, F = P(1 + i)n, the value (1 + i)n is called

A. discount factor B. interest factor -C. accumulation factor D. increase factor 165. Oueslion To find the present worth of a future amount in compound interest, we use the formula P = F(1 + i)-n . What do you calt the

factor

(1 +

i)'n ?

A. discount factor B. accumulation factor C interest factor D. reductioh factor 166. Queslion What refers to an equation stating that the sum of the values, on a certain comparison date, of one set of obligations is equal to the sum of the value of another set of this date?

A. Equality of value B. Equation of value C. Equality equation D. Similarity equation

What is an annuity whose payments extend over a period of time whose length cannot be foretold accurately?

A. Annuity ce(ain B. Annuity uncertain C. lncremental annuity D. Contingent annuity I68.Ouestion What do you call the time between successive payment dates of an annuity?

A. B. C. D.

Period interval

Annuity period Payment interval

Annuity term

169. Question The time from the beginning of the first payment interval to the end of the last of the annuity. one is called the

A. period B. term C. nature D. type 170. Question What iefers to the extinction of the debt by any satisfactory set of payments?

A. Liquidation B. Liability discharge C. Discharging of debt D. Amortization of debt I 71. Question What do you call a fund, usually by periodic deposits, to insure the accumulation of money to provide for a possible large payment?

A. B. C. D.

Escrow fund Sinking fund Mutual fund Corporate fund

298 I

72.

Question Bank

Queslion

What is the term for the borrowed principal usually mentioned in a typical

bond?

A. B. C. D. I

73.

Mrrlt,ipkr ()hoicc

Engineering Economics by Jaime R. Tiong

Bond rate Face value or par value Coupon rate Coupon value

Queslion

Any date on which a coupon of a bond becomes due will be referred to as a

When the price of the bond is less than the redemption value When the price of the bond is equal to the redemption value. When the price of the bond is either equal to or greater than the redemption value.

77.

Question Which of the following will happen if a bond is bough at a discoqnt?

C and-interest price D. coupon price I8l. 0ueslion

A U.

A. B. C. D.

maturity date term of the bond

0ueslion lf P is the price of a bond and V is its redemption value, what do you call the value P - V?

75.

Par value Face value Premium Bond discount

The unpaid interest on each coupon date will not be considered as a new investment in the bond. The difference between the coupon payment and the interest due is a pa(ial repayment of principal.

78. 0uestion ln the sale of a bond, the actual purchase price on any day is called

Queslion

When can we say that the bond is A.

purchased at a discount?

A. B.

When the price of the bond is greater than the redemption value. When the price of the bond is less than the redemption value When the price of the bond is equal to the redemption value. When the price of the bond is either equal to or greater than the redemption value.

76. 0ueslion When can we say that the bond is purchased at a premium?

the person dies B

When the price of the bond is greater than the redemption value.

C. D.

Face value Quoted price Accrued price Flat price

per period?

A B C. D

79.

Queslion What do you call the difference between the flat price of the bond and the quoted price ofthe bond?

Accrued interest

Bond rate

And-interest price

180. Queslion The quoted price of a bond is sometimes called

A B.

What is the term for the sum of depreciation charges to date?

Question

The difference between the value of an asset and its salvage or scrap value at the end of n year is called

Par value Face value

A. B C. D.

depreciation accrued value book value wearing value

185. Ouestion What is a life annuity?

quick assets current liabilities net credit sales average net receivables gross profit current liabilities gross profit net sales

I88.

0ueslion Which of the following represents the gross margin? A

Par value

A. B

The acid test ratio is also known as quick ratio. Which one represents the quick ratio?

183. Queslion

I84.

Yield Nominal rate Fixed rate Net rate

187. Question

A. Annuity bond B. Serial bond C. Treasury bond D. Government bond

A. Accrueddepreciation B. Applied depreciation C Accumulateddepreciation D. All of the above

indefinitely. It is the same as perpetuity

What term is usually used by the banks to represent the effective interest rate

182. Question What is a bond whose face value is redeemable in installments, with interest payable periodically as due on outstanding principal?

A sequence of payment intended for the life insurance of a person. A sequence of payment for a certain person which continues

186. Sueslion

coupon date

due date

l7tl.

A. B. C. D.

average investment average annual interest average annual interest average investment par value flat value flat value quoted value

299

A sequence of payment for a certain person which stops when

The yield of a bond is obtained by which of the following formulas:

'l'est

B C D

net income

owners'equity net credit sales average net receivables gross profit current liabilities gross profit net sales

300

Question Bank

189. Question A receivable turnover is calculated using which of the following formulas?

A. B.

C D.

^ U.

owners'equity net credit sales average net receivables gross profit current liabilities gross profit net sales

The percentage of each peso of sales that is net income is called

l9l.

Equilibrium market price Fair market price Real market price Exact market price

net income

190. Ouestion

A. B. C. D.

price-earning ratio profit margin profit margin ratio return of investment ratio

194. Queslion A principle that states that consumers will tend to spend an increasing proportion of any additional income upon luxury goods and a smaller proportional on staple goods; so that a rise in income will lower the overall share of consumer expenditures spent on stable goods (such as basic foodstuffs) and increase the share of consumer expenditures on luxury goods (such as motor cars)

A. B. C. D.

Placibo effect Luxury effect Engel's law Staple law

Question

Which one represents that priceearnings ratio? A

market price per share earnings per share

earnings per share market price per share

net credit sales average net receivables gross profit current liabilities

192. Ouestion The book value per share of common stock is the ratio of the common shareholdersl equity to

A average shares B. number of outstanding shares C. total subscribed shares O. authorized capital stock I93.

Mult,iplo Ohoico

Engineering Economics by Jaime R. 'tiong

Question

What refers to the price at which the quantity demanded of a good is exacfly equal to the quantity supplied?

198. Question A market where new entrants face costs similar to those of established firms and where, on leaving, firms are able to recoup their capital costs, less depreciation. A. B.

c D

Th{ory of values

Ec5nometrics Economatics Econoscience

c. D.

l96.Oueslion What refers to the fall in the general price level, frequently accompanied by

a reduction in the level of national income?

