PART 1 - Engineering Economics . Annuity ANNUITY PROBLEMS ECONOMICS - 1 (ME Bd. Apr. 98) 4 company issued 50 bonds of P1,000 face value each, redeemable at par at the end. of 15 years. To accumulate the Tunds required for redemption, the fim established 2 sinking fund consisting of annual deposits, the interest rate of the fund being 4%. What was the principal in the fund at the-erid of fhe 12th year? * A P95983.00 —C. PA1,453.00 + . 8. P38,378.00 D. P37,519.00 SOLUTION: tre t= 50,000 RR ; R RRR —_—_— oa2 3 12 43 44 45 Solving for 15 annual deposits: : p= RG+i-4 _ 50/4000) = 50,000 i 15. so00 = Ris 0.04 9 R=P2,497.00 After 12 years: 12, a Parone soe) = = P37,519.00 ECONOMICS - 2 (ME Bd. Apr. 98) A house and lot can be acquired a payment of P100,000 at the end of each year for a period of, end of 5 years from the date of pufchase. If money is 14% conipounded ahnually, vet is the cash price of the property? A. P810,100.00 C. P801,900.00 B. P808,811.00 D. P805,902.00 SOLUTION: \ R_ R RR {r<—_——__ R= P100,000 pe RUG+IP- 9 (+I down payment of P500,000.00 and a yearly i starting at thePAID 1 tenginnering Keonomies « Anowlty Vea ENE SAE De wanna 12954" 44) Sahn thes a 0 Coahs press = rem oper + D Cog Wren = SAE Cesta = PUNO * i 9 a a. ar 9) ) Wh wisedannsy tts We Weed Mh AIL gr yon ot Ms tei GINMAS -o% = oi Stet tr 1msrn va, on Seen wns pote, Wiha We ie winnaal ‘intra Ta tor Ue trae pines AD 81% hh 5% D ih wan " 1” e 1 "” pt y Pe sy UN vere sot A. oop oun: ny ihe sare WE VIO A (8 + GLMAT HOUSES) eh) 6S Hahn. 1° AIA LDWUNNI NS 4 UAE Be. ar) a in vans vba Des th punt woh nan wo wero Wilh Lay inn Monjea4 msn bata |S fae alan aa Haast ae, CO A a No hon teh oa Vy yh i (Mba td Ath ape ator iw PART 1 ~ Engineering Economies - Annuity ‘S32 PAID 7258 50000 B P242806.00 D 250,490.00 souTion: aor 29 2st aoe tae y + ¢ + 4+ 6 7 t a — % 4s ad ruts 9 Pe Gif y + fo jpn ZnO 28) 9 p42 906 00 (1 + 0.06)'*(0.06) Waics - § (ME Bd. Apr. 98) soto 1 an aye ns py tras oa 11 nls te Croce neat, go tal each eniployee will feoding 2 ‘bonus? oe P12;608.00" "C. 13,600.00" +” gf 12,610.00 D, P 12,300.00 SOLUTION: 7 jeo1at2=001 peo = 7 = 0(2000) = 160,000 sea000 = Rut +o0s) =H 0.0% R=P12,615.00 oONOMICS -6 (ME Bd, Apr 98) . Fount ayn aes erie wo ar os ri an account bearing erst at 5% compeund annual, ors See a eat atom soo ae sitaebesye ‘A. P36,041.73 ©, 730,941.73 B. 39,041.73 D. P25.941.73, SOLUTION; t 77 7 18 19 20 24 p+ —_|?_ = F100,000. PART 1 - Bygineering Poonomics « Annulty nye WN O14" wear or) Me paanett.ss sang ee ba can aR Seen vermont i Cash Price + 80,000 + \ eye sony » SREULEE. 5 p999 03500 (reat F Cash Paice > OS +3 (ME BA. Apr. 8) o>, Me auaciuneny a be KOAAghl AE PHLOLD cash, oF fr jown and Pee bent per yoo fo YS years. What isthe annua interes it fr tho time et A 46% © 571% & 38I% ©. 11.00% SOLUTION: Tso TSO (750 750 «(750 y yt yy t t t t i 14 ¥ y b= 10,009 - 2.000 Pe PSO = RD, ) ba (eee coo Bd Ty tedeit (+ 0.0861)'*(0,0861) 8,000 « 6,000 8.000 = Therefore: 1= 461% ECONOMICS - 4 (ME Bd. Apr. 98) eon plans to rebre exactly one year and want an account that wil Pay No) Aoatog0 a year for the nest 15'yeare. Assuming 2 6% annual elective interest rato, 25,000 4 yea! fs rid noo! 1s epost now? (The und wl be doposted ater TB years)| : PART 1 ~ Engineering Economtes.. annuity 83 ‘a. 240,000.00 c. 258,600.00 fh p247,000.00 0 250,400.00 SOLUTION: gt 29T a6T + 44 a a Sao yr pe Ld ae Ye po 720012 00611) = pa42.00 00 (1 + 0.08)'°(0.06) Jomics - § (ME Bd. Apr. 9 econ 5 8 0 wy mg el ima nfo ony fora yoo at 124 dina nites rte chmpoutded—* soot iaot ‘323900 bonus? : 3 ig that cach omplayee wil rood moni ge P1Z{608.00" , P12,600.00 B. P12,610.00 D. 12,300.00 soLUTIO! [20.1212 = 0.01 Riis 0" = T F = 00/2000) = P160,000 Ri + 0.01)" - 1) ~—901 160,000 R=P12,615.80 ECONOMICS - 6 (ME Bd. Apr. 98) ‘Apparent on the day tho child is born wishes to determine what lump sum would have {o'be paid Into an account bearing Intorost at 5% compound annually, in order to withdraw P20,000 each on the child's 16th, 19th, 201h and 21st birthdays A. P35,941.73, C. P90,081.73 8, P33,961.73 ©. P25,041.73 ‘SOLUTION: 5 $ 7 ie 10 1 R= P100,000 peos PART 1 - Engineering Economics - Annulty Ril 4 (tl 20000411 « 0.05) 9 i =P70919 (1+ 0.08)*10.08) =— p or rose _ 30,941.73 * ie 0.05)" ECONOMICS -7 (ME Bd. Oct. 98) What le the size of 10 equal annual payments to repay a loan of Pt ams 10%? Fick payments one year ater récahing te loan, = on TOMY A P1B.00445 ©. PI7.946.35 8. P6274 54 D. Pisisa7.4s SOLUTION: 2 ein aig yey yy ty ss 7s 3° 40 R= P16,274.54 ECONOMICS - 8(ME Bd. Oct. 99) ‘hand tool costs P200 and requies P'.21 labor cost per unit. A machine tool costs P3600 and reduces labor to PO.75. What is the break-even point (in years) at 5% for ‘an annual production of 4000 units? ‘A 1.999" ©. 234 yr 8 176 yr ©. 