標題:
F.3 MATHS (COMPOUNDED INTEREST)
發問:
susan borrows $10000 from a fiance company at an interest rate of 18%P.A companded monthly at the beginning of a month. If she repays $3000 at the each month, how many reapyments are needed to settle the loan?除左每個月計既方法之外,有冇d公式系一條過而唔洗每個月計?有 or冇都出句聲同埋比下你地計既方法比我睇.thzps::A = PR(1+R)n / [(1+R)n -... 顯示更多 susan borrows $10000 from a fiance company at an interest rate of 18%P.A companded monthly at the beginning of a month. If she repays $3000 at the each month, how many reapyments are needed to settle the loan? 除左每個月計既方法之外, 有冇d公式系一條過而唔洗每個月計? 有 or冇都出句聲 同埋比下你地計既方法比我睇.thz ps::A = PR(1+R)n / [(1+R)n – 1]<--呢條得唔得?
最佳解答:
Let amount borrowed = P. Monthly interest rate = r. Monthly repayment = p. Amount owe after 1 month = P( 1+r) - p. Amount owe after 2 months = [P(1+r) - p](1+r) - p = P(1+r)^2 - p(1+r) - p. Amount owe after 3 months = [P(1+r)^2 - p(1+r) - p](1+r) - p = P(1+r)^3 - p(1+r)^2 - p(1+r) - p. Amount owe after n months = P(1+r)^n - p(1+r)^(n-1) - p(1+r)^(n -2) - p(1+r)^(n-3) - ............. - p(1+r)^2 - p(1+r) - p. = P(1+r)^n - p[(1+r)^(n-1) + (1+r)^(n-2) + ...... + (1+r)^2 + (1+r) + 1] = P(1+r)^n - p[(1+r)^n - 1]/[(1+r) - 1]. After paying all the debt, amount = 0. That is P(1+r)^n = p[(1+r)^n - 1]/r. This is the formula to be used. Now P = 10000, r = 0.18/12 = 0.015, p = 3000. 10000(1.015)^n = 3000[1.015^n - 1]/0.015 0.05(1.015)^n = 1.015^n - 1 0.95(1.015)^n = 1 log 0.95 + n log 1.015 = 0 n = - log 0.95/log 1.015 = 0.022276/0.006466 = 3.445 months. 2008-10-07 10:20:01 補充： The formula stated in the question can be used to calculate the monthly repayment, that is A = p. It can also be used to calculate n or r or P.
其他解答:
F.3 MATHS (COMPOUNDED INTEREST)
發問:
susan borrows $10000 from a fiance company at an interest rate of 18%P.A companded monthly at the beginning of a month. If she repays $3000 at the each month, how many reapyments are needed to settle the loan?除左每個月計既方法之外,有冇d公式系一條過而唔洗每個月計?有 or冇都出句聲同埋比下你地計既方法比我睇.thzps::A = PR(1+R)n / [(1+R)n -... 顯示更多 susan borrows $10000 from a fiance company at an interest rate of 18%P.A companded monthly at the beginning of a month. If she repays $3000 at the each month, how many reapyments are needed to settle the loan? 除左每個月計既方法之外, 有冇d公式系一條過而唔洗每個月計? 有 or冇都出句聲 同埋比下你地計既方法比我睇.thz ps::A = PR(1+R)n / [(1+R)n – 1]<--呢條得唔得?
最佳解答:
Let amount borrowed = P. Monthly interest rate = r. Monthly repayment = p. Amount owe after 1 month = P( 1+r) - p. Amount owe after 2 months = [P(1+r) - p](1+r) - p = P(1+r)^2 - p(1+r) - p. Amount owe after 3 months = [P(1+r)^2 - p(1+r) - p](1+r) - p = P(1+r)^3 - p(1+r)^2 - p(1+r) - p. Amount owe after n months = P(1+r)^n - p(1+r)^(n-1) - p(1+r)^(n -2) - p(1+r)^(n-3) - ............. - p(1+r)^2 - p(1+r) - p. = P(1+r)^n - p[(1+r)^(n-1) + (1+r)^(n-2) + ...... + (1+r)^2 + (1+r) + 1] = P(1+r)^n - p[(1+r)^n - 1]/[(1+r) - 1]. After paying all the debt, amount = 0. That is P(1+r)^n = p[(1+r)^n - 1]/r. This is the formula to be used. Now P = 10000, r = 0.18/12 = 0.015, p = 3000. 10000(1.015)^n = 3000[1.015^n - 1]/0.015 0.05(1.015)^n = 1.015^n - 1 0.95(1.015)^n = 1 log 0.95 + n log 1.015 = 0 n = - log 0.95/log 1.015 = 0.022276/0.006466 = 3.445 months. 2008-10-07 10:20:01 補充： The formula stated in the question can be used to calculate the monthly repayment, that is A = p. It can also be used to calculate n or r or P.
其他解答:
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