Q. A sum of Rs. 1200 after 3 years at compound interest becomes a certain amount which is equal to the amount that results from a 2 year depreciation of Rs. 1887. What is the difference between the rates of interest and depreciation? (Rate of CI is 10% p.a.)

1.5%

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Solution

The correct option is B 2%
Sum after 3 years of investment $= 1200 (1 + \frac{10}{100})^{2} = 1597$
For depreciation
Amount $= P (1 - \frac{R}{100})^{T} 1597 = 1887 (1 - \frac{D}{100})^{2} \frac{D}{100} = d (1 - d)^{2} = .8463 d = 1 - .92 = 0.08 D = 0.08 \times 100 = 8 %$
Difference = 2%

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Q. A sum of Rs. 1200 after 3 years at compound interest becomes a certain amount which is equal to the amount that results from a 2 year depreciation of Rs. 1887. What is the difference between the rates of interest and depreciation? (Rate of CI is 10% p.a.)

The correct option is B 2%Sum after 3 years of investment =1200(1+10100)2=1597 For depreciation Amount =P(1−R100)T1597=1887(1−D100)2D100=d(1−d)2=.8463d=1−.92=0.08D=0.08×100=8% Difference = 2%

The correct option is B 2%
Sum after 3 years of investment $= 1200 (1 + \frac{10}{100})^{2} = 1597$
For depreciation
Amount $= P (1 - \frac{R}{100})^{T} 1597 = 1887 (1 - \frac{D}{100})^{2} \frac{D}{100} = d (1 - d)^{2} = .8463 d = 1 - .92 = 0.08 D = 0.08 \times 100 = 8 %$
Difference = 2%