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A bank offers a large sum fixed deposit plan of one-year period. The interest is calculated and compounded half-yearly. The annual interest rate for each 6-month peroid increases gradually. Moreover, the difference of the annual interest rates of two consecutive 6-month periods is 1%. Miss Wong joins the plan by investing $1 000 000. She receives an amount of $1 035 300 after one year. Find the annual interest rate for the first half year.
Ans: 3%
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F.4 Maths Quadratic Equation
2009-11-08 5:18 am
回答 (2)
2009-11-08 5:33 am
✔ 最佳答案
Let the annual interest rate for the first half year be r% (r/100)Then the annual interest rate for the second half year is (r + 1)/100
The total amount after 1 year
= $1,000,000 * (1 + r/100/2) * [1 + (r + 1)/100/2]
= $1,035,300
So (1 + r/200)[1 + (r + 1)/200] = 1.0353
(200 + r)(200 + r + 1) = 41412
(200 + r)(201 + r) = 41412
40200 + 401r + r^2 - 41412 = 0
r^2 + 401r - 1212 = 0
(r - 3)(r + 404) = 0
r = 3 or r = -404 (rejected)
Therefore the annual interest rate for the first half year is 3%
2009-11-08 5:55 am
本金=1 000 000
首6個月利率=n/2
尾6個月利率=[n+(0.01)]/2
[1000000×(1+n/2)]×{1+[n+(0.01)]/2}=1035300
(1000000＋500000n)×(0.5n+1.005)=1035300
500000n+250000n^2+1005000+502500n=1035300
250000^2+1002500n+1005000=1035300
250000n^2+1002500n-30300=0
2500n^2+10025n-303=0
用quadratic formula,n={-10025±√[10025^2-4×2500×(-303)]}÷2×10000
=[-10025±√(100500625+3030000)]÷20000
=(-10025±10175)÷20000
=150÷20000 or -20200÷20000
=0.03 or -4.04(rej.)
∴n=0.03
∴the annual interest rate for the first half year=3%
首6個月利率=n/2
尾6個月利率=[n+(0.01)]/2
[1000000×(1+n/2)]×{1+[n+(0.01)]/2}=1035300
(1000000＋500000n)×(0.5n+1.005)=1035300
500000n+250000n^2+1005000+502500n=1035300
250000^2+1002500n+1005000=1035300
250000n^2+1002500n-30300=0
2500n^2+10025n-303=0
用quadratic formula,n={-10025±√[10025^2-4×2500×(-303)]}÷2×10000
=[-10025±√(100500625+3030000)]÷20000
=(-10025±10175)÷20000
=150÷20000 or -20200÷20000
=0.03 or -4.04(rej.)
∴n=0.03
∴the annual interest rate for the first half year=3%
收錄日期: 2021-04-23 23:21:19
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