Title | Dorado Assignment 1 - This paper shows the answers to engineering economy problems, specifically compound |
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Author | Anonymous User |
Course | Bachelor of Science in Civil Engineering |
Institution | University of Mindanao |
Pages | 2 |
File Size | 78.9 KB |
File Type | |
Total Downloads | 144 |
Total Views | 696 |
Athasia Kaye D. Dorado CEE109(2122)PlatesCompound Interest By the condition of will, the sum of P2,000 is left to a girl to be held in trust fund by her guardian until it amounts to P5,000, when will the girl received the money if the fund is invested at 8% compounded quarterly? Given:P=₱2, F=₱5, i=...
Athasia Kaye D. Dorado
CEE109(2122) Plates
Compound Interest 1. By the condition of will, the sum of P2,000 is left to a girl to be held in trust fund by her guardian until it amounts to P5,000, when will the girl received the money if the fund is invested at 8% compounded quarterly? Given: P=₱2,000 F=₱5,000 i=0.08/4 -divide by 4 since compounded quarterly Sol: F = P(1 + i)n ₱5,000 = ₱2,000(1 + 0.08/4)4n ₱5,000 = ₱2,000(1 + 0.02)4n 2.5 = (1.02)4n log2.5 = log(1.02)4n log2.5 = 4n log1.02 4n = 46.27 n = 11.57 years
2. A student plan to deposit P1,500 in the bank now and another P3,000 for the next 2 years. If he plans to withdraw P5,000 3 years after his last deposit for the purpose of buying shoes, what will be the amount of money left in the bank after one year of his withdrawal? Effective annual interest rate is 10%. Given: P1=₱1,500 i=0.10 n=2 Sol: F = P(1 + i)n F1 = ₱1,500 (1 + 0.10)2 F1= ₱1,815 + ₱3,000 F1= ₱4,815 – the new principal F2 = ₱4,815 (1 + 0.10)3
F2 = ₱6,408.77 - ₱5,000 F2 = ₱1,408.77 – the new principal F3 = ₱1,408.77 (1 + 0.10)1 F3 = ₱1,549.64
3. If the interest rate of a certain account is 6.5%, compute the (a) single payment present worth factor; and (b) single payment compound amount factor at the end of 18 years. (a) Given: i=0.065 n=18 Sol: P/F = (1+i)-n P/F = (1+0.065)-18 P/F = 0.32 (b) Given: i=0.065 n=18 Sol F/P = (1+i)n F/P = (1+0.065)18 F/P = 3.11...