A. B. C. D.

lnflationary gap Dissavings Disinflation lnflation

I97.Ouestion A price for a product which just covers its production and distribution costs

with no profit margin added.

A. B C

Cost price Actual price Real price

Free market Competitive market Limited market Contestable market

199. Question What refers to a temporary grouping of independent firms, organization and governments, brought together to pool their resources and skills in order to undertake a particular project?

A. B. C. D. 200.

Original price

Consortium Cartel

Cooperative Union

Question

What refers to a market for buying and selling of raw materials such as tea, coffee, iron ore, etc.?

A. B. C. D.

Commodity market Raw market Natural market National market

Tost

301

302

Question Bank

Engineering Economics by Jaime R. Tiong

Chapter

ANSWERS

1. 2. 3. 4. 5. 6. 7. 8. 9.

B B A A A D D C C '10. D 11. A 12. C 13. A 14. D 't5. c 16. B 't7. A 18. C 19. C 20. D 21. C 22. D 23. B 24. B 25. A 26. C 27. B 28. A 29. A 30. D 31. C 32. C 33. B 34. B 35. C 36. B 37. D 38. C 39. A 40. c

41.C 42.C 43.C 44.C 45.B

46.4

47.C 48.C 49.B 50. D 51. D 52.

C B B D C

53. 54. 55. 56. 57. A 58. A 59. B 60. c 61. D 62. D 63. D 64. C 65. D 66. A 67. A 68. B 69. A 70. o 71. B 72. C 73. B 74. A 75. C 76. D 77. C 78. C 79. B 80. c

81. B 82. D 83. B 84. D 85. A 86. A 87. A 88. D 89. D 90. A 91. A 92. B 93. A 94. B 9s. c 96. B 97. C 98. B 99. D

121.

122.

123. A

124. 125. 126. 127. 128. 129. 130. 131.

163. B 164. C 165. A 166. B

168. 169. 170. 171. 172. 173. 174.

D C

143.

B B

145.

144 A 't46. A 147. A 148. C 149. C 150. A B

152. B 153. C 154. D 155. A 156. A 157. D 158. A 159. B 160. A

162. A

167

141. A

15't.

161.

135. A 136. C 137. A 138. C 139. D

142.

134. A

102. B 103. B 104. D 105. C 106. D 107. C 108. B 109. C 110. C 111. B 112. C 113. B 114. A 115. D 116. A 117. A 118. D 120.

133. C

140.

132. A

't00. D 't01. c

't19.

B B

't75.

C B D B B C C B

176. A

177. A 178. D 179. B 180. C 181. B 182. B 183. A 184. D 185. A 186. A 187. A

188. D 189. B 190. C 191. A 192. B 't93. A 194. C 195. B 196. C 197. A

't98.

199. A 200. A

What is Economics? Adam Smith (1723 - 1790), a Scottish professor of philosophy at Glasgow University, whose writings "An lnquiry into the Nature and Causes of the Wealth of Nations in 1776, formed the basis of classical economics defined economics as the study ofthe nature and causes of national wealth or simply as the study of wealth.

Alfred Marshall (1842 - 1924), an English professor of politieal economy at Cambridge University who depicted precise mathematical relationships between economic variables in his textbook "Principles of Economics" in '1890, defined economics as the study of man in the ordinary business of life. Arthur Cecil Pigou (1877 - 1959), an English student of Professor Albert Marshall, who succeeded him as professor of political economy at Cambridge University and devefoped the theory of welfare economlcs in his book 'The Economics of Welfare" in 1919, defined economics as the study of economic welfare which can be brought, directly or indirectly, into relationship with the measuring-rod of money. Lionel Robbins, an economist in the tradition of the English classical school defined economics as the science which studies human behavior as a relation between ends and scarce means have other alternative uses. Collins dictionary of Economics defined economics as the study of the problems of using available factors of production as efficiently as possible so as to attain the maximum fulfillment of society's unlimited demands for goods and services. While the standpoints of well-known economists and professors of economics

to the definition of the subject "economics" may vary, the main idea remained that in every human activity, in one way or another, ie connected with either money or time or both.

'ln its simplest definition, economics is the wise use of both money and time.

What is Monev? There are a lot of definitions of money. However the most aceeptable was that of Geoffrey Crowther who describes the major functions of money.

According lo him, money is anything that is generally acceptable as a means of exchange (i.e. as a means of settling debts) and at the same time acts as a measure and a store of value.

The money which is now used as a medium of exchange on local, national and international markets has evolved from shells, elephant tusks, animal skins to metallic money (i.e. coins) to paper money after the invention of printing press in 18tn century to finally bank money (i e. currency) with the development of the banking system.

With the currency system, a standard unit called "monetary unit" forms the basis of a country's domestic money supply ln the Philippines, the monetary unit is peso while dollar and pounds for the United States and United Kingdom, respectively. The physical form of the money supply (bank ncites, coins, etc.), the denomination of the values of monetary units (peso and centavos) and the total size of the money supply of the country are regulated through the policies and guidelines of what is known aslhe "monetary system". Money can fulfill the following functions:

A. As a medium ofexchange. Goods and services can be exchanged for money, which can be exchanged for other goods and servtces.

304

Question Bank

Engineering Economics by Jaime R. Tiong

As a unit of account. The units in which money is measured (pesos, dollars, pounds, etc.) are used as the units in which prices, financial assets, debts, accounts, etc. are measured.

'l'hr'orir.s rttttl l'\rt'ntulas 305

Cowumer goods and services are those

As a store of value.

satisfy much as

form in order to yield future consumption. Since money grows with time, the money held over a period of time as a store of value or purchasing power.

What is Enqineerino Economv?

goods and services. Exarnpres goods and services are factory builOings, machine tools, airplanes, ships, busesl etc.

"iJffJ"",

What is the difference between Necessitv and LuxuEfGoods and services are divided into two types, necessities and luxuries. Necessities refer to the goods and services that are required to support human life,

needs and activities.

Engineering economy may also be defined

Necessity product or staple product is defined as any product that has an

income-elasticity of demand less than one This means that as income rises,

proportionately What ar6 9o.nsumer and producer

Goods and Services?