156 yr SOLUTION: se ; p= 3400 P= present cost = 3600-200 = 3400.00 » PART 1 - Engineering Economics Anmlty ss acini Bs cn» IB = Alt pein 1+ 065P = 3400 = 10401 57'(005) 9.09230(1.05)" = (1.05) -1 0.90761(1.05) = 1 (hosp = 1.10179 Int.10179 = 3.086 years 8 pn. 05 machine c. paste 1467 a bie D. Ps.si18.67 B P4481.76 ‘SOLUTION: R= P1000.00 P= 1000 1+0.15)° - 1] as P = pasts.as56 PART 1 - Engin ECONOMICS - 10 (ME Bd. Apr. 2 ‘The maintonance of a car is Said 10 be P500.00. The maintenanco is said to increase of P50.00 each year and subsequent for 9 years. Tho monoy is estimated to 8% compounded annually, What is the present worth of maintenanco forthe full of 10 years? ‘A. PA,378.00 B. 4,064.00 SOLUTION: ‘eering Economtcs - Annuity 1000) ©, P4,658.00 D. P4.454.00 950 a wei, (+0. (1+0.08" 1) , a = 10 | P= 800) 0.08)"0.08] *” [ (1 +0.08)%0.08)" (1+ 0.08)"%(0.08) P = p4653.89 ECONOMICS - 11 (ME Bd. Oct. 2000) {A sinking fund consists of 15 annual feposits of 1,000.00 each, with interest ‘ded annually. What is the principal in the fund at Auad a th oo of 4% compour ‘Stonminal date? A ‘p25,115.00 c. p20073s9 8. P29.546.37 D. P28. 105.00 soLuTion: tr PART 1 ~ Engineering Economics Annulty 57 «12 (ME Bd, Oct. 2000) Feo eit anid off in 10 equal annual payments. The annual {A 10,000.00 1900 ov much Intec vl be paid Wo tha first two (2) youre? interest rato is Al be pa 3,005.10 c. P2971 ® pao 20 5. P2.920.10 R R SOLUTION: } } 4 000 x sealvng forthe annual payment: (+1 pe al Can. = pf teeta soem = 8 [amo R= P1,992.52 992.62 (1 + 0.15)! = P2,291,40 "992.52 (1 + 0.15)" = P2.635.11 0 Ht interest, R= 10,000/10 = 14 Intorest paid n the fist two years 2"@29t-40 + 2635.11) -2(1,000.00) = P292651 : ECONOMICS - 13 (ME Bd. Oct. 97) A stee! mil estimates that one of Its furnaces will require maintenance of 20,000.00 al the end of 2 years, 40,000.00 at the end of 4 years and P80,000.00 at tho end of 8 years. What uniform semi-annual amounts could it set aside over the next eight years al the end of each period to meet these requirements of maintenance cost if all the funds would eam interest al the rate of 6% compounded semi-annually? A. P7,897.35, C. P9,397.35 8. P6,897.35, ©. P8.897.35, SOLUTION:on 60 PART 1 - Engineering Economies - Annulty , ‘A. P19,907.00 ©. P13.875.00 B. P13,678.00 D. P13,487.00 SOLUTION: ti F Pees vid at ti = I heey, a r= alt n=2(12)=24 i= 0.122= 0.01 Fe sooltts 001 oot 13,487.00 ECONOMICS - 18 (ME Bd. Apr. 97) “The president of a growing engineering firm wishes to glvo each of 50 employees 3 olday bonus, How much is needed fo invest monthly for a year at 12% nominal | Interest rate, compounded menthiy, so that each employes will receive @ 1,000 i = parson PAST nego § Baad souumon: teeta t i= 0.122 =0.01 pe plead _ pllt+ 0.010)" - 4 1000 = R901 R= P78.848 per employee For 60 employees! R= 78.048(50) = P3,042.00 ‘ PART 1 - Engineering Economles - Annuity 61 economics - 19 (ME 84, Apr. 87) ECONO ond P1000 wo yr 200, The lems a oan we 7 Pere vith uniform payments. Ho just made bis seamed nmol jtorost fr Intoros! 101 ws much principal dove ho stil ovo? ¢. P10,117.00 ©. 11,790.00 on paymen ‘A. P12,000.00 @, P19.024.00 SOLUTION: eit pe i pits 0.10)" = 1 15000 5 0)%0.10) Re p244i.te Principal he stil owe: (+= 8 = 2.447.184 pe RO AT 0107070) 13,023.52 P= ECONOMICS - 20 (ME Bd. Apr. 97) EON jaye fs about fo receive the sum of P900.00 atthe end of each year for § foe Ope year prior to the receipt ofthe first sum, he decides to discount all § Ja ithe interest rate is 6%, what proceeds will he obtain? ‘A, P1,875.35 . 1,263.71 B. 2,287.03 DB. P1.450.72 SOLUTION: SPP LS yy yd Tt zon, = alle Po p= 300Ltt+ 0.06)° - 4} (1 0.06)°(0.06) ECONOMICS -24 (ME Bd, Oct. 26) mpany purchases P200,000 of equipment in year zero. It decide i Y Pu ent in . fos 10 use straight [na deprciaton over the expected 20 years feof the equipment. Tho interest rate be ge tax rale Is 40%, whal is the present worth of the depreciation ‘A. P3,500.00 B. P26,500.00 C. P39,700.00 1D. 4,000.00PART 1 - Engineering Economies - Annulty, “ k pres piste 6 Bas 6 ase sown: fron sou 1= 0.04/12 © 00033 RR RRR Rn ‘naam 88 steed 4 SPT ren( ett) $0,000" R. RePI204 ECONOMICS - 27 {in 18 years, P20,000 Is requited for a child's calloge exponese. How much thou be ceponted wach yor eartng onto day of ith 20 that is goa ba ma? (Resume that the fiat payment is made at bi, tha last paymanl Is on chil bethday, and that 5% interest fs paid payment on anit N0 A pae ©. Pasa 8 Pa2i D. Pred soLUuTion: tr 20.000 RRR RR x, yyy ' Tt ttt n= 19 (trom day of binh lo age 18) i= 005 “fete (1.05) - 4 0.05 20,000 = af R= P654.00 ECONOMICS - 28 Starting on January 1, year 1, P50 is deposited In an account paying 6% annually, Each Janvary 1 thereattor, up fo and Including January 1, yoar 10, another PSO wa Fadopostied’ Staring January 1, yoar 15 (ho date ofot withdrawal); fv union eee cainarewals aro mado, Tho last wihdrawal will exhaust the fund. How much wll be withdrawal each yoar? ‘A, PI97.52 " 8, P3aza4 ©. P6534 D. Pa53.45, PART 1 ~ Engineering Keonamter Annutty “ Mi sowution fa eect Hee Fre Fax 056.009 (1.009 * O92 02 (ae if. | Pe af with (14 0.06)" (1 pe 19752 ICS - 29 ECONOMICS 20 P1200 eri hy Wt yor Equprort timated mointonance Ie P1000 tho fst yar, bt i areata to £2000 TI, e00 och yoar bear, Usk 10%, fet tha pron worth of oe proc . A. P20,354 Cc. PID244 B. P2833 D. P21,050 2800 SOLUTION: ‘eon 1400 “pe so i | st P= prosent worth sv » 2000) profess], faites a |iro. eh PL eh aif oer = oo) (12° =1 y9 P so ] nf 4410 | ema (0) nO 1.01%0.1) P= 6144.507 + 5 Pr putssor «2000.44 98.55) + 12,000. 