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newnansm4

# newnansm4 - Chapter 4 More Interest Formulas 4-1(a 100 100...

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Chapter 4: More Interest Formulas 4-1 (a) R = \$100(F/A, 10%, 4) = \$100(4.641) = \$464.10 (b) S = 50 ( P/G , 10%, 4) = 50 (4.378) = 218.90 (c) T = 30 ( A/G , 10%, 5) = 30 (1.810) = 54.30 0 1 2 3 4 100 100 100 100 100100100 100 R \$0 \$50 \$ 10 0 \$15 0 S T T T T T 30 60 \$90 120 0 1 2 3 4 \$50 90

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4-2 (a) B = \$100 ( P/F , 10%, 1) + \$100 ( P/F , 10%, 3) + \$100 ( P/F , 10%, 5) = \$100 (0.9091 + 0.7513 + 0.6209) = \$228.13 (b) \$634 = \$200 ( P/A , i%, 4) ( P/A , i %, 4) = \$634/\$200 = 3.17 From compound interest tables, i = 10%. (c) V = \$10 ( F/A , 10%, 5) - \$10 = \$10 (6.105) - \$10 = \$51.05 \$0 \$0 \$100 \$100 \$100 B \$63 4 \$200 \$200 \$200 \$200 i = ? \$10 \$10 \$10 \$10 V
(d) \$500 = x ( P/A , 10%, 4) + x ( P/G , 10%, 4) \$500 = x (3.170 + 4.378) x = \$500/7.548 = \$66.24 4-3 (a) C = \$25 ( P/G , 10%, 4) = \$25 (4.378) = \$109.45 (b) \$500 = \$140 ( P/A , i%, 6) ( P/A , i %, 6) = \$500/\$140 = 3.571 \$50 0 \$25 \$0 \$50 \$75 2x \$50 0 x 3x 4x \$50 0 A = \$140 i = ?

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Performing linear interpolation: ( P/A , i %, 6) i 3.784 15% 3.498 18% i = 15% + (18% - 15%) ((3.487 – 3.571)/(3.784 – 3.498) = 17.24% (c) F = \$25 ( P/G , 10%, 5) ( F/P , 10%, 5) = \$25 (6.862) (1.611) = \$276.37 (d) A = \$40 ( P/G , 10%, 4) ( F/P , 10%, 1) ( A/P , 10%, 4) = \$40 (4.378) (1.10) (0.3155) = \$60.78 4-4 (a) \$0 \$25 \$50 \$75 P \$10 0 F P \$0 \$40 \$80 \$12 0 A A A A
W = \$25 ( P/A , 10%, 4) + \$25 ( P/G , 10%, 4) = \$25 (3.170 + 4.378) = \$188.70 (b) x = \$100 ( P/G , 10%, 4) ( P/F , 10%, 1) = \$100 (4.378) (0.9091) = \$398.00 (c) Y = \$300 ( P/A , 10%, 3) - \$100 ( P/G , 10%, 3) = \$300 (2.487 – 2.329) = \$513.20 (d) \$0 \$0 \$50 \$ 10 0 \$15 0 x \$25 \$50 \$75 W \$10 0 Y \$30 0 \$20 0 \$10 0

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Z = \$100 ( P/A , 10%, 3) - \$50 ( P/F , 10%, 2) = \$100 (2.487) - \$50 (0.8264) = \$207.38 4-5 P = \$100 + \$150 ( P/A , 10%, 3) + \$50 ( P/G , 10%, 3) = \$100 + \$150 (2.487) + \$50 (2.329) = \$589.50 4-6 x = \$300 ( P/A , 10%, 5) + \$100 ( P/G , 10%, 3) + \$100 ( P/F , 10%, 4) = \$300 (3.791) + \$100 (2.329) + \$100 (0.6830) = \$1,438.50 Z \$10 0 \$10 0 \$50 P \$10 0 \$15 0 \$20 0 \$25 0 x \$30 0 \$40 0 \$50 0 \$40 0 \$30 0
4-7 P = \$10 (P/G, 15%, 5) + \$40 (P/A, 15%, 4)(P/F, 15%, 1) = \$10 (5.775) + \$40 (2.855) (0.8696) = \$157.06 4-8 Receipts (upward) at time O : PW = B + \$800 (P/A, 12%, 3) = B + \$1,921.6 Expenditures (downward) at time O : PW= B (P/A, 12%, 2) + 1.5B (P/F, 12%, 3) = 2.757B Equating: B + \$1,921.6 = 2.757B B = \$1,921.6/2.757 = \$1,093.70 4-9 F = A (F/A, 10%, n) \$35.95 = 1 (F/A, 10%, n) (F/A, 10%, n) = 35.95 P \$0 \$50 \$60 \$70 \$80 O \$800 \$800 \$800 B B B 1.5B

