# HW2 Dianna Padilla - (A(P-A 1 11 A=(3000 A.01(1.01)^11(1.01...

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HW 3 4.3. (a) Arithmetic Gradient present Find C. C= C (P/C, 10%, 4) ) 1 ( ) 1 ( 2 1 i i i n n in + + - - i=.1 n= 4 C= 25 C(4.378) = 25(4.378) = \$109.45 (b) Find A P(A/P, 10%, 6) 500(.1771561/ .771561) = 500(.22961) = 114.8 = \$115 (c) Find F 25+ 25(a/G, 10%, 4) 25 + 25(1.38) P= 25(P/G, 10%. 5) F= 25[(1 + (.1))^5 - (.5 – 1)] / .01 = 11.051 25 * 11.051 = \$276.275 (d) Find A = (A/G,10%,4) 40[1.1^ 4 - (.4 - 1)]/ (.1 * 1.1^4 - 1 )) = .0641/ .04641 = \$55.25 4-4. (a) (P/A, 10%, 4) + (P/G, 10%, 4) W= 79.25 + 109.45 = 188.70 (b) (P/G, 10%, 4) (P/F, 10%, 1) X= 100 [ 1 – (1.4)*(1.1^-4) / .10^2] * 1.1^-1 X = 398.01

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(c) (P/A, 10%, 3) - (P/A, 10%, 3) Y = 746.06 – 232.91 = 513.15 (d) (P/A, 10%, 3) - (P/F, 10%, 2) Z = 248.69 – 41.32 = \$ 207.37 4-11 F = 100 (F/A, 10%, 4) = 100(4.641) = 464.10 464.1 (F/P, 10% , 1) = 464.1 (1.1) = \$510.51 510.51 (A/P , 10%, 4) = \$161.07 4.37 Calculating the payments……
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Unformatted text preview: (A/(P-A) , 1%, 11 ) A= (3000 - A) * ([.01(1+.01)^11] / [(1+ .01) ^11 -1] ) A = .09645(3000-A) = 289.36 - .09645 A 1.09645A = 289.36 A=263.90 The new buyer will pay 1000 + the 6th payment + the value of the remaining 5 payments: Payoff = 1000 + 263.90 + (P/A, 1%, 5) Payoff = 1000 +263.90 + 263.90 [(1.01^5 - 1)/ (.01 * 1.01^5)] Payoff = 1000 + 263.90 + 1280.82 = 2544.72 4-86 (a) P= (P/A , i, n) A(P/A, .75%, 36) 2800/ 31.447 = 89.04 (b) 36 - 9 = n P= A (P/A, ¾ % , 27) 24.360 * 89.04 = 2169.01 4-90 (a) P= (P/A , i, n) A(P/A, .75%, 24) 100/ 21.889 = 4.57 (b) 24- 13 = 11 = n P= A (P/A, ¾ % , 11) 10.521 * 4.57 = 48.08 4-116 A= \$1000 r= .05 n= 5 years nominal interest rate r = 4 * 5% = 20% i = 1 + (.2/4) = 1.05^4 = 1.215 – 1 = .215 = effective interest rate is 21.5%...
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## This note was uploaded on 04/27/2008 for the course EIN 4354 taught by Professor Tufecki during the Spring '08 term at University of Florida.

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HW2 Dianna Padilla - (A(P-A 1 11 A=(3000 A.01(1.01)^11(1.01...

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