771.006 (1.10)PART 1 - Engineering Economics - Annuity 6s union: “ at at af oat ot” n= 12 months (1 yr) tid ! t ? pe a(Qert Tt TR 0% Pr t00 too= 946/220" =1 (1+ i" (14 2 wosr= (See = (an) Tog: |= 2% (monty) | 1057-1057 | Soin forthe annual effective interest | ig=(1+)?= 12 (1 40.02)" = 1 0.2682 = 26.82% | ECONOMICS - 34 te for a payment plan of 30 equal payments of P39.30 P2000 would have been an outright purchase? ©. 34.23% D, 26.82% What is the effective interest ra pper month, when a lump sum of A 20.34% 8. 12.38% SOLUTION: ‘n= 30 months. (emt 22.396 (eae 5 22,396 = 22.398, ‘Sohing for the annual effective interest. eC ?-1 = (1+0.02)"- 1 = 0.2682 | | | } “Try: 122% (monthly) e288 PART 1 - Engineering Eeonomles. Annulty 69 EgRNONC eau a OES nh. Avprannualy. 2500 units are produced annualy. What is tne Payback pero at 10K atzys C. 557918 8 324 yrs D 834yr6 SOLUTION: ‘Savings = 0.06(3500) = P210 Net savings = 210-40 = P170 ° weit) BP gh 0 wo eR, , (aR yyey — ro | Ot TZ ¢ 5.77 et 0.41176(1.1)"=(1.4-1 0.588(1.1)"= 7 taped? In? sserys int. ECONOMICS-36* Econ otiment fk pesos is made”at tho end af each year fr three yous. al an Aan invessy of ote per yoar compounded annually. Wha wil the peso value of the Tateravestment be upon the deposit of the third payment? R OTex ©. 1.295% B 227m D, 3276 ' x F OLUTION: x + + = p( emt) - [oor ron(te Pat) = OS F=3278x ECONOMICS - 37 The folding cash flow diagram represents an investment of 400 pesos and a revenue of x pesos at the end of years one and two. Given a discount rate of 15% compounded annually, what must x be for this set of cash flows to have a net resent worth of approximately zero? ‘A 206 C. 255, x t 8 257 D. 248 Y + souution: é t i Pr 40066 ECONOMICS «30 PART 1 - Engineering Heonomtcs + Annulty A fast acting broke on a fnet-turning latho ts estimated to saya sovon seconds por piece produced since the operator Ihave To walt a Fong For the lathe ‘Assuming i then yoar ifo, no salvaga valu, and an 8% Interoat rl puch pric of tho brako? ‘yo tho maxing ‘A P3006.63) 6 OLUTION: [P 14,000 plot of land ean Pe a(t) eh nn? 10,000 » (200 wey onus svgs = 19 Pe Ri . {1 0.00) neue (og 00 MICS «34 (pad at tho rato of P18,00 por hour) dos not {0 stop. 40,000 piocos aro produced annually. ‘what should G, PA633.45 D. pr972.45 Tos y 40,000pe. , te “pe yr S600sec ‘Annuinl savings © P116567 Pe purchase price ofthe brake: tbo purchasod for P4000 down and P1,200 per year for 12, ae What is tho annual Interost rate being charged? c. A 6% a3 3 1aa% 5238 SOLUTION: | pe 14,000-4,000 = P10.600 goo og? e so ; sf Fa id ;? 3 << PART 1 ~ Engineering Zeonomtes- Annulty 67 10.1% ( { eooen! =1 339 (chstentoaen 0.333 © 8.333 1 ‘Thoroforo = 6.1% rst year, P500 the second year, ics -92 EconoW! 96000. far. Use 10% to find the year en et i eg co hould be replaced. Fatih the car st 10 ay 0. 19 400+(a-1)100 SOLUTION: 600 500 ro 3 ein-4],¢[/O+i-1__ von (tat] of te i R= 400 P= 6000 62100 = goof -=1] . 100 [P= 4. _0 98.0001 af 4) Gr Oa? G.4P(0-0. toen= 13 g,000 = 400 | 1-9-1) , 100 [-C9> 1 13 " nF 0., GaPO? 1.50.0) 6,000 = 6179.04 ‘Thotefore the car should be replaced in year 14 since it started at year 1. ECONOMICS 33 loan company advertises that P100 borrowed for one year may be repaid by 1 methlyirsialments of P8.46, Assuming the diferenco xtwoon fe amount repaid borrowed Is interest only. What is the effective annual interest rate bang cared? neten a issu 5 taePART I - Engineering keonomles « Annulty reer) (ety (ayy (1180.15) 6.04 70 won| ECONOMICS - 38 Mr. Richardson borrowed P15,000 two years ago. Tho terms of the loan A POAT ©, P1700 8. P12,000 ©, P1024 SOLUTION: Forn 10 rp ay" hv } oR tem -4 wom ReP2adiie ‘Amount sill own: ne 10-268 gap 2 2441.18 (1.90.9, | 13,029.52 ECONOMICS - 39 ‘A fim considering ronting a trailor at P300 per month. Tho unit !s noadad for vo yoars. Tho leasing company offers a lump sum payment of P24,000 al tho ond of fro yours na an alternalivo payment plan, but 1s wll fo dlgcour ts Moyo. The frm placs a valu of 10% (olfecive anna ale) on ivostmont cpt. How tage should te dscount bo n oder tobe accoplabl as an oauvaont? ‘A. P7S0 ©. Paz) 0. pan Seg lt SOLUTION: ff f t pe R(t aii Solving the equivalent monthly Intorost, ) keaeired 24000 PART 1 ~ Engineering Economics . Annulty " ote (att 1 0.007074 n= 6(t2) = 00 (1,007074)" 1 ‘o.007074 Discount * 24,000 - 22008.77 = P4,031.22 rye 300 ] eer ECONOMICS - 40 Foe oalinont of 350,000 /s mado, 0 bo followed by payments of P200,000 each foarte tee year, What f the annual ato of rtum en Investment for Bis project? A "dn 921% | BA1.7% 0. 