₹ 6,100 was partly invested in Scheme A at 10 | vedclass

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(C) Let the amount invested in Scheme $A$ be $₹ x$.Then,the amount invested in Scheme $B$ is $₹ (6100 - x)$.For Scheme $A$ (Compound Interest):$CI = P

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$4000$

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(C) Let the amount invested in Scheme $A$ be $₹ x$.Then,the amount invested in Scheme $B$ is $₹ (6100 - x)$.For Scheme $A$ (Compound Interest):$CI = P \left[ (1 + \frac{R}{100})^T - 1 ight] = x \left[ (1 + \frac{10}{100})^2 - 1 ight] = x \left[ (1.1)^2 - 1 ight] = x (1.21 - 1) = 0.21x$.For Scheme $B$ (Simple Interest):$SI = \frac{P 	imes R 	imes T}{100} = \frac{(6100 - x) 	imes 10 	imes 4}{100} = \frac{40(6100 - x)}{100} = 0.4(6100 - x)$.Given that both schemes pay equal interest:$0.21x = 0.4(6100 - x)$.$0.21x = 2440 - 0.4x$.$0.21x + 0.4x = 2440$.$0.61x = 2440$.$x = \frac{2440}{0.61} = 4000$.Therefore,the amount invested in Scheme $A$ is $₹ 4000$.

₹ $6,100$ was partly invested in Scheme $A$ at $10 \%$ p.a. compound interest (compounded annually) for $2$ years and partly in Scheme $B$ at $10 \%$ p.a. simple interest for $4$ years. Both the schemes pay equal interests. How much was invested (in ₹) in Scheme $A$?

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