MA373 S10 Test 1 Solutions

MA373 S10 Test 1 Solutions - Wt $73 5/0 7257 1 The nominal...

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Unformatted text preview: Wt $73 5/0 7257/ 1. The nominal interest rate over the last year was 9%. The rate of inflation over the last year was 5%. Calculate the real rate of interest over the last year. A. 3.7% t c 0 1+4 Lt: e 140%. B. -4.0A ”bf [‘05 c. 3.8% D. 4.0% ..,. “Lg“,wlfi ,(958 E. 14.5% 2. A loan of is being repaid with level annual payments of 1000 for 20 years. The outstanding balance of the loan immediately after the 12 payment ls 6,569.02. Calculate the annual effective interest rate on the loan. A. 4.6% W, . ()L E - P a W B. 5.1% y 00 D a w C 75‘? 65Q9.0Z=‘ /000 @3737 5/ C 87 D. 10.8% U S Q ”Jaw/4AM? E. 14.1% wwwww wwwflwm PM)”: #2on PV“ ...- £05029 02* Mr 17V " L‘Mfl’ 3. Brandon purchased a car with a car loan of 15,000. The loan has an annual effective interest rate of 12%. The loan will be repaid with monthly payments over 5 years. Calculate the amount of each payment. A. 325.94 We and 3M“ L:- .m» B. 329.04 ‘ U33 c. 330.35 \A/Q ”1575’4 k}: 5 fine 6., D. 333.67 Pafi'mw‘fi) fife mam/71%. E. 4,161.15 Am) I?” a + [ ”5,2" M" #7:; .3; /./2~(/ ,g > ‘/. cm". (l.’% «l: .OOQL/887Q3 ”TE " ~ — A M51703 Cf/J/QQMICU‘DQH W “Mia-M ' e ”‘2 00 N; [.00 a dim w /b/ 0 WV 3’ 3483793 .. .Wagfmmw—a; ”M X (D \ I ”“ ( pomuaavng 3;“ W PV A), O) .009'14513‘793 4. Under a trust fund, Will is receiving an annuity and a perpetuity. The annuity pays 2000 at the end of each year for n years. The present value of the annuity is 17,702.74. The accumulated value of the annuity at the end of n years will be 60,648.57. The perpetuity is 5000 at the beginning of each year. The same interest rate is used for all calculations. Calculate the present value of the perpetuity. A. 27,000 3, H7017? 2000 am 3‘ y a We: <8, 85» 37 c. 54,000 M 71:7 3000 D. 62,500 I!” 02000 g m a: [90 was / E. 67,500 "”7 ' . 6v; 3 Mi... e 30'. 3242,85” " A aooa l ~ vie” " W L... 3 “2.7 +0 550' $935157 30‘? a :3") “ z; .03 WWW 5. Cathy invests 1000 into an account earning simple interest at a rate of10%. Taylor invests 1000 lnto an account earning compound Interest at an annual effective rate of i. in the 11th year, both accounts earn the same effective rate of interest (in). Calculate the amount that Taylor will have at the end of 20 years. A. 1628.89 2535.52 2653.30 3000.00 6727.50 6. Amanda loans 2000 to Claire. Claire repays the loan with a payment of 1200 at the end of one year and another payment of 1200 at the end of two years. Amanda reinvests Claire’s payment at the end of year one in an account earning 10% interest. Using the bottom line approach, calculate Amanda's total annual return on the loan. :.1::: How we uc/Qx w I l/ Ammda c.12,2% her/()6? a/t» 3» years 0. 13.1% / ”/00 [700 E. 13.6% 0 I} . 2/ ~1\ ‘100(\\l>+l200 s; £35243 ll >3 0 o O @(fifidrru 7. findfinvests an amount of money in an account earning interest at a rate of (St = 0.01t. At the end of 4 years, Brandon has 5000. How much does Brandon have at the end of 10 years? 0 A. 3,285.23 5’70 X B. 4,176.35 RM} 6. 5,986.09 Ll. /O D. 7,609.81 E. 8,857.81 SID 51-3011: at 5000 63 gm ,ol‘b Jab" q» / $2900 6 2.)“? 4 50000 63 .S’v—‘Og 5,1,, £2,300 63 HP} 8. Jenna invests 10,000 in an account. Using the Rule of 72, she expects to have 40,000 In 16 years. How much will she really have in 16 years? A. B. C. D. 20,000 20,224 38,960 39,703 40,000 jaunt} Expgcm firm (110:0er m Quaaflx.