57.1% | souution: | Ws f a | ee af (re ) ef | ne3 + + ¢ st +33 260,000 20 one ot } fpr ssne00 (ewes | 17s ( we ) ty 932.1% | ies 764 Therefore |= 92.1% ECONOMICS - 44 Martha deposited P10 por month Is an account paying 6% inlerest compounded monty fr 24 months. Thereafter, sho mado no copost of witravate fr fe yeas. How much must have accumulate In Martha's account afer the seven-year | A. paass.t6 ©. pso.g2084 8 Paseo78 D. Prrsieco | soLuTiON: Fu Fu { f Pop epee n= 60 ee dele or oe alepagT 1 -Bngineering Beonomlcs - Annuity ne 24 oaths (E09) feooeiz=0 ee a(2F>] 4.0057 009 008 Fue = P2543.19 For another 5 years: n= 5(12) = 60 Feu=P(1 +i)" = 2543.19(1,005)" = P3,430.99 ECONOMICS - 42 ‘A young engineer borrowed P10,000.00 at 12% ‘annum for the first 4 years, What does he have to ‘order to pay off the loan? interest, and paid P2000.00 pay at the end ofthe fith yearn ‘A. P3,926.00 C. P3,037.00 | 8. P5674.00 D, Past9.28 | ‘SOLUTION: t . SPP On ep ([{eit=1], a +eeudl aa | a VTS, fuse | 10,000 = 2000 | ——__—__ P= 10, {i -12)*(0.12) t ieee \ A=pesi7.72 -CONOMICS -43 ‘A bond pays P500.00 interest per year and has a face value of PS,000.00 atthe end ‘ot 8 years, when it has to be redeemed. fits current discounted price is P3,900.00, ‘what true interest could be eared on the bond? A. 15% C. 13% \ sure SESS yt yy oS Jeesee = present value of bond PART 1 ~ Engineering Zeonomtes. annuity 7 eon em] Tar 3,900= ont a sy |= 482% apne 2 et |" Tene 3900 = 3912.12 “Thetetore choose i= 15% ICS «44 Oh bid of ot ye trey Letx= bof fish to break-even oss SH Pe af (ih ) “ES < 41.457" —4 P= 473,000 00 = (6.7x)| M 473.06 cael at) x=P582283 ECONOMICS - 45 ‘AP 10,000 direct-reduction loan is taken out. IIs to be repaid atthe rate of P200 per ‘month with annual effectivo interest rate of 19.56% charged against the unpaid balance. What principal mains to be unpaid after the third payment? A PS4cO, C. PO7ES B, Posty D, Poss SOLUTION: Solving forthe month interest bea sips Oagss= (rer 15045 (monty)_oaX~—~— pant 1 ~ Brgineering Beonomles - Annuity “ she ne the praspa aftr 3 months: Suelo ots = PHOse78 svi for ho tal amount pai afer 3 yoars for P200!month aie = P609.045 0.015 Amount unpaid 3—F = 10,456.78 - 609.045 = P9s47.73 [ECONOMICS ~ 45 ote of machinery has an inal cost of 40,000 and results In an Increase in A Pesrmaintenance costs of P2000. If the mackinery saves the company P10,000 per year, in how many years will the machinery pay for itself if compounding is Eensdered? (1= 7%), 4 years ©. 6 years B. S years D. 7years } SOLUTION: . $ s $ s | Annual saving See ee SF) ‘Anal sevings =P3,000 +e eds { wr soe =R - i r (ae) {Pe aoooo | ‘40000 = 8000 ( {1077 =1 (10077007) 0.35(1.07)"= 1 t (1.07)" = 1.538465, t tnt. s3846 n= ESSE 6.357 yor 017 years ECONOMICS - 47 1 ‘A tractor costs PS0,000 and has an expected ile of ten years. The salvage value is | ‘estimated to be P2000 and annual operating cost are estimated at P1,000. What is ‘he approximate rate of retum on the investment if the annual revenue is 10,0007 At 13% | 8. 12% 0. 14% | c= 2000 SOLUTION: Re pe (teat), ? em) OF R= 10,000 1000 R = P9000 PART 1- Engineering Economics Annuity 7 2,000 ay (iy -4 = 9000) U1) —" 50,000 ( rr" Ty 12 12.7% 50,000 = 50032.43, ‘Therefore the answer is: C ECONOMICS -48 CO 000 bark fan Is to be pad in equal yearly payments over 15 years at an A Pas annual intrest rate of %. Whal percentage of he fst payment sapped tothe principal? n O% ©. 39% B. 36% ©. 51% sain oe : ‘Solving forthe annual payments for 16 years: = R(t eo (140.07) 5,000 = R| O07 = aso Sao * R= P2744.06 ‘Actual interest after one year: 1=25,000(0.07) = P1750 ‘= 2744.86 ~ 1750 Percentage = 27855 1 = 36.24% ECONOMICS - 49 ‘You have decided to take 2 car loan from a bank that charges a nominal rate of 9% ‘compounded monthly. The car costs P16,000 aller down payment, Including all ‘options, taxes and other expenses. You agree to pay the loan of P16,000in a series OF 60 esate payments Whal tne month A. P121.80 payment? ©. P33280 B.-P26667 D P1440.00 soLuTION: RRRRR R yyy ‘ i= 0.0011 + ---% P= 16.000PART 1 ~ Engineering Economies - Annuity cues 0a 76 (1.0075) ~ 4 : 16,000 = R| —— (i .0075)°"(0.0075) R= P932.13 ECONOMICS - 50 (ME Board Oct. 1995) {1 P500.00 Is invested al the end of each year for 6 years, 1 Br 74s, what Is tho total peso amount available upon’ tho depoek cy gee ale at the deposit of the sixth ‘A. P3,210.00 C. P3,000.00 B. P3,577.00 D. 4,260.00 SOLUTION: (140078 ~1) 0.07 P3,576.64 | Fe oxo ECONOMICS - 81 (ME Board Apr. 1996) in five years, 18,000 will be needed to pay for a bulding renovation. In order to. | ‘genorato this sum, a sinking fund consisting of three annual payments Is established now. For tax purposes, no further payments will be made after three years, What payments are necessary If money is worth 15% per annum? ‘A. P2870 . P5100 B. P2019 D. P2670 SOLUTION: P_ 18,000 Fa= Teoh” Geoasy 7P1361059 . ne fost] 13,610.59 = RI R= .P319.54 I~ PART 1 - Engineering Economies. annulty ” ECONOMICS $2 (ME Board Oct 1990) EcoMeed PA000 por yoar for 4 yoars 10 G0 to colege, ¥ Nerd Da ry el 9 fo Pir end of your 171 1th, Yh and 20th Birthday, how much meney wl bein at coount atthe end of tho 21et year? ‘x P2500 . P1700 8. P3400 D. P4000 SOLUTION: LP ee A “ _ teyyy Si {r - 5000————>] Fe ‘Amount at the end of 16th year: § = P(t +i)" = 5000(1 + 0.07)" = P14,760.82 | pidsih=t rete (07) -4],_A A = P1,699.85 ECONOMICS - 53 (ME Board Oct. 1995) ‘A fecal frm Is establishing a sinking fund for the purpose of accumulating a Sufficient capital to rele Its outstanding bonds at maturity. The bonds are redeemable in 10 years, and their maturity value is P150,000.00. How much should be deposited each year if the fund pays interest atthe rate of 3%? A PI2,547.14 Cc. P14,094.85 B, P13,08458 D. P16,648.87 SOLUTION: the» 12000 A ptege 4 oe oles 150,000 = aaa = 0.03 R= P13,084.58zn Ps - ' "ART 1 - Engineering Economics - Annuity ECONOMICS - 54 (ME Board Oct. 1996) ‘A machine costs P20,000.00 today and has an estimated sera 2000.00 ater 8 years. Inflation fs 8% per year The flac. anrua) tra fale cared on money invested is 6%. How much money needs to be set aside ‘each year’ ‘ach year to replace the machine with an identical model 8 years from now? ‘A. P2,980.00 ©. P3921 8. 