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標題:
A maths question
發問:
Tom borrowed $100 000 from a bank at 12% p.a., interest being compounded monthly. He repaid a fixed amount of money at the end of eack month. If he wanted to repay for not more than $200 000, how much should he pay each month? How much would he pay in total? Plz help me!! You can answer in Chinese or English!! 更新: Actually I am only F.2 Can someone answer simply? Thx~
最佳解答:
其他解答:
The FIXED amount is $1,261.99 158 x 1261.99 = $199,393.72
A maths question
發問:
Tom borrowed $100 000 from a bank at 12% p.a., interest being compounded monthly. He repaid a fixed amount of money at the end of eack month. If he wanted to repay for not more than $200 000, how much should he pay each month? How much would he pay in total? Plz help me!! You can answer in Chinese or English!! 更新: Actually I am only F.2 Can someone answer simply? Thx~
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
From the given, we can see that the monthly interest rate is 1%. Suppose that Tom repays $C every month, then: Amount he still owes after his first repay = 100000 x 1.01 - C Amount he still owes after his second repay = (100000 x 1.01 - C) x 1.01 - C = 100000 x (1.01)2 - 1.01C - C Amount he still owes after his third repay = (100000 x (1.01)2 - 1.01C - C) x 1.01 - C = 100000 x (1.01)3 - (1.01)2C - 1.01C - C So, we can see that after his nth repay, he still owes: 100000 x (1.01)n - (1.01)n-1C - (1.01)n-2C - ... - 1.01C - C = 100000 x (1.01)n - C[(1.01)n-1 - (1.01)n-2 - ... - 1.01 - 1] = 100000 x (1.01)n - [(1.01)n - 1]C/(1.01 - 1) = 100000 x (1.01)n - 100[(1.01)n - 1]C = 100000 x (1.01)n - 100 x (1.01)nC + 100C = 100 x (1.01)n x (1000 - C) + 100C If he pays back all the debt after the nth repay, then: 100 x (1.01)n x (1000 - C) + 100C = 0 100 x (1.01)n x (C - 1000) = 100C (1.01)n = C/(C - 1000) n = {log [C/(C - 1000)]}/log 1.01 By trial, we can find out that C should be 1260 so that the total amount that he pays will not be more than $200 000. Then n = log (1260/260)/log 1.01 = 158.6 Thus a total of 159 months he will need for repayment. For the first 158 months, he has paid 1260 x 158 = 199080 After that, amount he still owes for the last month is: 100 x (1.01)158 x (1000 - 1260) + 100 x 1260 = 757.9 Thus, total amount he would pay = 199080 + 757.9 = 199837.9 2009-07-12 11:50:39 補充: Making C as the subject, we have: C = 1000 x (1.01)^n / [(1.01)^n - 1] So when n = 158, C = 1261.99 and the total amount paid = 1261.99 x 158 = 199393.72 2009-07-12 11:50:50 補充: When n = 159, C = 1258.72 and the total amount paid = 1258.72 x 159 = 200136.6 Therefore he should choose to repay for 158 months. Thanks for the hints from nelsonywm2000.其他解答:
The FIXED amount is $1,261.99 158 x 1261.99 = $199,393.72
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