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?
F.3 MATHS (COMPOUNDED INTEREST)
2008-10-05 1:45 am
回答 (1)
2008-10-07 6:15 pm
✔ 最佳答案
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.
收錄日期: 2021-04-25 22:40:32
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