3,290.00 D. Passioog SOLUTION: yey ‘ $s oTrs is 3 Yee 2000 = Replacement cost after 8 years: F = 20,000(1 + 0.08)" - 2,000 = P35,018.60 i aft] R = P3,292.26 ECONOMICS - 55 (ME Board Oct. 1995) coon anoney Must you Invest today in order to withdraw P1,000 per year for {0 years ifthe interest rate is 12%? ‘ et, progoneo 8 pass000 6; panos soLuTion: ° SELF IL SS yeyey ety TTS = 0 :| ‘o.12(t + 0.12)" P = P5.650.22 part 1 - Engineering Economics. Annulty 79 Jr yr, 1996) ee MRA np st tm mie Ff" no machine wil generate a savings of a oe as Samal ee, C. 1,67 years: & FO ? yey id, years 07 perteor B. 3.17 years a LUTION: . t = a Je~ 25,000 187" | asoon = 1500) Sap (0.10 )1.18)" = 3(7.187"-3 (isy = 1.4286 in4.4266 «2.46 yea 0 nt yeas i EGONOMICS -57 (ME Board Oct. 1996) ECONO Hsing P'100,000 In annual rent for offce space af the beginning of each lnsteag the rext 10 years, an enginoering firm has decided take out a 10-year Yee 05 loan for @ new bulding af 6% Interest. The fim wil Invest P100,000 of Foo es and eam 16% annual interest on that amount. What wit be the Uimerence between the firm's annual revenue and expenses? ‘A. The firm will need P'17,800.00 extra B. The firm will break even C. The firm will have P21,500.00 left over D. The frm will have P13,000.00 extra SOLUTION:‘a S 0 PART 1 - Engineering Economics - Annulty B= 195,067.96 \nnual interest eamod for P4 , re 100,000 investment= 10 = Ditererce = 1358678600000 tgo00 = Pirgsrea rerefore, the firm needs P17,867,96 extra ECONOMICS - 58 (ME Board Apr. 1897) years and POO 0 oe ai 20,000 for 2 reais. 40,000 at the end of 4 ; ears. Compu fe the sem annual moun est eld fr ths equpmon Novey woh To compounded anual © P5,099.00 D. Pr0.0%2.00 ‘A. P7,954.00 B. P9.088,00 SOLUTION: For i= 10% compounded annually: P = 20,000/(1.1)' + 20,000/(1.1)° + 40,000/(1. 4) + 80,0001(1.1)" P = P99,351.87 Solving for the equivalent interest of semi-annual Effective interest annually = Effective Interest semi-annual (Georet = (1eif-1 12 diate = 4.88% semi-annual = pfs ; pe af ai ] | vt soasiar= p[ 04s 1] (1,.0488)'°(0.0488)_ R= P9089 ECONOMICS -59 (ME Board Apr. 1997) : Foon che lecount rte is 15%, what fs the equivalent unform annual cash : _, 200000, $0000 , 751000 P= 100000 5 ase OF Shea som aca aes? yearO - $100,000 year 1 - $200,000 year2- $50,000 | yeu” Sis aed «. s124200 § Sindo0o & $gsa00 SOLUTION: es oi # tt Pt tf yok oR Sy parr 1 Engineering Beonomies- Anny > = rage » [RE 361.0" (1.157'00.15), p= $150,126.54 p= sos10n.94 the equivalent uniform annual cash ics -60 (ME Board APT. 1997) Year2 - 300,000 conor ‘discount rate is 11.5%, what is ue fot Ning sueeam of C2Sh flows? fet efoto g.goa Year? ~ (7200.000 198.000 Yes viapan © £5000828 B. P255,124.24 1. P525,421.20 SOLUTION: 435,000 200,000, + ats? = 100,000 + 2 P= 100000 + 7 45) | af (itt? = 810,068.62 = a aR 115) R = P255,124.24 p = P618,068.92 ECONOMICS - 61 (ME Board Oct. 1994) IT yu oblained a loan of P1M at the rate of 1 Sider to build a house, how much must you pay monthly to amor period often years? 2 percent compounded annually in tize the loan within @ AP11,995 cc. P13,995 B.P12,995, 0. P14,995 SOLUTION: Solving for the interest rate per month: ons tenet RR RRR z = 00094887 yyy + ee ee ee {= 0.94887% per month, n= 10(42) = 120 fr = 1,000,000s2 PART 1 - Engl pe alte et ring Economica - Annulty iy 4,000,000 « | (14 0.0094887)'24 (0.0094007)(1 + 0.009488)" ‘ R= P13,904.64 ECONOMICS - 62 (ME Board Apr. 1993) Rainer Wandrow borrowed P50,000 from SSS, in the for Interest at Bpercant, compounded quarry. pave instalments for ton yours. Find the quarterly payments. A.P1,345.58 ‘©. 1,827.79 B.PIS43.87 P1895: rm of calamity foan, with blo in equal quarterly SOLUTION: i= 0.084 = 0.02 10. 50000 = R (1+ 0.02)" 1. 