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(C) Let each instalment be ₹ $x$.The present value of the instalments must equal the borrowed amount.$\frac{x}{(1 + \frac{4}{100})^1} + \frac{x}{(1 +

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$2028$

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(C) Let each instalment be ₹ $x$.The present value of the instalments must equal the borrowed amount.$\frac{x}{(1 + \frac{4}{100})^1} + \frac{x}{(1 + \frac{4}{100})^2} = 3825$$\frac{x}{(26/25)} + \frac{x}{(26/25)^2} = 3825$$\frac{25x}{26} + \frac{625x}{676} = 3825$Multiply by $676$ to clear the denominators:$25x 	imes 26 + 625x = 3825 	imes 676$$650x + 625x = 3825 	imes 676$$1275x = 3825 	imes 676$$x = \frac{3825 	imes 676}{1275}$$x = 3 	imes 676 = ₹ 2028$Thus,each instalment is ₹ $2028$.

$A$ man borrows ₹ $3825$ to be paid back with compound interest at the rate of $4 \%$ per annum by the end of $2$ years in two equal yearly instalments. How much will be each instalment (in ₹)?

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