In the beginning of 2000,Ken borrowed $40000 from a bank at an interest rate of 16% p.a. compounded quarterly.
(a) How much compound interest did Ken owe the bank in the beginning of 2004?
(b) If Ken repaid $10000 at the end of each year,how much did he owe the bank in the beginning of 2004?
(Give your answer correct to nearest $1)
Percentage Taxation
2013-08-21 5:47 am
回答 (4)
2013-08-22 1:46 am
✔ 最佳答案
See email.the answer should be $32919,and the steps are alsox included in the email.
2013-08-21 8:32 am
Hehe,感你仲咩唔答Cool?
2013-08-21 15:20:10 補充：
Well,actually,I love changing names,do you love it also?
2013-08-21 15:21:28 補充：
And I am using Cool's name ,(the owner) to ask other questions.
2013-08-21 15:20:10 補充：
Well,actually,I love changing names,do you love it also?
2013-08-21 15:21:28 補充：
And I am using Cool's name ,(the owner) to ask other questions.
2013-08-21 7:33 am
好野呀～
呢種係典型的利率應用題～
如果你將來會讀關於 financial mathematics，那麼 part b 的計算就會用到我們叫的 annuity 的概念。
而分期還款（如供樓、供車）的時間表可以做出一個 amortization schedule。
呢一刻你未必有興趣知呢ｄ野，但係將來或者有機會會再見到～
：）
2013-08-21 13:08:58 補充：
望月一都 已經回答你啦~ ^^
BTW, 明明你就係 Cool, 唔好當大家唔知~ =P
呢種係典型的利率應用題～
如果你將來會讀關於 financial mathematics，那麼 part b 的計算就會用到我們叫的 annuity 的概念。
而分期還款（如供樓、供車）的時間表可以做出一個 amortization schedule。
呢一刻你未必有興趣知呢ｄ野，但係將來或者有機會會再見到～
：）
2013-08-21 13:08:58 補充：
望月一都 已經回答你啦~ ^^
BTW, 明明你就係 Cool, 唔好當大家唔知~ =P
2013-08-21 6:03 am
a.
Amount of compound interest owed:
40000[1+(0.16/4)]^16 - 40000 = $34919, correct to the nearest $1
b.
Amount owed: (all correct to the nearest $1)
At the beginning of 2000: $40000
At the beginning of 2001: 40000[1+(0.16/4)]^4 - 10000 = $36794
At the beginning of 2002: 36794[1+(0.16/4)]^4 - 10000 = $33044
At the beginning of 2003: 33044[1+(0.16/4)]^4 - 10000 = $28657
At the beginning of 2004: 28657[1+(0.16/4)]^4 - 10000 = $23525
2013-08-20 22:05:01 補充：
試下畫條 timeline 幫自己計算
Amount of compound interest owed:
40000[1+(0.16/4)]^16 - 40000 = $34919, correct to the nearest $1
b.
Amount owed: (all correct to the nearest $1)
At the beginning of 2000: $40000
At the beginning of 2001: 40000[1+(0.16/4)]^4 - 10000 = $36794
At the beginning of 2002: 36794[1+(0.16/4)]^4 - 10000 = $33044
At the beginning of 2003: 33044[1+(0.16/4)]^4 - 10000 = $28657
At the beginning of 2004: 28657[1+(0.16/4)]^4 - 10000 = $23525
2013-08-20 22:05:01 補充：
試下畫條 timeline 幫自己計算
收錄日期: 2021-04-13 19:38:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130820000051KK00385