0.02(1 + 0.02) Re P1827.79 FecONOMICS - 63 (ME Board Apr. 1991) FeONoMchiased cat wih a cash price of P360,000, He wes 20 negotiate Aa Pec aow nim to pay only a down payment of 20 pers ‘and the vi ne payable in equal 48 ond ofthe month instalment al Peed) lrest per balance payday ne pald the 20th naalment, he decided (0 Pay ne remaining Sanat meirosse o. prs838s Sipwazse pion? soLUTiON: gigs p= 2anc00 - PART 1 ~ Engineering Economies. Annulty 83 vit pe al! wr p< 280000 - 020(380,00) = 260,000 (1.05) - 4 = [08 280,000 [sien Re P8,225.00 (140.057 = 1 = ozs) 20 = nw = 02 sooo Po = P186,927.25 ics - 64 CHO Sno i cmt a a Nn nt to San $B equal month mio ired monthly payments? payat noe the requ Fon much lb Be gree MN 105.78 8. P16,185.78 B, PY7.185.78 ‘SOLUTION: } + + + + R ae sats Ts 7 p=2z38000 ‘Down Payment = 0.15(280,00) = 42,000 Balance = 280,000 - 42,000 = P238,000 P= ooze] 299,000 =| 1+ 0.018)" =1 1+ 0.015) (0.018) R = P15,185.78 Econowtcs 5 (Me Board Ao, 1296) team annul amount should bo deposited wa posited each year in order to 100,00 at the end of the 5th annual deposit if money eams 10% incest ‘A. P15,708.83 c. P18 x . P18,494.45 8. Pi7963.45, D. P16,379.75 SOLUTION: . {fF 100.000 G+iP teace aes yyy TTT,84 PART 1 ~ Engineering Economies - Annuity 100,000 = | {10% =1 010 R = P16,379.75 ene +66 (ME Board Apr. 1990) 1,000 pesos Is dopostod enth yoar fr 9 year, how much annul , ly can a person get annually fram the bank avory year for 8 years elarting 1 spectro, Cex clmeneyistapeeant nny nero ‘A PI4675 ©. P30,675 8. PAG,G75 0. P25,675, SOLUTION: fe nso 40% 407 407. JOT RR RR (140.14) <4 +» = 1000] 110.44" = 4 Fe = 10) [ ‘a Fo P160853.46 (40.4 ‘oiait+ 0.147 | | rosssae = Re P34, 675.18 CONOMICS - 67 (ME Board Apr. 1990) ‘A now company developed a program in which the employees wll be allowed fo purchinsod shares of stocks of the company al the end of is fith year of poration, whon tho company is thought to have gained stability already, ot par value oP 100 pur shore. Believing in the good potential of the company, an employee Geaidod guvo in a bank the amount of 8,000 at the end of every year which wil earn Hesiin om intorosl, compounded yoarly. How much charos of stocks wall he be able 2 | 1 dhs ond in oar of his yor depo | 1A goa chores ©. 470 shores 8.07 sharos Bonz shares | | . SOLUTION: (14 0.099 Fe sooo| ot Fe pa7,877.69 Numbor of sharos = 47,877.69/100 = 478 shores PART 1 ~ Engineering Keanomles - Appulty 7 Ics -68 ECONOMICS ot monthly Inatallnont ut P1000 at cA ated. ‘Tha whens ni ‘Aanan pure A sna et arena av aya ntwenty your, Wa so raat ar 10 a tO i area pabeatt ©. PH62.15 n Piatt 0. Punt? SOLUTION: ne ERD '. j= 22 001 yyyyd 4 ew hE SE ner oe : Jo- 100.000 vente pl ie 0.007? — 4 aaa | econ f= P1101.006 ECONOMICS - 69 (ME Board Apr. 1996) FeO OM rend paymonts ovtanding avr elghlyoare aro a fas: 0.201 feats tyes, 720,000 for tno vac yoor, PO.000 for Who Did your, wr eat from the fourth through the Bit your, Find the oyeatent 1P40,000 for each ran at worth of these payments if. tho annual intareat Is BY ‘A. 44,800.00 G, 35,050.00 8, P30,563.00 SOLUTION: OO yyyyyd sf f won «nema 0.00(1.08 7 Solving forthe present worth P = 10009 | 20000 | 50000 , 150,700.40 (6.08 * (1.087 * (1,087 “hoa 7 P192.879.9986 PART 1 - Engineering Economies - Annulty yo2sro.9a = | 108" <1 0.08(1.08)° R » P33,563.04 ECONOMICS - 70 (ME Board Oct. 1996) Re Ayala borowws. 100,000.00 at 10% elfectvo annual intorost. Ho, must pay Baek the Toan over 30 yoars with uniform monthly payments duo on tho first day of ‘each month. What does Mc. Ayala pay each month? ‘A. P870.00 C. P878.00 B. P846,00 1. P839.00 | SOLUTION: | RRRRRR yy tyre aoe P= 100.000 Solving for the equivalent monthly Interest: u= (eit G10 = (14-1 1 = 0.007974 (per month) | For annuity dve: po men[lates crs 0 Jonaoo = r+”) —00reraree = Terarac. 007974) R = P839.19 ECONOMICS -71 It you obtained a loan of P500,000 at 12% compounded annually n ordor fo BUY 8 Sa. what amount must you pay yearly to amortize the loan within a period of § » years? . P138,704.87 1, P346,567.23 quiapa) $e 1.12)°(0.12), 00000 = al ‘ R = P138,704.87 souon Pg Pa ty fs ff PART 1 ~ Engineering Economtes - Annulty 87 jomics -72 ECON’ 'P100,000 from SSS with 10% compounded quarterly payablo in 00 Gruz, bor instalments for 18 years. Find the quarterly payments, oust pztS04 P20 46 Ae Sawa _ qftetes beeeae pee ait TTT j= 0.10/4 = 0.025 P= 100.000 n= 154) = 60 1025/4 orf 1025 = | 100.000 [eee R= pan86.34 ECONOMICS -73 Ecom000 fa dopositod in the Bank every month with 1.5% Interest, what is tho ‘ocumulated amount after 10 yours? A. P234 765.23 ‘©, P465,900.90 _ BLP Ss85,890.65 0. P331,288.19 SOLUTION: n= 12(10) = 120 EE FSF SF ge afoot yyyyt ‘S TI" he = (1.015)'29 ECONOMICS -74 Aman wos i wihdrow the ‘an amount of P5,000 per month rats 1% mony, whet amount shad ho doposk own tre bank? ‘A. P190,537.52 A x C.P140,537.52 150,537.52 ©. P160,537.52 rH ed58 PART 1 - Engineering Economics - Annuity pART 1 - Engineering Econom(ca - annulty 89 iso Fisizy He oe coe) HB]. pr4aa00 os = 012 mount to bo withdrawn onaually for another 6 yoars is 8 = 123) = 36months 1] For te f 8 = pied (on? p= Rift —1) , sooo soon wei qoN%(0.00 | 1 P = PI50537.52 Pe RR 140,389.88 = al; " }OMICS 75 tno sale of a car made for 9 cash prot of POCO. a f dealer the aca a 2 Y A car dealer advertise Mine reqared down payment #5 20%, and te beanee is R= 34,146.43 Seer imorest rave of 1 5% per month. How 3 Bayable wn 24 equa! monthly instore much wa be the ‘monthly payment? | Tana epee | cose ann a pg yt me at opaens | Seater am a asso paar eeree 8 Ejereeheaene ite teed sear a soot --3 | B, P1007.24 .PG1,747.21 o a cy | {P = 280,000 | soLuTiON: . ‘so¥vng for Uno annul payment: [iso =1 pe af wi own Bown p00 70000 = 280,000 (1.015) = 1 | 280.000 = a Serta 20° 1015) 0 = nf 20 nae (é 20)'9(0:20) R = 23,952.27 The present worth of py = 29862.27| 020 1.207" (0.20) the remaining 4 years Is: yyment_= 0.2 (350,000) = 70,000 } R = P13,978.75 | | CONOMICS - 76 fed In the bank each year for 10 years, how much annuity “0 & rs starting 1 year after he 1 P8,000 Is deposit "get anally from the bank every year for 6 Yo Moin deposit is made. cost of money Is 12% 7 Pe = P61,747.19 ‘A.P24,146.43 C, P44,146.43 } 1, P54,146.43 \ ECONOMICS -78 A father deposit P50,000 with 16% Interest annual , ly to pay tho tuition feo of 10,000 evary yoar of his son. How fong could the amount pay the tuition I he will SOLUTION: ‘After 10 years, the amount in the bank is: paying 1 year from now? tr Btjeee 5.3 yoors, eT_st_sT_st 8T Rg RR ). 