€‘é lam W mémo 5/0 alga/2% I E Wé’ififl. 075 74-) J8afigfim R; 915% yea/«A dewfla 01.wa W 925 U Mil WW HM.) muugx 00:“ 619 “9le L‘WQ" mom {/ 43-13%:- mooo (I. 09"”; 36}, 703 @ 9. Neal will receive a payment at the start of each month for the next four years. At the start of the first month, Neal will receive a payment of 200. Each payment thereafter will be 10% higher than the prior payment. In other words, the second payment will be 200(1.1), the third payment will be 200 1.1 2, etc. ( ) t L( It); ’ 6) 2?, Neal takes each payment and invests in a fund earning 8% compounded monthly. L02) 0 6 ”7.; i If; Calculate the amount that Neal will have at the end of4 years. ‘11 V A. 148,979.62 0 ‘ CD a“ 9 $00 I ' 0 2° 2"” 21" B. 149,972.82 PMWWW”W ' .. i" l Lt“? 2/ c. 204,946.22 0 \ J D. 206,312.52 2 ‘évwstd’P—‘WM E. 613,511.93 wwmwwflfl y/‘fi 61"”; ,/ ,zm/ M +33 yaw” 43 J 1+7 ‘ + 200 (Li) /I CW5) 5200 (in) i—goo (lib law it ' Firs 47 N 5 X 7“ Them 0% 9,3 ’39 TZ/Lm ... (flét‘éle MIST” (200 (1.4— 7;) “92000. WMWWWW l g M .,. my. W ~ / Vol/l IQ I .) f, g.» Use the following information about an investment fund for Questions 10-11. Fund Value Before Contributlons Contributions 20,000 10. Calculate the exact annual doilar weighted return on this fund. A. 10.2% \/0 (A. WV”) 5:)“ W ”Mg“ 9. 21.5% Ca/QQJJIQfig . c. 27.3% o. 45.8% /§:é7 E 62.0% all?) r» 1,0, 000 804,: “851000 Use the following information about an investment fund for Questions 10-11. 1/z n - Fund Value Before Contributions Contributions 18,000 36,000 4 000 25,000 36,000 11. Calculate the annual time weighted return (im) on this fund. )QOOC’) mis( Hflrw r; < 070000 ‘gm’a’w’ A. 10.2% B. 21.5% C. 273% D. 45.8% E. 62.0% W —3,000 5,000 0 a Qi‘er? «L. ¢ (”LL/9" V Di 12. Allison agrees to pay Kelli 2000 now. She also agrees to pay Kristin 3000 now. At the end of 1 years, Kelli will pay Allison 2500. At the end of 2 years, Kristin will pay Allison 3500. Using the bottom line approach, calculate Allison’s annual return. A. 6.0% ,4” 153m TDMS 02000 ml: 5"” 3. 8.0% #30 c. 9.4% wag’ 3000 A; ( D. 12.3% ”QgéeiWB '3“ng ax “I? E. 25.0% nge’Né’s 3500 6d {7 . _ ()0 b’li’F/MWWMM M . "MW-Mm] | 51:66) 0 £34990 [,9 K, " l .. M + 55790 TE” 5000 ( Hf)" 2Q)“ ) m: {513mb f O X‘zz‘; ‘ +K/ I sac) $1 0 0 ol "=1- QQDG 79590 x2” a.“ 2.4.13ch “”13“ C O x $1 3563?) b +2500 3: ”3‘00)L”'W5b00>('53w3 I- Q» X 1% 3652300) C’fljm ‘ “ (.2 \ 3 30/2; Neal 50 V” \l 6) 13. Cohen Industries is building a new plant. .The expected cash flows from the plant are as follows: “— _ 3 5 mllllon -—— Calculate the net present value of this project at a 12% annual interest rate. . '4 A. -4.00 'll‘ "‘ .5 ,, mIIon AWFV: “/0 J/VqLSV‘L’MSU ¥7\/ B. —0.49 million . .1...“ c. 0.49million wk“ V“ M2...” D. 3.94million E. 4.00million ‘5 ’ 0' km @ 14. You are given in“ = 6%. _._ mwmwmm“w°°fikk—am, «nu- X is the Iquarterly effective interest rate. L“‘> m.p:.um.-an~=,m (5% Yisdm. Catcuiate 10000(X—Y). K 4% Q 6"‘5 \W A. 444.05 ’ "(I '2» [V B. 443.30 N») a; ‘N) ‘ 5,” I ) L s )v‘" m C. 1.49 If; D. 2.24 b 3, I I0 'V S"— E. 8.95 r: (1 7L '72) .,, tr («(0») ‘29; a”) "“"‘ w» 14/ (I +' ‘7'"; ”W Q uf Dr dz..- .» ‘ —-* < I + ‘73“ ‘ q I 3, , 0 /ooo¢>( c:>1€§70fl7§"‘?S " ”#436 30 15. If a(t) = 1 + 0.01t +0028, calculate v(10). A. 0.3226 8. 0.4762 (2. 1.1000 \/ [{J 5" 0. 2.1000 E. 3.1000 “1”“ a (+3 ...
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