Mathematics of Investment
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MATHEMATICS ofINVESTMENTBY: DIGNA PUZON ANTONIOSET A(Simple interest)How much is needed to settle a loan of ₱7,500 at 9 2/5 % due in 2 years and 6 months?Let P = ₱ 7.500 r = 9 2/5% or 9. 4% or 0.094 t = 2.5 yearsSolution:Step 1. I = prt = (7,500) (.094) (2.5) = ₱1762.50Step 2. F= P + I = 7,500 + 1762.50 = ₱ 9262.50 (amount needed to settle loan in its due date)Find the interest on ₱ 9,000 at 10 ½ % from April 12 to November 12, 2016.Let: P = ₱ 9,000 r = 10.5 % or 0.105 t = from April 12 to November 12, 2016 I=? Solution: Step 1. 2016 11 12 2016 _ 4 12 7 0 (equivalent to 7 mos.) Step 2. I = prt = (9,000) (0.105) ( )[pic 1] = ₱551.25 (interest gained within 7 mos.)How long will it take ₱ 12.000 to accumulate to ₱ 17,280 at 11% simple interest rate?Let: F = ₱ 17,280 r = 11% or 0.11 P = ₱ 12,000 t =?Solution:Step 1. Find interest. I = F – P = ₱ 17,280 – ₱ 12,000 = ₱ 5280 (accumulated interest)Step 2. Find time. t = (derivation of formula)[pic 2] = [pic 3] = [pic 4]= 4 yrs. (time needed to accumulate given amount )What principal will accumulate to ₱ 12,812.50 on the third month if the interest rate is 10 %?Let: F = ₱ 12,812.50 r = 10% or 0.10 t = 3/12 or 0.25
Solution: F = P (1 + r t) P = (derived formula from above)[pic 5] = [pic 6] = [pic 7] = ₱ 12,500 (principal that will accumulate)Compute the payable amount for ₱ 30,000 loan at 10.5% after 90 days.For ordinary interestLet: P = ₱ 30,000 r = 10.5% or 0.105 t = [pic 8]Solution: [pic 9] = (30,000) (1 + (0.105) ()[pic 10] = (30,000) (1.0625) = ₱ 30,787.50 (payable amount after 90 days)For exact interestLet: P = ₱ 30,000 r = 10.5% or 0.105 t =[pic 11]Solution: [pic 12] = (30,000) (1 + (0.105) ()[pic 13] = (30,000) (1.025890411) = ₱ 30,776.71 (payable amount after 90 days)Determine how much should be paid for a ₱15,000 loan after 2 years and 6 months at 15.5% simple interest.Let: P = ₱15,000 r = 15.5% or 0.155 t = 2.5 yrsSolution: [pic 14] = (15,000) (1 + (0.155) (2.5))