Good or commodity is defined as any tangible economic product (soap, cir, shirts, tools, machines, etc.) that contributes direcfly or indirecfly to the satisfaction of human wants. Seryice is defined as any tangible economic activity (hairdressing, insurance, banking, catering, etc.) that contribute directly or indirecfly to the satisfaction of human wants.

Econo either "produ

nd seruices as

services,'or s',.

such products. products includ bread'and rice, clothing, etc.

Luxuries are those goods and services that are de-sired by human and will be acquired only after all the necessities have been satisfied. I.uxury producr is defined as any product

that has an income-elasticity of demand greater than one. This means that as

income rises, proportionately more income is spent on such products. Examples of luxury products includes consumer durables like electric appliances, expensive cars, holidays and entertainment, etc. Necessities and luxuries are relative terms because there are some goods and seryices which may be cJnsidered by one

person as necessity but luxury to another person. For example, a man living in the Metropolis finds a car as an absolute necessity for him to be able to go to his workplace and back to his home. lf the same person lived and worked in another city, less populated with adequate means of public transportation available, then a car will become a luxury for him.

@ Situalhons? The term "market" refers to the exchange mechanism that brings together the sellers and the buyers of a product, factor of production or financial security. lt may"also refer to the place or area in which buyers and sellers exchange a well-defined commodity.

l'rt'f tt'l t tnqrclition (also known as ttlttttti:;lit' t'omltetitionl refers to the market situation in which any given product is supplied by a very large number of vendors and there is no restriction against additional vendors from entering the market Perfect competition is a type of market situation characterized by the following:

Many sellers and manY buYers: Since there is a large number of sellers and a large number of buyers, each seller and buYer will become sufficientlY small to be unable to influence the Price of the Product transacted.

Homogeneous Products: The products offered bY the comPeting sellers are identical not only in physical attributes but are also regarded as identical by the buyers who have no Preference between the products of various producers.

Free market-entry and exit: There are no barriers to entry or impediments to the exit of the existing sellers.

Perfect information: All buyers and all sellers have complete information on the prices being asked and offered in all other Parts of the market

Biyer or consumer is defined as the basic consuming or demanding unit of a commodity. lt maY be an individual purchaser of a good or seryice, a household (a group of individuals who make joint purchasing decisions)' or a government.

Selleris defined as an entity which makes product, good or service available to buyer or consumer in exchange of monetary consideration.

The price of any commodity or product will depend largely on the market situation. The following is a tabulation of the differenl market situations:

E. Absence

of all economic friction:

There is a total absence of economic friction including transport cost from one Part of the market to another. This market situation Provides an assurance of complete freedom on the part of both the vendors and the buyers though the latter benefits more from the reduced prices brought about by the competition while more and better services are afforded by the vendors or players in the industry.

306

Question Bank

Engineering Economics by Jaime R. ,l,iorrg

Monopoly is the opposite of perfect competition. This market situation is characterized by the following:

One seller and many buyers: A market comprised of a single supplier selling to a multitude of small, independenfl y-acting buyers. Lack of substitute products: There are no close substitutes for the monopolist's product. Blockaded entry: Barriers to entry are so severe that it is impossible for other sellers to enter the market.

There exists a perfect monopoly if lhe single vendor can prevent the entry of all other vendors into the market. The monopolist is in the position to set the market price.

A natural monopoly is a market situation where economies of scale are so significant that costs are only minimized when the entire output of an industry is supplied by a single producer so that supply costs are lower under monopoly than under perfect competition and

differences are created through advertising and sales promotion.

C. Difficult market entry:

High barriers of entry which make it difficult for new sellers to enter the market.

What is demand? Demandis the need, want or desire for a oney to purchase demand is always d abitity to pay,, want or need for

the product.

'l'hr.orir,n irntl l,'ot'utttltts 307 lf the selling price for a product is high, more producers will be willing to work harder and risk more capital in order to reap more profit. However if the selling price for a product declines, capitalists will not produce as much because of the smaller profit they can obtain for their labor and risk.

Therefore, the relationship between price and supply is that they are directly proportional, i.e. the bigger the selling price, the more the supply; and the smaller the selling press, the less is the supply. Figure 2 below illustrates the price-supply

supply and demand at a given Price

Price

relationship. The demand for a product is inversely proportional to its selling price, i.e. as the selling price is increased, there will be less demand for the product; and as the selling price is decreased, the demand will increases. Figure 1 below illustrates the relationship between price and demand.

Figure 3 When there is additional supply without an additional demand, a new and lower price is established as shown in Figure 4.

Previous supply

oligopoly.

Oligopoly exists when there are so few suppliers of a product or service that the action of one will inevitably result in a similar action by the other suppliers. This type of market situation is characterized by

the following:

A Few seller and many buyers: The bulk of market supply is in the hands of a relatively few sellers who sell to many small buyers. B

Homogeneous or differentiated

products: The products offered by the suppliers may be identical or more commbnly, differentiated from each other in one or more respects. These differences may be of a physical nature, involving functional features, or may be purely "imaginary" in the sense that artificial

What is the Law of Supplv and Demand? Figure 4

Figure

The law of supply and demand maY be stated as follows: Under conditions of perfect competition, price at which any given product will be supplied and purchased is the price that will result in the supply and the demand being "

Assuming a linear function for the relationship between price and demand, it shows that at point 1, the selling price, pr is high, thus there is less demand for the product as compared to point 2, where the demand, D2 is great because of lower selling price, P2.

What is Suoolv? Supply is the amount of a product made available for sale.

When there is additional demand without an additional supply, a new and higher price is established as shown in Figure 5

the

equal." The relationship between price-supplydemand may be illustrated by combining figures 1 and 2. For general application, the curves for supply and demand are no longer represented in linear function. Figure 3 illustrates the price-supplydemand relationship, showing equal

308

Question

lJank

l!rrginot'rirrg l,it:orrornir:s by .laime

I.1..

