Chapter 1 Money and Interest 1. A man wishes to accumulate P3,722 after 5 years, 8 months and 28 days. How much should be deposited by the man in a bank if the ordinary simple interest is 15% per annum? Given: F = 3,722.00 I = 15% or 0.15 n = 5 years, 8 months, 28 days Required: P=? Solution: π = 5 + 8 ππππ‘βπ π₯ 1 π¦πππ 12 ππππ‘βπ 1 π¦πππ + 28 πππ¦π π₯ 360 πππ¦π n = 5 + 0.67 + 0.078 π= πΉ 1+ππ 3,722.00 = 1+(0.15)(5.745) π = β±2,000.00 2. A man deposited P2,000 in a bank at the rate of 15% per annum from March 21, 1996 to October 25, 1997. Find the exact simple interest. Given: P = 2,000.00 i = 15% or 0.15 n = March 21, 1996 – October 25, 1997 (Mar 22, 1996 – Mar 21, 2997 = 365 days) March – 10 (excluding Mar 21) April – 30 May – 31 June – 30 July – 31 Aug – 31 Sept – 30 Oct - Required: P=? Solution: π = 5 + 8 ππππ‘βπ π₯ n = 5 + 0.67 + 0.078 1 π¦πππ 12 ππππ‘βπ 1 π¦πππ + 28 πππ¦π π₯ 360 πππ¦π π= πΉ 1+ππ 3,722.00 = 1+(0.15)(5.745) π = β±2,000.00 3. A bank charges 1.5% per month on a loan. Find the equivalent nominal rate of interest. Given: I = 1.5% or 0.015 n = 12 Required: j=? Solution: π π = π1 ππ π = (π)(π1) j = 0.015 x 12 j = 0.18 or 18% 4. A financing company charges 1.5% per month on a loan. Find the equivalent effective rate of interest. Given: i = 1.5% or 0.015 n = 12 Required: πΌπ = ? Solution: πΌπ = (1 + πΌ)π1 − 1 πΌπ = (1 + 0.015)12 − 1 πΌπ = 0.1964 ππ 19.56% 5. A nominal rate of 12% compounded monthly is equal to an effective rate of ___________. Given: j = 12% n = monthly or 12 Required: πΌπ = ? Solution: π πΌπ = (1 + π)π − 1 πΌπ = (1 + 12% 12 12 ) −1 = (1 + 0.01)12 − 1 = 1.1268 − 1 πΌπ = 0.1268 ππ 12.68% compounded monthly 6. Convert 16% compounded semi-annually to equivalent nominal rate which is compounded daily. Given: π (1 + 365)365 − 1 = (1 + π 365 (1 + 365) 365 √(1 + π ) 365 = (1 + 365 = π π = 0.000421793 2 0.16 2 365 √ 1+365 = 1.000421793 365 0.16 2 2 ) 729 625 ) −1 j = 0.153954 j = 15.39% 7. Find the accumulated amount of P1,000 after 5 years when deposited in a bank at the rate of 16% compounded monthly. Given: P = 1,000.00 j = 16% π1 = 12 n = 5 years or 60 months Required: F=? Solution: F = P (1 + π π π1 ) 0.16 60 ) 12 F = 1000 (1 + F = 1000 (2.2138) F = β±2, 213.81 8. How long in years will a certain sum of money to triple its amount when deposited at a rate of 12% compounded monthly. Given: F = 3X P=X j = 12% or 0.12 π1 = ππππ‘βππ¦ ππ 12 Required: n=? Solution: F = P (1 + 3π π =X (1+ π π π1 ) 0.12 π ) 12 π 3 = (1 + 0.01)π Ln 3 ln 1.01 =π ln 1.01 ln 1.01 n = 110.41 months or 9.2 years 9. How much should be deposited in a bank at a rate of 12% compounded continuously for 5 years if its accumulated amount is P9,110.60? Given: j = 12% or 0.12 n = 5 years F = β± 9,110.60 Required: P=? Solution: F = Pπ ππ πΉ P = π ππ = β± 9,110.60 π (0.12)(5) = β± 5,000.00 10. An effective rate of interest which is 12.75% is equivalent to what percent if compounded continuously. Given: ππ = 12.75% = 0.1275 Required: j=? Solution: ππ = π π − 1 π π = ππ + 1 = 0.1275 + 1 = 1.1275 j = 0.12 or 12% 11. How much is expected to be received by a man that makes a loan of P851.06 which is payable at once if the bank gave him a discount of 6%? Required: π π = ? Given: F = β±851.06 d = 6% or 0.06 Solution: d= π· πΉ = 0.06 = πΉ−π· πΉ 851.06−π 851.06 P = β±800.00 12. Find the cash price of a generator which was bought in installment basis that requires a down payment of P50,000 and payment of P30,000 after one year, P40,000 after two years, and a final payment of P76,374.34 after four years at a rate of 15% per annum. Given: i = 15% or 0.15 Using 4 as focal date: Solution: P(1 + π)4 = P(1 + π)3 + P(1 + π)2 + π X – 50,000(1.15)4 = 30,000(1.15)3 + 40,000(1.15)2 + 74,374.38 1.74900625X – 87,450,3125 = 45,626.25 + 52,900 + 74,374.38 1.74900625X = 174,900.59 + 87,450,3125 262,350.4025 1.74900625X = 1.74900625X X = β±150,000.00 13. A man made a loan of P100,000 at a rate of 15% per annum and promise to pay it according to the following manner, P30,000 at the end of first year, unknown payment at the end of second year and a final payment of P76,374.38 at the end of fourth year. Find the unknown payment made by the man. Given: i = 15% or 0.15 Using 4 as focal date Solution: P(1 + π)4 = P(1 + π)3 + P(1 + π)2 + π 100,000(1.15)4 = 30,000(1.15)3 + X(1.15)2 + 76,374.38 174,900.625 = 45,625.25 + 1.3225X + 76,374.38 1.3225X = 174,900.625 - 45,625.25 - 76,374.38 1.3225X = 52,899.62 1.3225 X = β±40,000.00 14. Find the present worth of the following payments, P5,000 after 1 year, P4,000 after 2 years, P8,000 after 4 years at the rate of 12% per annum. Given: i = 12 or 0.12 Using 4 focal date Solution: P(1 + π)4 = P(1 + π)3 + P(1 + π)2 + π X(1.12)4 = 5,000(1.12)3 + 4,000(1.12)2 + 8000 1.57351936X = 7,024.64 + 5,017.6 + 8000 20,042.24 1.57351936X = 1.57351936 X = β±12,737.21 15. Find the amount of the following payments at the end of 5th year. P3,000 at the end of 1st year, P4,500 at the end of 2nd year, and P6,000 at the end of 4th year if money worth 12% per annum. Given: i = 12% or 0.12 Using 5 focal date Solution: F = P(1 + π)4 + P(1 + π)3 + P(1 + i) F = 3,000(1.12)4 + 4,500(1.12)3 + 6,000(1.12) F = 4,720.55808 + 6,322.176 + 6,720 F = β±17,762.73