13 years tay dt ui FF one Se oe ee re nooo ro a(t yyy ey 4 “Tt P= 0.00090 PART 1 ~ Engineering Economtes - Annuity PART 1 - Engineering Economlca- Annulty 91 3.16)" = | ECONOMICS - 81 50,000 » 10.000 15) saiorrowed P100,000 with 2% interest monthly and promise: ° 18) A mats ASG qual monthly Instant staring T yer ham now, Dessabna a n= 10 yoors nt ron pa .68 c.P7.55146 8. PO.551.65 0 Passi.6s ECONOMICS -79 . 7 A P10,000 Is deposited in the bank annually with 12% Interest rate. How kc SOLUTION: ‘monoy bacome P'1,000,0007 ¥ ena a ‘A. 12.34 years, ©.2263yoars att B. 1065 years, D. 45.22 years, Pu SOLUTION: enone oa [pr oosee FPS be | soving to PF 9% FoF 0! * Sov for 400000 1.02)" yy yyy ‘ Pu = P124,337.43 css -5 For the manthly payment: + + . ah ! | per or] ayy 4 (+i Fe a 7 | { 02) 4,397.43 = RI stp = | ? (1.027(0.02) ¥ 1,000,000 = sono| 274] R = P5,551.66 n = 2263 years x ECONOMICS - 82 : ‘A man invested P1000 per month on a bank who offers 6% interest. How much can hh get after 6 yoars? SONOMICS - 80 ded semi-annually 200,000 with 16% inorest compoun 7A P60.000 c. 72,540 an eoyoe very ernie {for 10 years starting today. What amount should he B. P69,493 D. P62'570 Bay every & 8 mone “¢. P24,861.52 | SOLUTION: A pieestee 5. P26.861.52 Shing re oquvaert mnt nore RR a B . 0.06 = (1+i ae SOLUTION: EE ie “ooteser ”* n= 5(12) = 60 = 200,000 12 ote2= 008 hmm Fe p[(ieut-t n= 2(10) = 20 F T ai -4), t pe a[tstt)en F + | coenr 1 0.004867 09)" -1-] , Fe 200,000 = alg 7 (0.08), oie 10,061.52 Tee OMICS ~ a9 R= PIB, Ieney Is wer 4%, Worth What '$ 01500 morithiyg Pa¥Tent at the end of each quarter will replace ——92 PART 1 ~ Engineering Keo Annulty A P1800 & pHon BPI SI D Ptnos SOLUTION: PP GP GP GP 2 For ®t year fy yd For monthly payment fe Goan = 0.00339 P p= soo|—!t 00333)" (7.00333) p= S72 RAR RRR For quartrly payment R ie 0.048 = 0.01 yyy 1 © piteut=2 ---4| Per Pe 5072 se72 © RJ (1.01* (0.09) Re P1605 | ECONOMICS - 64 | What amount would have to bo Invoatod at tho ond of each yoar for tho nox! O yout tiny compoundad sem-annually in ordor to havo 5,000 atthe end of tho tna? ‘A. P541.06 . P542.64 B. PS53.62 D.P268.25 SOLUTION: 1 = 0.0472 = 0.02 | ns 20) = 16 re afoeth= oR font =1 ong al 0.02 R= 726025 ROONOWC ° 85, 420,000 if pald In cash. Tho mixor mar HOONOMES 25, copra mira \ Hana th ee ae sy: ay eth PART 1 - Engineering Economies . Annutt innuley 22 mount of ch annual pment al ayiments mada at to. Ihe beginning of each year, vat Peta a wes mH nent 4 oon a eet (1.08) ~4 = RR] 08) 1 raoo00 = Re [aeerass| s3a7R = 120000 R= PO977.55 ECONOMICS - 66 coo iacl eal (or somi-annual payments of P40,000 forthe next 10 yoars and on Acari paymont of P250,000 at tho ond of that timo. Find tho equivalent ae dtnat ine contract at 7% compounded semiannwally? jeaupvsiont:canh, wk Paaa,s02 25 (c'p599.509.00 B: o04,130.00 © p7s2777.00 SOLUTION: SELLS Pye tt 1 1 a i a P 280,000 For somi-annua payment: T= 00772 = 0,038 n= 10) = 20, Pe {tei r) f sone taney 1 + (7.035 )%0, Py = 860,496.19 oe) P, = —F__ 250,000 (+i (1,035)? Pa = 125,641.47 7 PePap = Pe pasa fy 268406.19 + 125,641.4724 PART I~ Engineering Economics - Annuity ECONOMICS -87 Mr. Rankine bought a piece of p ce of property for P100.000 down paymen eae eran eae P.O cig yar fm Tew ihe eres rate is / compounded semiannually. what is reser : 7x Bia 999.08 eM eiee as MA Ne te Pome B. P148,555.02 ©, Pre6.608 01 SOLUTION: se 400,000 ho 0,12/2 = 0.06 For the deferred semi-annual payments: n= 10 (iyi -t perce! [ eitt | 0 py = 2000[ 106)" =1] (106)'° (0.06) p, = $0000:89 5 4999.08 (1.06) p = P+ down payment = 43999.08 + 100,000 P = P143,999.08 ECONOMICS - 88 ECONOMICS cptalned a loan of PA0,000 at he rate oh 6 compounded annualy a An emeleyeg ainause, How much must he pay meri [2 ‘amortize be Toan within a riod of 10 years? pence’. PUI0.24 c. pi7643 B. P234.12 D, P135.98 SOLUTION: ee R RR ' “the monthly interest will be: ye ett 4 ee tet $ tt “to fe 0.00487 eu tae nein pe Rf aert (1.00467) -1 10,000 = | Gpo4e7)™ 0.00487) R= P110.22 | | \ \ PART 1 ~ Engineering Economies - Annuity 95 ECONOMICS - 82, 126,000 credit for his old car when es buying a new mode! costing ea ce e ssh pret aba tere a fe et by payer by payrerin rate of @¥ compounded Meni? scrpgs 400 e'p33.650 6 P2850 B. P35.340, LUTION: : : so & FF PF iT FoF 375,000 pr 128000+6 j= 0.06/12 = 0.005 ‘The balance wil be: prs 375000 - 125000 -C p = 250000—C fmere: © = the cash