'l'rorrg

'l'hcorics and

Chapter 2 "

was invested Suppose a debtor loans money from a creditor. The debtor must pay the creditor the original amount loaned plus an additional sum called interest. For the debtor, interest is the payment for the use of the borrowed capital while for the creditor, it is the income from invested capital. Simple Interest (denoted as l) is defined as the interest on a loan or principal that is based only on the original amount of the loan or principal. This means that the interest charges grow in linear function over a period of time. This is usually used for short{erm loans where the period of the loan is measured in days rather than years. lt can be calculated using the following formula:

where: P = principal / loan | = interest i = interest rate n = period

The future amount of the principal may be calculated by adding the interest (l) to the principal (P).

F=P+l

F=P+Pin Thus,

Exact simple interest is based on the exact number of days in a given year. A normal year has 365 days while a leap year (which occurs once every 4 years) has 366 days. Unlike the ordinary simple interest where each month has 30 days, in the type of simple interest, the number of days in a month is based on the actual number of days each month contains in our Gregorian calendar To determine the year whether leap year or not, one has to divide the year by 4. lf it is exactly divisible by 4, the year is said to be leap year otherwise it will be considered just a normal year with 365 days. However, if the year is a century year (ending with two zeros, e g. '1700, I 800... ), the year must be divided by 400 instead of 4 to determine the year whether or not a leap year Hence, year 1600 and year 2000 are leap years Under this method of computation of interest, it must be noted that under normal year, the month of February has 28 days while during leap years it has 29 days. Again, the values of n to be used in the preceding formulas are as follows:

n=- d

There are two types of simple interest namely, ordinary simple interest and exact simple interest.

ln a bank loan, a person borrows P 100 for 1 year with an interest of 10%. The interest as computed is P 10. The bank deducts the interest, which is P10 from P 100 and gives the borrower only P 90. The borrower then agreed to repay P100 at the end of the year. The Pl0 that was deducted represent the interest paid in advance. ln this case, discount also represent the difference between the present worth (i.e. P90) and the future worth (i.e. P 100). Discqrnt

= Fduevorth - peoentworttt

fhe

rate of discount is the discount on one unit of principal per unit time.

d=1-:for leap years

366

DISCOUNT

Ordinary simple interesl is based on one banker's year. A banker year is composed of '12 months of 30 days each which is equivalent to a total of 360 days in a year. The value of n that is used in the preceding formulas may be calculated as

CASE 2:

for normal years

F = P(1+ in)

n=- d

A person has a bond or financial security that is not due yet but payable in some future date, desires to exchange it into an immediate cash. ln the process, he will accept an amount in cash smaller that the face value of the bond. The difference between the amount he receives in cash (present worth) and the face value of the bond or financial security (future worth) is known as discount. The process of converting a glaim on a future amount of money in the present is called discounting.

Let d = rate of discount

365

Consider the following cases where discount is involved:

309

Chapter

360

where: d = number of days the principal

I,brlnulas

Compound interest is defined as the interest of loan or principal which is based not only on the oiiginal amount of the loan or principal but the amdunt of loan or principal plus the previous accumulated interest. This means that aside from the principal, the interest now earns interest as well. Thus, the interest charges grow exponentially over a period of time.

Compound interest is frequently used in commercial practice than simple interest, more especially if it is a longer period which spans for more than a year. The future amount of the principal may be derried by the following tabulatlon: Period

Principal

lnteresl

Total Amount

P+Pi Pll + i'l P('!+i)(1+i) =

P(l

P(1 + i)i

P(1 +

i)z

P(1 +i)zi

= P (1 +i)z P(1 +i)z('l +i) = P(1 + i):

P(1 + i)"

The tabulation above shows that the future amount (total amount) is just the value p(1 + i) with an exponent which is numerically equal to the period. It is also observed that compound interest is based on the principles of geometric progression and using such method, the total amount after each period are as follows:

Firstterm, ar = P(1 + i) Second term, a2 = P(1 + i)2 Third term, a3 = P('l + ;;3 and so on. Solving for the common ratio, r:

310

Question Bank

Engineering Economics by Jaime R. Tiong

a^ f =-L

where:

a,l

I (1

worth factor

Using the formula for nth term of a G.P. ?n

:3tr"^t

The concept of continuous compounding is based on the assumption that cash payments occur once per year but

an =P(1 +i)n

There are two types of rates of interest, namely the nominal rate of interest and the effective rate of interest.

compound iryterest is

FUTURE AMOUNT, F:

F = P(1+ i)n

Nominal rate of interest is defined as the basic annual rate of interestwhile effective rate ofinteresf is defined as the actual or the exact rate of interest earned on the principal during a one-year period.

but for m periods per year

r =p(t*!E)'*

Cash Flow

F = P(1+

let

m,J

x./

ln this statement, the nominal rate is 5% while the effective is greater than 5% because of the compounding which occurs four times a year. The following formula is used to determine the effective rate of interest:

rx(NR)N

.=,[1,.i)'l'.' r-i, [r * 1)' x

-->o\

" where: m = number of interest period per

therefore,

year

PRESENTWORTH,P

i = interest per period

.NR

F =PENR)N

(1+i)n

where the payments are made at the end of each period beginning from the first period. Derivation of formula for the sum of

ordinary annuity: Let A be the periodic or uniform payment and assuming only four payments:

A ; :

where: p = priniipal e = 271828... NR = nominal rate N = number of

years e(NR)N - continuous ,/ compounding / compound amouht

factor

J8r D)

A(1+

iro -+

A(1+

factor

b. 0123n

1. Ordinary annuity is a type of annuity

but

+ i)n = single PaYment compound amount

Annuity ls defined as a series of equal payments occurring at equal interval of time. When an annuity has a fixed time span, it is known as annuity certain.The following are annuity certain:

For example: A principal is invested at 5% compounded quarterly.

F=Pl1+fl

i)n

where: P = Principal i = interest per period (in decimal) n = number of interest periods (1

eNR(N)

Rate of interest is the cost of borrowing money. lt also refers to the amount earned by a unit principal per unit time.