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From the 10% interest table, n = 16. 4-10 P = A (P/A, 3.5%, n) \$1,000 = \$50 (P/A, 3.5%, n) (P/A, 3.5%, n) = 20 From the 3.5% interest table, n = 35. 4-11 F = \$100 (F/A, 10%, 3) = \$100 (3.310)= \$331 P’ = \$331 (F/P, 10%, 2) = \$331 (1.210)= \$400.51 J = \$400.51 (A/P, 10%, 3) = \$400.51 (0.4021) = \$161.05 Alternate Solution: One may observe that J is equivalent to the future worth of \$100 after five interest periods, or: J = \$100 (F/P, 10%, 5) = \$100 (1.611)= \$161.10 \$100 \$100 \$100 F P’ J J J
4-12 P = \$100 (P/G, 10%, 4) = \$100 (4.378)= \$437.80 P’ = \$437.80 (F/P, 10%, 5) = \$437.80 (1.611) = \$705.30 C = \$705.30 (A/P, 10%, 3) = \$705.30 (0.4021) = \$283.60 4-13 Present Worth P of the two \$500 amounts: P = \$500 (P/F, 12%, 2) + \$500 (P/F, 12%, 1) = \$500 (0.7972) + \$500 (0.7118) = \$754.50 Also: P = G (P/G, 12%, 7) \$754.50 = G (P/G, 12%, 7) = G (11.644) G = \$754.50/11.644 = \$64.80 P \$10 0 \$20 0 \$30 0 P’ C C C \$500 \$500 0 G 2G 3G 4G 5G 6G P

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4-14 Present Worth of gradient series: P = \$100 (P/G, 10%, 4) = \$100 (4.378)= \$437.80 D = \$437.80 (A/F, 10%, 4) = \$4.7.80 (0.2155) = \$94.35 4-15 P = \$200 + \$100 (P/A, 10%, 3) + \$100 (P/G, 10%, 3) + \$300 (F/P, 10%, 3) + \$200 (F/P, 10%, 2) + \$100 (F/P, 10%, 1) = \$200 + \$100 (2.487) + \$100 (2.329) + \$300 (1.331) + \$200 (1.210) + \$100 (1.100) = \$1,432.90 E = \$1,432.90 (A/P, 10%, 2) = \$1,432.90 (0.5762) = \$825.64 4-16 D D D D P 0 \$10 0 \$20 0 \$30 0 \$30 0 \$20 0 \$10 0 \$20 0 \$10 0 \$20 0 \$30 0 E E P
P = \$100 (P/A, 10%, 4) + \$100 (P/G, 10%, 4) = \$100 (3.170 + 4.378) = \$754.80 Also: P = 4B (P/A, 10%, 4) – B (P/G, 10%, 4) Thus, 4B (3.170) – B (4.378) = \$754.80 B = \$754.80/8.30 = \$90.94 4-17 P = \$1,250 (P/A, 10%, 8) - \$250 (P/G, 10%, 8) + \$3,000 - \$250 (P/F, 10%, 8) = \$1,250 (5.335) - \$250 (16.029) + \$3,000 - \$250 (0.4665) = \$5,545 4-18 Cash flow number 1: P 0 1 = A (P/A, 12%, 4) Cash flow number 2: P 0 2 = \$150 (P/A, 12%, 5) + \$150 (P/G, 12%, 5) Since P 0 1 = P 0 2 , A (3.037) = \$150 (3.605) + \$150 (6.397) \$10 0 \$20 0 \$30 0 \$40 0 4B 3B 2B B P

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A = (540.75 + 959.55)/3.037 = \$494 4-19 F = ? n = 180 months i = 0.50% /month A = \$20.00 F = A ( F/A , 0.50%, 180) Since the ½% interest table does not contain n = 180, the problem must be split into workable components. On way would be: F = \$20 ( F/A , ½%, 90) + \$20 ( F/A , ½%, 90)( F/P , ½%, 90) = \$5,817 Alternate Solution Perform linear interpolation between n = 120 and n = 240: F = \$20 (( F/A , ½%, 120) – ( F/A , ½%, 240))/2 = \$6,259 Note the inaccuracy of this solution. n = 90 n = 90 F’ F A = \$20
4-20 Amount on Nov 1: F’ = \$30 (F/A, ½%, 9) = \$30 (9.812) = \$275.46 Amount on Dec 1: F = \$275.46 (F/P, ½%, 1) = \$275.46 (1.005) = 276.84 4-21 The solution may follow the general approach of the end-of-year derivation in the book.

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newnansm4 - Chapter 4 More Interest Formulas 4-1(a 100 100...

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