Engineering Economy 3rd Edition by Hipolito Sta. Maria (Solution's Manual)

Engineering Economy 3rd Edition by Hipolito Sta. Maria (Solution's Manual)

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CHAPTER 2 Interest and Money-Time Relationship Solved Supplementary Problems Problem 2.1 What is the annual rate of interest if P265 is earned in four months on an investment of P15, 000? Solution: Let ‘n’ be the number of interest periods. Thus, on the basis of 1 year (12 mo.), the interest period will be, Hence, the rate of interest given by the formula, i = , is computed as i = ( )( ) = 0.053 or 5.3% Thus, the annual rate of interest is 5.3% 2-2. A loan of P2, 000 is made for a period of 13 months, from January 1 to January 31 the following year, at a simple interest of 20%. What is the future amount is due at the end of the loan period? Solution: ( ) ( ) ( ) Answer: 2-3. If you borrow money from your friend with simple interest of 12%, find the present worth of P20, 000, which is due at the end of nine months. Given: Future worth: F = P20, 000 Number of interest period: n = Simple interest i = 12% Solution: ( )
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( ) ( ) Answer: 2-4. Determine the exact simple interest on P 5, 000 for the period from Jan.15 to Nov.28, 1992, if the rate of interest is 22%. Given: P= P 5, 000 i= 22% Solution: January 15= 16 (excluding Jan.15) February = 29 March = 31 April = 30 May = 31 June = 30 July = 31 August = 31 September = 30 October = 31 November 28 = 28 (including Nov.28) 318 days Exact simple interest = Pin = 5000×318∕366×0.22 = P955.74 Answer: P955.74 2-5. A man wishes his son to receive P 200, 000 ten years from now. What amount should he invest if it will earn interest of 10% compounded annually during the first 5 years and 12% compounded quarterly during the next 5 years? Given: F= P 200, 000; For compound interest: i= 10%; n=5 For compound interest i= 12%∕4= 3 %; n= 5×4=20
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Solution: P 2 = F (1+ i) -n = 200000 (1+0.03) -20 P 2 = P 110, 735.15 P 1 = P 2 (1+i) -n = 110,735.15 (1+0.10) -5 P 1 = P68, 757.82 Answer: P68, 757.82 2-6. By the condition of a will the sum of P 25, 000 is left to be held in trust by her guardian until it amounts to P 45, 000. When will the girl receive the money if the fund is invested at 8% compounded quarterly? Given: P = P 25, 000 i = 8% 4= 2% F = P 45, 000 Solution: F = P (1+i) n 45000= 25000 (1+0.02) 4n 45000∕25000= (1.02) 4n 1.8= (1.02) 4n In (1.8) = 4nln (1.02) 29.682 = 4n n = 7.42 years Answer: 7.42 years 2-7. At a certain interest rate compounded semiannually P 5, 000 will amount to P 20, 000 after 10 years. What is the amount at the end of 15 years? 0 1 2 3 4 5 6 7 8 9 10 P1 P2 200,000 0 1 2 3 4 5 P1 110 735.15
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Given: P = P 5, 000 n 1 = 10 F 1 = P 20, 000 n 2 = 15 F 2 =? Solution: At n 1 = 10, F 1 = P 20, 000 F 1 = P ( ) ( ) P 20, 000 = P 5, 000 ( ) ( ) = 14.35% At n 2 = 15 F 2 = P ( ) ( ) F 2 = P 5, 000 ( ) ( ) F 2 = P 39, 973.74 Answer: P 39, 973.74 2-8. Jones Corporation borrowed P 9, 000 from Brown Corporation on Jan. 1, 1978 and P 12, 000 on Jan. 1, 1980. Jones Corporation made a partial payment of P 7, 000 on Jan. 1, 1981. It was agreed that the balance of the loan would be amortizes by two payments one of Jan. 1, 1982 and the other on Jan. 1, 1983, the second being 50%larger than the first. If the interest rate is 12%. What is the amount of each payment? Given: =12% Solution: 9,000 12 ,000 7 ,000 X 50% X+X 1978 1979 1980 1981 1982 1983
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P 9, 000 ( ) ( ) ( ) ( ) P 9, 000 ( ) ( ) ( ) ( ) X = P 9, 137.18 X= P 13, 705.77 Answer: P 9, 137.18; P 13, 705.77 2-9. A woman borrowed P 3, 000 to be paid after years with interest at 12% compounded semiannually and P 5, 000 to be paid after 3 years at 12% compounded monthly. What single payment must she pay after years at an interest rate of 16% compounded quarterly to settle the two obligations?
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