The basic equation for future worth of

What are the Tvpes of lnterest Rates?

compounding is continuous throughout the year.

an =P(1 +iX1 +i)n-l

ts:' '

What is the concept of continuous comDoundinq?

r=1+i

ChapEer

fhe present worth of continuous compounding is

present ^ =single payment

+ i)n

P(1 + i)2

P(l+

Theorios and Formulas 31 1

m NR = nominal rate of interest

Annuity is based on the principles of compound interest. Hence computation of the sum of annuity may be done using the formulas for geometric progression. Solving for common ratio:

aa1

.,/

I =-

A(1+i) A

r='l +i Note: i = NR if the mode of compounding is annually

Solving for the sum:

312

Question Bank

'l'ltcorics irrttl l,'ormrrlas 313

Engineering Economics by Jaime R. Tiong PRESENT WORTH OF PERPETUITY:

n[1r*i1"-rl r But F= L

l?34

I I I I--TI

12341

n-1

AAAA

Substituting the value of F:

A a. SUM OF ORDINARY ANNUITYT

where:

ltr*if-rl L i(1

, AAAA2 i.........................- F ELf

nltr+if

i)n

=uniformseries present worth

factor

Annuity due is a type of annuity where the payments are made at the beginning of each period starting from the first period.

P=+ (n-2)G

where: i = interest per period A = uniform payment

Chapter

-r"l

where: i = interest per period n = number of periods A = uniform payment

ltr*if-r'] uniform series compound * = i

amount factor

P -'...'.....'.-.

.'......-.......i

:.........,..........

.,.,.....,>, , F

The total present worth P is equal to sum of Pr and P2, mathematically, P = P1+P2

Deferred annuity is a type of annuity where the first payment does not b'egin until some later date in the cash flow

Unifurm arithmetic gradient is a series of disbursements or receipts that increases or decreases in each succeeding period by constant amount.

Also, the total future worth is F

:FJ +Fz

Analyzing the uniform arithmetic gradient cash flow only:

PRESENT WORTH OF ORDINARY

ANNUITY:

PRESENTWORTH:

A+(n-2)G

Let A

AAAA

When an annuity does not have a fixed time span but continues indefinitely, then it is referred to as a perpetuity. The sum of a perpetuity is an infinite value.

Using compound interest formula: F

D_ (1

i)n

The above cash flow can be converted into its equivalent annuity and uniform arithmetic gradient whose present worth are Pr and P2, respectively

P = present worth of the following uniform arithmetic gradient

Please refer to the cash flow in the following page

314

Question Bank

'l'lroorics

Engineering Economics by Jaime R. Tiong

n-l

-l,.,*,,

(n-2)G

where: G = arithmetic gradient change in the periodic amounts at the end of each period

i)3

JU +

31 5

(1+i)"]'

(1+i)2

n_ n

trlas

r=eitr*iI-r_nl rL ,

rrrl l'irlrrt

i)n

F cliI 1 (1iri)

P(1. )-P=G

4 ..'

P(1 r

EQUIVALENT PERIODIC AMOUNT

i)a

Let A = equivalent periodic amount (n

+ i)n-1

234

2)G<.

(n-'l)G('1

^G2G3G (1 + i)' (1 + i)' -=-f

Substitute values in Eq. 3

i)n

..............-T-f

' ',Pi G[t-rt*il'n I i

'..

+ i)*

(n-2)G (n-l)G "'-(1 -

+il'r

(i*D"

4)--

(nJ)G

(1+i)"

]

.\-n 1-(1+i)" n 1 '-iL i (1+i)n]

o_Gl

(1+i)' I

So v ng for the common rat

The above cash flow is equivalent to the figure below:

t---

a. (1+il2 \ , a1 I -(1+i) 1

(1+i)

-,r-rTl-lr

01234...n-ln where: G = arithmetic gradient change in the periodic amounts at the end of each period

AAAAAA

B. FUTUREWORTH:

Multiply Eq. 1 by (1 + i):

"""""''"""i

-""""""'

(1+i) 1

Factor out G:

n I

.,*..-.-........

................-'i

Using formula for annuity:

Let F = future worth of the uniform

arithmetic gradient

1 2

3 4 ... n-l

(1+i)3 _ 1

+ i)2

Subtract Eq. 1 from Eq. 2 Solving for the sum, S: ./

"'('-'/) ltr

'l n ._clr-(1+i)n -(1.,f ' -TI

316

Question Bank

Engineeling Economics by Jaime R. Tiong

Substituting:

'l'lrr.orrcs

PRESENTWORTH, P

ot=X

B. FUTURE

?z= *

0123

33=x3

G1t*r1"2 : -\'r'' G(1+r)z iclt -.................i (1-i)3

a. a1

i(l

i)n

(1+

i[.]L(1+

ni I n=9[ri (1+i)"-t] L

G(l +

r)n-2

G(l + r)n

I,", i

iA=c[]^ I (1+if_jl

'1.':l

Uniform geometrio gradient is a series consisting of end-of-period payments, where each payment increases or decreases by a fixed percentage.

r)n-2

+ifr -

G(1t)rt

-rn)

,(t-x") / \ -x

when x

G G(1 +r), G(1 +r)2 D_ ' -(1+D- (i+D2 * (1+i)p +"' "'r, G(t+

*r)3

1-r

(1 + i)n

.l:

G(1+r)"2

a,(1

^ i

{..................

Substitute in Eq.

+rf

p=9f tr)[t--l"

1+r\1rilf t-x

, I

]

,=erL{l 1+i( 1-x

P=G

*1'. =1*' 1+i F = P(1+ i)n

o x(1-x" ,- 1+r (1-x)

G(1 + r)n-1 (1

c(t.+

Sum of geometric progression, S:

: 1

l=X

(1 + i)n-1

Solving for the common ratio, r:

(1 + i)2

lrrtl l"orrurrlits 317

=-9-[L{],, (1+i)l 1-x l'

x" r = c1r * iY-'[1- ) 1-, ( J

Whenx=1,r=l

J F = P(1+ i)n

Multiply by

where:

1+r 1+i

*1.thusr +i

G = first payment at the end of the first period (1 + r) = rate of increase or decrease

Whenr=i,thenx=1.

