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Question 65
The compound interest on ₹40,000 at 6% per annum is ₹4,944. What is the period (in years) for which the amount is invested?
Solution
$$P((1+\frac{R}{100}))^{n}-P$$=4944
$$40000((1+\frac{R}{100}))^{n}-1)$$=4944
$$((1+\frac{6}{100}))^{n}-1)$$=4944/40000
$$(\frac{53}{50})^{n}-1)$$=4944/40000
$$(\frac{53}{50})^{n}$$=44944/40000
$$(\frac{53}{50})^{n}$$=$$(\frac{212}{200})^{2}$$
$$(\frac{53}{50})^{n}$$=$$(\frac{53}{50})^{2}$$
n=2 years
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