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# MAT119 M5WA5 ANSWERS

AU

GPA 3.8

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Date Created: 11/15/15

MATH 119 Written Assignment 5 Section 51 2 35000 at 6 for 9 months P 35000 r 06 t 2 12 I Prt I 3500006 2 1575 127 Simple interest is 1575 4 1875 at 53 for 7 months P 1875 r 053 t l 12 I Prt 7 I1875053 5797 7 Simple interest is 5797 8 8940 at 9 loan made on May 7 and it is due September 19 31 7 30 31 31 19 135 days P 8940 r 09 t E 365 I Prt I 894009 51 29759 3657 Simple interest is 29759 16 3475 loan at 75 for 6 months A P I P Prt A 3475 34750755 z 360531 Future value is 360531 18 24500 loan at 96 for ten months APIPPrt A 24500 24500096 z 26 460 Future value is 26460 22 48000 for eight months money earns 5 A48000 t r05 12 A 48000 1 105 127 Present value is 4645161 4645161 26 Sixmonth 18000 Tbill with a discount rate of 1925 a Discount 18000019255 17325 Price 18000 17325 1782675 17325 17826755r r 17325 17826755 z 01944 The actual rate is about 1944 30 P 72589615 r 98 t i 365 APan A 72589615 1 098 1 3657 A 7258961510092188 A 73252269 The amount paid was 73252269 36 Interest 61875 r 675 P 10000 I I Prt Wh1ch means t Pr t 61875 100000675 t 61875 g 675 12 The loan is for 11 months 44 P 8000 r 7 t Fee 100 A P1rt100 A 80001 07 100 A 800010525 100 A 8520 The actual interest is 8520 8000 520 I I Prt Wthh means r Pt 520 z 0867 r 9 8000 The actual interest rate is about 876 Section 52 8 P1000 i06 n10110 APa0 A 100010610 A 100010610 A 100017908477 A 179085 10 P 15000 i 023 n 112 22 A P1 i A 15000102322 A 15000102322 A 15000164916436 A 2473747 14 P22000 i05 n8 A P1 i A 220001 058 A 220001058 A 22000147745544 A 3250402 Interest is 3250402 22000 1050402 18 P 2763035 139 0115 n 394 156 A P1 i A 27630351011515396 A 27630351011515396 A 27630351195275 A 3302587 Interest is 3302587 27 63035 539552 22 P 9000 A17118 n16 A P1i 17118 90001i16 17118 1i16 9000 17118 16 19 1i 16 9000 104099981i iz04141 24 P 13328 i 0205 n 102 20 A P1 i A 13 3281 0205 A 133281020520 A 1332815005836 A 20 000 26 P 10106 i 022 n 252 50 A P1 i A 10106102250 A 10106102250 A 1010629685529 A 30000 32 r 47 and m 2 rE 13 l 2 rE 102352 1 rE 104755 1 rE 04755 4755 36 A 8500 i 06 n 9 PAanr P 8500106 9 P 8500591898 P 503114 44 If money can be invested at 6 compounded annually which is larger 10000 now or 15000 in six years Using present value A15000 i06 n6 P A1 iquot P 15000106 6 P 150007049605 P 1057441 Since this is larger than 10000 the option 15000 in 6 years is larger 46 P 80000 139 025 n 54 20 A P1 i A 80 00012520 A 80 OOO1638616 A 13108932 The interest Will be 131 08932 80 000 51 08932 52 P10000 i03 n326 A P1 i A 10 0001036 A 10 000119405229 A 1194052 She needs to contribute 1194052 66 It is best to use the smallest numbers for the calculation So to find the number of years it Will take to double from 1 to 2 you have A2 P1i04 A P1 i 2 104 In 2 ln104 ln 2 nln104 ln 2 n ln104 n 177 years Section 53 4 R 20000 i 045 n 12 104512 1 045 S 20 6958814 045 S 30928064 S 20 000 8 R20000 i015 n12448 48 S 20000 1015 1 015 S 20 OOO10434783 015 S 139130439 10 R 500 12 n1220 240 1 240 1 Q 12 S 500171264 0041667 S 20551683 S500 14 S 65000 139 03 n 452 9 1039 1 03 65000 R1015911 R 639820 65 000 R 20 R 1200 i 2 n 154 60 s 100000 L 60 10000012001 43 1 Z This is very difficult to solve algebraically so you need another method Graphing Will give a close approximation of the answer So graphing both sides you have I 0035 004 0045 03905 0055 The two sides of the graph cross at approximately 0425 so interest rate of 425 is needed 26 R1050 i035 n8 9 S1050 035 1 1050 035 S 1088692 1050 S 983692 28 R25000 i06 n113 13 S250001 0606 1 25000 S 472 05344 25 000 S 44705344 34 S12000 i051 n616 7 12000R 1051 1 R 051 10517 1 051 12000 R71668 R 167438 12000 R 7 075 42 R80 i n146 12 46 S80112 1 80 075 12 S 424830 80 S 416830 44 R 2435 i n 82 16 103 1 S 2435 2435 03 S 5298947 2435 S 5055447 The account then accrues interest for another five years P 5055447 i 03 n 52 10 A P1 i 505544710310 A 50554471343916 A 6794098 Section 54 2 R 890 i06 and n 16 16 P890 1 106 06 P 89010105895 P 899425 8 P1200 i056 andn14 14 P 1200 1 1056 056 P 120095295954 P 1143551 12 P45000 i andn1112132 1 139132 053 12 45000 R 45000 998630432 R 45062 16 R 10000 i 04 and n 15 The lump sum is the same as the present value 15 P 10000 1 13904 04 P 100001111838743 P 11118387 20 There is a twopart investment an annuity that pays 450 every six months for eight years and the 20000 face value of the bond Which Will be paid When the bond matures 8 years from now The purchaser must be Willing to pay the present value of each part of the investment assuming 59 interest compounded semiannually The interest that is paid each half year is I 200000455 45000 Present value of annuity R450 zig andn2816 141 P450 16 059 2 P 450126092444 P 567416 Present value of 20000 in 8 years P A1 03916 P 200001 l6 P 200006280271 P 1256054 567416 1256054 18 23470 28 P140000 i03 andn15 R140000 3903 15 1 103 R 14000008376658 R 1172732 34 P96511 zig andn2512300 R 96511008099498 R 78169 40 Locate the entry of the table that is in the row labeled Payment Number 10 and in the column labeled Portion to Principle to observe that 8624 of the tenth payment is used to reduce the debt 44 The yearly payment for 207000000 is 207000 000 6900000 R 6900000 139 0578 and n 30 1 105720 6044 0578 J S 6 900 000 6 900 00 S 6 900 000 6 900 000139097174948 S 6 900 000 95 977 051 S 102877051 R 35000 41418 50 1 1 17 3 420 0743 J 12 Total payments equal 101241418 4970160 Interest is 4970160 35000 1470160 The monthly payments are 41418 and the interest paid is 1470160

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