F= Gn (1 +i)n (1

i)'

=Gn('!+0n-l

Chapter

Substituting values in Eq. 1: Gn o' 1+i

Geometric progression

Bondis a financial security note issued by businesses or corporation and by the government as a means of borrowing longterm fund. lt may also be defined as a long-term note issued by the lender by the

318

Question Bank

'l'htrorion and

Engineering Economics by Jaime R. Tiong

Bonds do not represent ownership of a business or corporation and therefore not entitled to share of the profits. The bondholder has no voice in the affair of the

business However the bondholder has

Typor of Depreclatlon:

Let Vn = value of bond n years before redemption

borrower stipulating the terms of repayment and other conditions.

more stable and secured investment than does holder of common or preferred stock of the business or corporation.

Vn=Pl+P2

-+Eq

Using formula for present worth of annuity

B. a

z'[{r*

i)n

i(1

I'hysicol Deprecialion is due to the reduction of the physical ability of an equipment or asset to produce results. Functional Depreciation is due to

the reduction in the demand for the function that the equipment or asset was designed to render. This type of depreciation is often called

Lender

Bonds are issued in certain amounts known as the /ace value or par value of the bond. When the face value has been repaid, normally at maturity, the bond is said to be redeemed or retired. Bond rate is the interest rate quoted in the bond

llorrnulns 319

-r]

obsolescence,

Methods of Gomputing Depreciation: Using the formula for present worth of compound interest:

ln this method of computing depreciation, it is assumed that the loss in value is directly proportional to the age of the equipment or asset.

P^= ' 11+i1n p.= C ' (1+i)n '

The following illustrates the normal life cycle of a bond.

Value of a bond is the present worth of all the amounts the bondholder will receive through his possession ofthe bond. The two payments that the bondholder will

Derivation of the formula for the value of a bond:

Substituting in Eq.

STRAIGHT LINE METHOD

Annual depreciation charge, d

,-Co-Cn u--

z'[(r*i)" -r]

(1+i)n

c *(1+i)n

where: Co = first cost Cn = cost'after "n" years (salvage/scrap value) n = life of the property

where: Z = par value ofthe bond r = rate of interest on the bond per period C = redemption price of bond i = interest rate per period n = number of years before redemption

Chapter

Lender

Book value at the end of "m" years of using, Cr: Cm

Depreciation is the reduction of fall in the value of an asset or physical property during the course of its working life and due to the passage of time.

-Dm

where: D, = total depreciation after "m" years D, = d(m)

B. SINKING

FUND METHOD

ln this method of computing depreciation, it is assumed that a sinking fund is established in which funds will accumulate for replacement purposes.

Annual depreciation charge, d

320

Question Bank

where: Co = first cost Cn = cost after "n" years (salvage/scrap value) n = life of the property Book value at the end of "m" years of using, C.:

;9, ='%tD.,

= Co

ZYears

METHOD

!co

k=1-im

years

n(n+1) =-n-

--',_

0.4507(10,000)

5,493.00

= 4,507.00

0:4507(5,493) = 2,475.69 0.4507(3,017.30) = 1.359.90 0.4507(1,657.40) 746.99 0.4507(910.41) 410.32

2 1,556

076[(1+o

1)3

-1]

Dg=

0.1 Ds = 5,'t 50.61 16

= =

3,0'17.30

1,657.40 910.41 500.09

Note; S/rght difference is a result of rounding off of values.

10,000_500

D. Using SYD method

I,Year

Tabulation of book value: Period

Depreciation

Book value

2 3

zyears

4 5

1,556.076[(1+0.05 Ds=

1,900.00 '1,900 00 1.900.00 1,900.00 1.900.00

Using sinking fund method:

zyears

8,100.00 6.200.00 4,300.00 2.400.00 500.00

1.556 08 3.267 76

5,150.6',l

Book value 10,000 00 8.443.92 6.732.24 4.849.39

7.221.75 9,500.00

500.00

0 1

Assume 10% interest rate

o-(Co-C")i ('l + i)n

-'l

Tabulation of book value: Depreciation

year =

q =(co

Ds = 9'500 00

Period

- 1Dt;+tl (5X1+

= '' u

-1]

0.'1

'10,000 00

d1=(co-c.)---n

and so on. ..

'10,000.00

d = 1,900

zyears

Book value

Respective depreciation charges:

Third year:

Depreciatibn

Period 0

METHOD

tr:2 d3 = (co - c.)-

Tabulation of book value:

6-'9u-C"

!co

k = 1- 0.5493 k = 0.4507

Using straight line method:

*='t-,8

d2=(co-c.)#

o[(r*i)'-r]

Dz=3,267.7596

First cost, Co = P 10,000 Salvage value, Cn = P 500 Lifeof property, n = 5 years

D. SUM-OF-YEARS' DlclT (SYD)

Second year:

oz=

ln order to establish the comparison between the depreciation methods mentioned above, let us consider the following data:

The value k is the constant percentage. Hence k must be decimal and a value less than 1. ln this method, the salvage or scrap value must not be zero.

First year:

\lUo

Dz=

Matheson Formula:

Uring decllning balance method

o, +'-'+ dr)

DIFFERENT METHODS OF DEPREGIATION:

ln this method of computing depreciation, it is assumed that the annual cost of depreciation is a fixed percentage of the book value at the beginning ofthe year. This method is sometimes known as constant percentage method or lhe Matheson Formula.

k=1-"8

)

321

6k=1-d+

COMPARISON BETWEEN THE

C. DECLINING BALANCE

(a,r +

o[tr*if -1] L

Sum of year's digit,

where: D. = total depreciation after "m" years

n_-

Book value at the end of "m" years of using, C,.

l'\lrmulus

'l'hooriott rtnd

Engineering Economics by Jaime R. Tiong

'2.778.29

-"")tk

=(1o,ooo-5oo)*

dr = 3'166'67

d2=(co-.^)# d2

=(1o,ooo-5oo)+

dz = 2'533'33

322

Question llank d3

Dngineeriug Economics by Jaime R. ,I.iong

=(co_c")+

depreciation charge while the sinkingJunr) method has the smallest annual depreciation charges.

