A sum of Rs. 1000 after 3 , text{years} at co | vedclass

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Question

(A) The amount of $Rs. 1000$ at $10 \% \, p.a.$ after $3 \, 	ext{years}$ becomes:$\Rightarrow A = P(1 + \frac{R}{100})^n = 1000(1 + \frac{10}{100})^3

Vedclass · Accepted Answer

$1.7$

Vedclass · Answer

(A) The amount of $Rs. 1000$ at $10 \% \, p.a.$ after $3 \, 	ext{years}$ becomes:$\Rightarrow A = P(1 + \frac{R}{100})^n = 1000(1 + \frac{10}{100})^3 = 1000(1.1)^3 = 1000 	imes 1.331 = Rs. 1331$.Now,$Rs. 1728$ depreciates at rate $R_d \%$ per annum for $3 \, 	ext{years}$ to become $Rs. 1331$:$\Rightarrow A = P(1 - \frac{R_d}{100})^n$$\Rightarrow 1331 = 1728(1 - \frac{R_d}{100})^3$$\Rightarrow (1 - \frac{R_d}{100})^3 = \frac{1331}{1728} = (\frac{11}{12})^3$$\Rightarrow 1 - \frac{R_d}{100} = \frac{11}{12}$$\Rightarrow \frac{R_d}{100} = 1 - \frac{11}{12} = \frac{1}{12}$$\Rightarrow R_d = \frac{100}{12} = 8.33 \%$.The difference between the rates is $10 \% - 8.33 \% = 1.67 \%$,which is approximately $1.7 \%$.

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