F5 math problems

2007-09-23 5:33 am

1.An insurance company promotes a ten-year life insurance with zero cost. In the scheme,$10000 is deposited by the insurer on the same insurance date of the year for ten years.After ten years, a total of $100000 is returned to the insurer.Assume that the interest rate is 4%p.a.,which is compounded annually.Find the actual cost to the nearest dollar of the life insurance policy.
(Ans:$24864)

回答 (1)

[匿名]

2007-09-23 6:00 am

✔ 最佳答案

At the beginning of the 1st year
Deposit=$10000
At the end of 1st year
Deposit=$10000*(1+4%)=10400
At the beginning of the 2nd year
Deposit=$10400+$10000=$20400
At the end of 2nd year
Deposit=$20400*(1+4%)=$21216

So, according to the pattern,
At the end of 3rd year,Deposit=$(21216+10000)*(1+4%)=$32465
At the end of 4th year,Deposit=$(32465+10000)*(1+4%)=$44163
At the end of 5th year,Deposit=$(44163+10000)*(1+4%)=$56330
At the end of 6th year,Deposit=$(56330+10000)*(1+4%)=$68983
At the end of 7th year,Deposit=$(68983+10000)*(1+4%)=$82142
At the end of 8th year,Deposit=$(82142+10000)*(1+4%)=$95288
At the end of 9th year,Deposit=$(95288+10000)*(1+4%)=$110061
At the end of 10th year,Deposit=$(110061+10000)*(1+4%)=$124864

So,
the actual cost=Total Deposit after 10 years - $100000
=$124864-$100000
=$24864

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