!year

'l'huorior and l'lrrntulrrn 323

=(1o,ooo-5oo)*

ds = 1,900.00

What is Deoletion Cost?

d4=(Co-c")a

Depletion cosl is the reduction of th€ value of certain natural resources such as mines, oil, timber, quarries, etc. due to the gradual extraction of its contents

Lyeat

=(1o,oo\-5oo)#

Methods of Computing Depletion Charge for a year:

d4 = 1,266.61

-c"[=ii Lyeat l, 0,000 L soo) *

ds =

(co

ds =

Tabulation of book value: Period

Depreciation

3.166.67 2,533 33

'1,900 00

1.266.67 633.33

6.833.33 4,300 00 2,400.00 1.133.33 500.00

Book Values obtained by Various Depreciation Method 10,000

U.S. = units sold The depletion cost during any year is:

lC' T_U

ts I \v.e., rt

Method The depletion charge (or simply depletion) under this method, is computed as follows: Depletibn = Fixed % of gross income

SYD Method

6,000

or Declining

4,000

Balance Method

Depletion = 90 :%iof 1et ta;able incci4e

2,000

Note: Use the smaller depletion charge

500

Fixed Percentages Allowed for Ce(ain Natural Resources:

The diagram shows that ltrc declining balance method has the largest annual

5Yrs

llrcttk-even analysis is used to determine the break-even cost, which is the cost at which the total income is exactly equal to the total expenses incurred in the business.

1o 5

The diagram below is the most convenient chart used in break-even analysis. This is known as the breakeven chart. lt represents the graphs of fixed costs, variable costs, the expected

income, etc.

ure and the cement, long time

Ke!

CQ=Frntcrd+Cdd structure or the annual rest ofthe first cost

A p d a

Ltr

ing and maintenance

costs, or

Straight Line Method

8,000

or forever, or

B. Percentage or Depletion Allowance

Sinking Fund Method

Chapter I

l.C. = initial cost of property T.U. = total units in property

Depletion __-. cost _ =

* Based on Gross lncome

Unit or Factor Method

Let:

'10.000 00

Sulfur, Cobalt, Lead, Nickel, Zinc. etc Oil and Gas wells Gold, Silver, CoPPer, lron ore Sodium Chloride -Coal. Gravel, Sand, CleY-

following formula: Book value

Percentage'

This method is dependent on the initial cost ofthe prope(y and the number of units in the property. The depletion charge during any year is calculated using the

ds = 633.33

Chapter

Maximttttt

Natural Resource

Production

p = profit L = loss

324

Question Bank

(lloannry 325

Engineering Economics by Jaime R. Tiong amalgamation (syn merger) lhe creation of a new company by a combination of business units

asset a property or item owned by an indlvldual or organization which has a monetary valuo

auction the selling of goods or producls and amortization as applied to capitalized cost, the

distributi charges the redu inegular

I cost by

periodic

in depreciation or y either periodic or gram

annual general meeting (AGM) the required by law yearly meeting of shareholders of corporation or organization

accounls the financial slatements

annual ratb of return is the ratio of the annual an

individual or organization prepared from a syslem of recorded financial transactions

acid-test hatio (syn. current ratio) the accounting measure of individuals, or company's ability to pay its short{erm liabilities

actuary the chief executive of an assurance company whose job is to calculate premtums

administered price the price for a product whish is set by an individual producer or group of producers

salaries, lease of space and offices occupied, electricity and water utilities, mailing costs and equipment costs

agent an individual or organization who acts iir

of a

net profit to the initial capital investment

leg compan trading unOerlyi

ndividual or

audit the

sheet and

ccount and and records

auditor an accountant duly licensed by PRC tasked to check the accuracy of a joint stock company's ledger account and annual

report and accounts, and to present an independent report to shareh0lders on

annuity a series ofpqual payments occurring at

wheiher the accounts present

equal interval of time. Types of annuity are:

fair vies of the comPanY's affair

Ordlnary annuity, Annuity due, Deiened annuity and Perpetuity

annuity due

type

annuity where the payment is made al the beginning of each

period starting from the first period

anticompetitve prac{ice the practice which effect the restricting, distorting and

auth

client and buys

products/properties

return

a corporation can by the SecuritY and Exchange Commission (SEC)

automatic saving the unspent income of a person if his net income is too large to spent

products and capital goods

average cost price the selling price for a good oi service which is equated to the average COSt

product between two or more markets in order to take profitable advantage of any differences in the prices quoted in these

aggregate supply the total amount of domestic goods and services supplied by

both

increase in the value of a currency against other curencies under a floating exchinge_ rate system

for

consumei

which human operations are substituted by mechanical or electronic oPeration or both

average cost is the ratio of the total cost to the total number of units

arbitrage the buying or selling of goods or

markets

articles of incorporation the legal constitution of a corporation which is submitted to the Security and Exchange Commission (SEC)

receives deposits of money from the public and fulfils its obligation to return that money to the depositor when withdrawn

bank deposit an amount of money deposited bank loan credit made available to individuals or corporations bY commercial bank

bank note the paper currency issued by central bank which forms part of the country's moneY suPPlY

bank rate the rate of interest charged by central

bank

to its bonowers, mainlY the

commercial banks

banke/s yeat ayeat. which is equivalent to-12 mont-hs of 30 days each and is equal to 360 daYs

bankruptcy (syn. insolvency) the

average revenue is the ratio of the revenue to the unit of commoditY sold

B BIR Bureau of lnternal Revenues

bad debt money owed which is unlikely to be paid due to the customer being insolvent

condition

individual or company is insolvent, i.e. unable to meet its debts on

wnere

appreciation (ant, depreciation) the increase in the value of an assel. lt also refers to lhe

sells

commission

businesses, including

analysis to real world economic situations

a true and

maximum amount

eliminating competition in a market

applied economics the application of economic

bank a commercial institution which

with the bank

examined bY a qualified auditor

automation the process

administration costs the cost of maintaining a firm within which goods and services are produced. Administration costs includes

behalf

services bY comPetitive bidding

statemenl of a flrm's assel and llabilities on the last day of a iradlng Period

brlrncr rhcct an accounting

due dates.

barGr exchange of goods or services without the use of moneY

bearer bond a financial security which are not registered under the name of a particular holder but where possession serves as proof of ownershiP

bilaterat monopoly

market situation where

there is only one seller with only one buyer

bilatenl oligopoly a market situation

where

there are few sellers and with few buyers

billion refers to one million million (1012) in the United Kingdom and Germany - and one

thousand mlttion (10g) in the United States

FAQs

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Principle 1: A dollar earned today is worth more than a dollar earned in the future. Principle 2: The only thing that matters is the difference between alternatives. Principle 3: Marginal revenue must exceed marginal cost. Principle 4: Additional risk is not taken without the expected additional return.

