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CCF08182010_00020 - Solutions for Exam l—Morning With...

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Unformatted text preview: # Solutions for Exam l—Morning Section 75 56. With simple interest, F : P(1 + i) = ($300)(1 + (0.12)(3)) = $408 Answer is B. 57. P = P0(1 +7)“ 2P0 : P0(1 +4)" log2 =n10g(l +2”) 10g2 log2 7‘ :10g(1 + i) = log1.05 = 14.2 yr (14 yr) Answer is D. 58. P = (net cash flow)(P/A, 6%,71) $40,000 = ($10,000 — $2000)(P/A, 6%,7L) (P/A, 6%, n) = 5 Interpolating from the 6% economic factor table, n = 6.1 yr. Answer is B. 59. A 10—yr period with interest compounded semi— annually is 20 periods. _r_8%_ O ¢_%_7_4/° P = F(P/F, 4%, 20) = ($10,000)(0.4564) = $4564 ($4560) Answer is B. 60. The present sum, P, is $500. The interest rate, i, is 6%, and in 3 yr there will be three interest periods, n. Compute the future sum. F : P(F/P, i%,n) : P(1 + 2')” : ($500)(1 + 0.06)3 = $595.51 ($600) Answer is B. 61. Use the uniform series compound formula. 2 F = A(F/A, 4%,n) = 14%) 5 7 28.00) (_<1+%g55> 1) = $2762.82 ($2760) Answer is C. 62. Use the uniform series capital recovery factor to solve for disbursements. A=P(A/P,z'%,n) :P((1,$_:):):_1> _ (0.08)(1 + 0.08)5 _ ($5.00) ((—>—) = $1252 ($1250) Answer is D. 63. Use the single payment present worth formula. 1 (1 +4)“ 1 = ($800) (W) = $658.16 ($660) Answer is C. P = F(P/F,z'%,n) = F 64. The distributed load is replaced by F = ($) (100 E) (4 m) = 200 N m The force is applied at (2/3)(4 m)=8/3 m from point A. 2MB : 0 8 (10 m — g m) (200 N) — (10 m)RA = 0 RA = 146.67 N (150 N) Answer is D. 65. First, find the reaction. Rleft,y : Rright,y = W 2 = 322.5 N Cut the truss vertically through the second section from the left. 25N SON Professional Publications, Inc. ...
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