The difference between the C.I. and S.I. on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest % p.a. ?10%15%20%25% (2024)

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Standard VIII

Mathematics

Question

10%

20%

Let the rate is R% p.a.
[18000(1+R100)2−18000]−[18000×R×2100]=405
$18000[(100+\frac{({R}{^2}{10000})-1-(\frac{2R}{100})]=40518000[(100+R)^2-10000-200R]=405\frac{9 R^2}{5}=405R^2=\frac{405\times 5}{9}=225R=15$
Rate of interest=15%

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The difference between the C.I. and S.I. on an amount of Rs. 18,000 in 2 years was Rs. 405. What was the rate of interest % p.a. ?10%15%20%25% (2024)

Let the rate is R% p.a.[18000(1+R100)2−18000]−[18000×R×2100]=405$18000[(100+\frac{({R}{^2}{10000})-1-(\frac{2R}{100})]=40518000[(100+R)^2-10000-200R]=405\frac{9 R^2}{5}=405R^2=\frac{405\times 5}{9}=225R=15$Rate of interest=15%

Let the rate is R% p.a.
[18000(1+R100)2−18000]−[18000×R×2100]=405
$18000[(100+\frac{({R}{^2}{10000})-1-(\frac{2R}{100})]=40518000[(100+R)^2-10000-200R]=405\frac{9 R^2}{5}=405R^2=\frac{405\times 5}{9}=225R=15$
Rate of interest=15%