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Is economics for engineers hard? ›

Its difficulty depends on grasping the basics. If you got the basics down, everything will be chill but if your basics are not solid, you will end up getting confused on whats going on and why its going on !! My advice is to use this book Engineering Economy by Sullivan.

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Eugene L. Grant

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Cost Analysis and Engineering Economy

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The 5 basic problems of an economy are as follows:

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What are the 3 major economic questions? ›

Economists address these three questions: (1) What goods and services should be produced to meet consumer needs? (2) How should they be produced, and who should produce them? (3) Who should receive goods and services? The answers to these questions depend on a country's economic system.

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Petroleum Engineers

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Top 10 Highest Paying Engineering Jobs of 2022

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So in the spirit of organizing how to think about Engineering leadership I am in this article putting forward the five "essential elements" of an engineering leader: People, execution, strategy, technology and self awareness.

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Some of the concepts are scarcity, supply & demand, incentives, trade-off and opportunity cost, economic systems, factors of production, production possibilities, marginal analysis, circular flow, and international trade.

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What is the single most important concept in economics? ›

Supply and Demand

The relationship between supply and demand sits at the heart of most economic theory, for a simple reason: They are inextricably linked. The law of supply and demand can be explained as follows: When supply of a good or service exceeds its demand, prices will fall until an equilibrium is reached.

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Students study the principal economic objectives of stable prices, economic growth, full employment and the Balance of Payments.

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Government has adopted a macroeconomic policy framework which will deliver job creation, better export performance, more investment, greater efficiency and equity of government spending and enhanced human resource development.

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In Engineering Economic Analysis, where: Present Worth = A(P/A, i%, n) P = present value A =...

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Engineering plays a key role in supporting the growth and development of a country's economy as well as in improving the quality of life for citizens. As such, there is an important link between a country's engineering capacity and its economic development.

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He introduces the Cashflow Quadrant, which, as the name indicates, has four sections: Employees, Self-employees, Business Owners, and Investors. Each quadrant has advantages and disadvantages, and as the author explained, they're not created equally.

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PRINCIPLE 5 - CONSIDER ALL RELEVANT CRITERIA: Selection of preferred alternative (decision making) requires the use of a criterion (or several criteria).

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Which branch is daddy of engineering? ›

Mechanical branch is father of engineering so we teach father of engineering..

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Who are the three fathers of economics? ›

The Big Three in Economics: Adam Smith, Karl Marx, and John Maynard Keynes: Adam Smith, Karl Marx, and John Maynard Keynes.

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What are the five main types of engineering economic decisions? ›

The term engineering economic decision refers to all investment decisions relating to engineering projects.
The five main types of engineering economic decisions are (1) service improvement, (2) equipment and process selection, (3) equipment replacement, (4) new product and product expansion, and (5) cost reduction.

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What is demand in engineering economics? ›

Demand is a principle of economics that captures the consumer's desire to buy the product or service. The demand is calculated as the price the consumers are willing to pay for the product or service.

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What is future worth in engineering economics? ›

Conversely, the future value of today's equipment is given by the formula: F = P(1 + i)ⁿ where F is the future worth of the equipment, P is its present value, i is the average rate of inflation or the interest rate, and n is the number of years.

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What are the three types of engineering economy problems? ›

Simple problems. Intermediate problems. Complex problems. Where is Engineering Economic Analysis most suitable?

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What are the three 3 measures of worth used in the analysis of engineering economy? ›

There are several different measures of worth used to select an alternative for a specific set of estimates, some of these are: Present worth (PW), Future worth (FW) and Rate or return (ROR).

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Why is Marr important in engineering economy? ›

The Minimum Attractive Rate of Return (MARR) is a reasonable rate of return established for the evaluation and selection of alternatives. A project is not economically viable unless it is expected to return at least the MARR.

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Should an engineer study economics? ›

Engineers must understand the economic viability of their projects. Fundamentally, engineering economics involves formulating, estimating, and evaluating the economic outcomes when alternatives to accomplish a defined purpose are available.

Does economics go well with engineering? ›

1. Engineers have a natural aptitude for economics. “Engineers are ideal economics students because of their ability to think logically and their strong mathematical skills,” says Carl.

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Is economics a good minor for engineering? ›

Adding an Economics Minor to your Engineering degree will allow you to better understand the business side of Engineering. This can help you to pursue management or logistics roles in your chosen field.

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Is it hard to pass economics? ›

A college-level economics class can be challenging because you need to grasp new concepts like supply and demand, scarcity, diminishing returns, and opportunity costs. It requires you learn new vocabulary and to use critical thinking skills.

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Who is the father of engineering economics? ›

Eugene L. Grant

What is the best minor with economics? ›

Suggested Minors for students who are majoring in Economics might include Mathematics, Statistics, Politics, Environmental Studies, Business, IDS, or Psychology to name a few. A Minor in Economics is also available for students majoring in other Departments or Programs.

Is economics a good major to get a job? ›

A degree in economics can provide a variety of job opportunities in nearly every industry, from local government and the sciences to finance and insurance. While some of these jobs might require additional certifications or education, others are available to those with a bachelor's degree.

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Can I major in economics if I bad at math? ›

According to the United States Bureau of Labor Statistics (BLS), four important qualities economists should have are analytical skills, critical-thinking skills, speaking skills and writing skills. You cultivate strong skills in all of these areas even if you are “bad” at math.

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Does economics have a lot of math? ›

There are many diagrams in economics, but there is not a large amount of math. A proviso: The amount of math in the economics curriculum varies across colleges and universities. Some economics departments do not require their students to learn much math or statistics, but others do.

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Is economics a STEM major? ›

Nonetheless, recognizing economics as a STEM major welcomes a math-based option for learners with broader interests. They can acquire the skills and lessons that STEM content promotes but implement them in different fields such as entrepreneurship and finance.

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