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標題:
F.4 Maths Quadratic Equation
發問:
我唔識做呢條數，幫幫忙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... 顯示更多 我唔識做呢條數，幫幫忙 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% 詳細寫出部驟，thx!!
最佳解答:
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%
其他解答:
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%DF665233EBFAA7EC
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