that will be paid. 0 2so000-c = 12500 1-005) —— 1 ° (.005)" ECONOMICS -$0 termine the present worth of an of Dulaminn he pagent worth of r annual payment 9 P2500 at the end of each year SOLUTION: one 5 paseoaan Bi prsweoe o[asas epee e- ¢ far] ee |r 1203 4 ~The P = P11.983.06,96 PART 1 ~ Engineering Economics - Annulty ECONOMICS - 91 A man borrowed P200,000 from a bank at 12% compounded monthly, which fy payable monthly for 10 years (120 payments). Vf the first payment is 10 bo mage Sher 3 months, hew much is the monthly payment? m paee9 42 c Poors 10 8 Pesan t0 B. pa.t2e 2 soLUTION Phd i eee i 20 bo p= 200,000 oa = 001 Py = P(1sif = 200000( 1.01 Py = zotea0 (if -4 perl oi ] | 20 zooao = r{ 1.00 —4 | (1.01)'7° (0.01) R= 2927.09 NOMICS -92 fan agreed to pay the foan he is borrowing from the bank in 6 equal end-o-tho, PART 1 - Engineering Economies - Annuity 97 cs -93 8 epare the future of his 10 Year old son. Determine the ron shes 1 propre tn fur Cin intveat 5.41% per Brum Yo amour a - time his son will be 18. Prenoanal et E prassss Plana § buat sOLUTION: RRR R ++ to TTF $ JP 120.000 ates glen ‘The monthly in terost will be: ‘iy “4 (18-10) = 8yrs.(12) = 96 months (1.0044) “|e 0.0044 R = P1007.2% ECONOMICS - 94 coor p0D pesos Is deposited each year for 9 years, how much annuity ean a person tre tenualy trom the Bank every year for 8 years starting 1 year after tha Sth deposit femade, Cast of money Is 14%. rr Ye pret 1% por annum compeurded oun ad Feet c. po4e75.10 Pa eo ie payne tne bank, row much money be § basenss 5 Pizzseze man borrowing from the bank? . Se ee c, pasa4ss75 | sownon: A fatoiees ©. paso,o00s9 tn 7 ne6 SOLUTION: © PP PO | 401_40T 107 101 JOT. RR RR eee | Th htt Oa a a | crt — +e ere ye eee oe ne cote | reafterat lg \ : Fe = 10000] (1.141° oped 0.141) PER wait =4] ; om 160,853.46 7 ‘or the withdrawal: R = 71,477.70 ‘6 drawal: use -1] 2 0,853.46 = p|_(1.14)° pe 74477.70 eo = P250,000.59 aa aoe (1.187 (0.18), . (1.14) 0.14) R= Page75.1, : akee part 1~ Engineering Economics - Annuity jomics -95 ECONOMICS 5g tho salo of a car modol for a cash price of P260,000. Mualiment, tha required down payment is 15%, and the balan i rt monty inatotiment olan rea aloof 5% per month Hoy pave po tno roqured monly paymonts? arg aren B. P14,234.67 D. P12,857.34 R eyyd 4 vom? SOLUTION: aben eaten Down paymont = 0.15 (260000) = 42,000 Balanco = 280,000 ~ 42000 = 238,000 pe afte a ih ! 230000 = R|—tuorsitt =| (1.015)"" (0.018), R= P15.108.77 ECONOMICS - 96 Team Wola Cruz borrowed P2,400 al 1% per month payable in 24 equal end of he flor he has Jronth payments. How much of the Joan remains unpaid immediately a ppald the 12th payment? A, i270 c. P1436.87 8. P1356.67 B. P1528.3427. SOLUTION: R RR RR RRR R ty yey ' ly payment: Ssalving forthe mont PART 1 - Engineering Eeonomtes - annut les - Annulty af tsu=1 wir (1.0074 -1 = {io an = § [aa R= 112976 “The balance wil be the presont worth at the 12" paymont (00)? -1 ay = 112.976 | AS Paw 112 [re =| Pr = P1271.55 ECONOMICS -97 . EcONCMrchased 8 car with a cash price of P350,000. He was able to negotiate Aman pura gnaw ho pay ony 0 down payment of 20% and tha ba ye sae: of nth toon + nes par ment On he sual tle ne decided to pay th romain bance. What's Bay he paid Gaye Fring bolonce tat he paid? ‘234,546.34 c. p196,927.25, B. P534,524.63 B, 20d 543.34 SOLUTION: own payment = 0.2 350000 ) = 70000 Balance = 35000-70000 = 280000 ‘The monthly installment will be: ps aor 1 (14iP i 280,000 a[ ee (1.015) (0.015) R = P8225 per month, ‘Atterthe 10” payment, tho balance will bo its present worth that time.“AKL 1 Bagineertng Economics - Annuity ART I - Engineering Economtes - Annuity 101 yovigs va sland orton eon tavolaped a program in wich tho employees il bo allowed ty Mets 255 sack tt erany aie ond ah yoo eran, logwviny tn thought fo have gained stabitty alroady, at par value af P100 000 = 4000) — eee “2° YRMnpin the good polar th company am employe decked to ee re assy (0.0266) 20x yeh the amount of P8,000 at the end of avery yor which will earn for him 9X ~ 410,000 = 110,000 ieweet, compounded yearly. How many shares of stocks will ho bo ablo ty Therefore: jurehased al the end of the fifth year of his yearly deposits? sai = 1202.68) = 30.60% A. 470, C. 476 . 8474 B. 478 100 JGONOMICS-100 4 surplus Income of P1000 per yoar which he plans 1 SOLUTION: F Seca hv? 2 CRA Purana 1% par ann fo mo deost over § rr) Am Bank Wl eo rvgstr cole at th end of 13 yas eR” FF Fie Poors. compute now ©. P23,256.87 eet PyggTT A Pao 5. par.26721 8 Mé ” F = gooo| 0% —1 Ft cre 0.08 | : Ifo F = pa7er7.68 FE LF Numbor of shores = PA7877:99. = 47a shores vit -- \ 100 share me] 4 ECONOMICS - 99 Because of tho peso devaluation, a car costing P150,000 Is to bo purchased frrough a finance company instond of buying cash. | the buyer is required to pay tae eBo“as down payment and PS,000 cach month for four yoars, what isthe a interost rate? 000) (1:18) ‘A 30.60% ©. 22.88% 0.18 8. 32.66% D. 43.88% f= panies ‘SOLUTION: FEES HF nF oF wo vo yy ey n= 4(12) = 48 ‘Solving for the monthly interest rate: © pfteit=t me al ll | { { JP= 110,000 ann » sine 40% = 1000 | Po tie