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### solution to problem set 1

Course: ENGINEERIN 152, Spring 2012
School: York University
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Word Count: 341

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and Answers Hints to Sample Problems #1 3.4 End of Year Principal Repayment Interest Payment 0 1 2 3 \$1,671 \$1,821 \$1,985 \$900 \$750 \$586 Remaining Balance \$10,000 \$8,329 \$6,508 \$4,523 4 5 \$2,164 \$2,359 \$407 \$212 \$2,359 \$0 3.5 P = \$12,000( P / F , 5%, 5) = \$12,000(0.7835) = \$9,402 3.8 (a) P = \$5,500( P / F ,10%, 6) = \$3,105 (b) P = \$8,000( P / F , 6%,15) = \$3,338 (c) P = \$30,000( P / F , 8%, 5) =...

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and Answers Hints to Sample Problems #1 3.4 End of Year Principal Repayment Interest Payment 0 1 2 3 \$1,671 \$1,821 \$1,985 \$900 \$750 \$586 Remaining Balance \$10,000 \$8,329 \$6,508 \$4,523 4 5 \$2,164 \$2,359 \$407 \$212 \$2,359 \$0 3.5 P = \$12,000( P / F , 5%, 5) = \$12,000(0.7835) = \$9,402 3.8 (a) P = \$5,500( P / F ,10%, 6) = \$3,105 (b) P = \$8,000( P / F , 6%,15) = \$3,338 (c) P = \$30,000( P / F , 8%, 5) = \$20,418 (d) P = \$15, 000( P / F ,12%,8) = \$6, 058 3.13 P= \$32,000 \$43,000 \$46,000 \$28,000 + + + = \$114,437 1.08 2 1.08 3 1.08 4 1.08 5 3.20 \$ 30, 000 = \$1, 500( F / A, 7%, N) ( F / A, 7%, N ) = 20 There are two different methods to proceed. The first one uses the above equation. From the table we find (F/A, 7%, 12)=17.89 < 20 and (F/A, 7%, 13)=20.14. Hence N=13 years. The second method is to solve the above equation: ( F / A, 7%, N ) = 20 (1 + i ) N 1 = 20 1.07 N 1 = 20 0.07 = 1.4 i 1.07 N = 2.4 N log(1.07) = log(2.4) log(2.4) N= = 12.94 13 years log(1.07) 3.22 (a) A = \$10,000( A / P, 5%, 5) = \$2,310 (b) A = \$5,500( A / P , 9.7%, 4) = \$1,723.70 (c) A = \$8,500( A / P , 2.5%, 3) = \$2,975.85 (d) A = \$30,000( A / P, 8.5%, 20) = \$3,171 To find the corresponding factors in (b)-(d), we need to calculate them. For example in (b) 0.097 1.097 4 A = \$5,500( A / P,9.7%, 4) = \$5500 = \$1, 723.70 1.097 4 1 3.26 F = F1 + F2 = \$5,000( F / A, 8%, 5) + \$2,000( F / G, 8%,5) = \$5,000( F / A,8%,5) + A \$2,000( / G, 8%, 5)( F / A,8%,5) = \$50,988.35 In order to find \$2000(F/G,8%,5) we have to calculate (F/G,8%,5) in above manner since there is no (F/G,i,N) column in Interest Factor Table. Alternatively, we can use the following formula to compute (F/G,8%,5) with the help of the table. ( F / G ,8%,5) = ( P / G ,8%,5) ( F / P,8%,5). 3.31 20 P = An (1 + i ) n n =1 20 = (2, 000, 000)n(1.06)n 1 (1.06) n n =1 20 1.06 n ) = (2, 000, 000 /1.06) n( 1.06 n =1 = \$396, 226, 415 3.34 (a) ( F / P, i, N ) = i ( F / A, i, N ) + 1 (1 + i ) N 1 +1 i = (1 + i ) N 1 + 1 = (1 + i ) N (1 + i ) N = i (b) ( P / F , i, N ) = 1 ( P / A, i, N )i (1 + i ) N = 1 i = (1 + i ) N 1 i (1 + i ) N (1 + i ) N (1 + i ) N 1 = (1 + i ) N (1 + i ) N (1 + i ) N (c) ( A / F , i, N ) = ( A / P, i, N ) i i i (1 + i ) N i (1 + i ) N i[(1 + i ) N 1] = i = (1 + i ) N 1 (1 + i ) N 1 (1 + i ) N 1 (1 + i ) N 1 i = (1 + i ) N 1 (d) ( A / P, i, N ) = i [1 ( P / F , i, N )] i (1 + i ) N i = N N (1 + i ) 1 (1 + i ) 1 N (1 + i ) (1 + i ) N = i (1 + i ) N (1 + i ) N 1 3.36 P (1.08) + \$200 = \$200( P / F , 8%,1) + \$120( P / F , 8%, 2) + \$120( P / F , 8%,3) + \$300( P / F ,8%,4) P = \$373.92 3.45 (2), (4), and (6) 3.49 (b) 3.51 2 P = P (1 + i ) 5 log 2 = 5 log(1 + i ) i = 14.87% 3.54 (1 + 0.06) N 1 0.06 31 = (1 + 0.06) N log 31 = N log 1.06 N = 58.37 59 years \$1, 000, 000 = \$2, 000( F / A , 6% , N ) 500 =
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