Mr. Thomas invested an amount of Rs. 13,900 d | vedclass

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(A) Let the amount invested in Scheme $A$ be $Rs. x$ at $14 \%$ $p.a.$Then,the amount invested in Scheme $B$ is $Rs. (13900 - x)$ at $11 \%$ $p.a.$The

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$6400$

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(A) Let the amount invested in Scheme $A$ be $Rs. x$ at $14 \%$ $p.a.$Then,the amount invested in Scheme $B$ is $Rs. (13900 - x)$ at $11 \%$ $p.a.$The formula for simple interest is $SI = \frac{P 	imes R 	imes T}{100}$.According to the problem,the total interest after $2$ $years$ is $Rs. 3508$.$\frac{x 	imes 14 	imes 2}{100} + \frac{(13900 - x) 	imes 11 	imes 2}{100} = 3508$Multiply the entire equation by $100$:$28x + 22(13900 - x) = 350800$$28x + 305800 - 22x = 350800$$6x = 350800 - 305800$$6x = 45000$$x = 7500$So,the amount invested in Scheme $A$ is $Rs. 7500$.The amount invested in Scheme $B$ is $13900 - 7500 = Rs. 6400$.

Mr. Thomas invested an amount of $Rs. 13,900$ divided in two different schemes $A$ and $B$ at the simple interest rate of $14 \%$ $p.a.$ and $11 \%$ $p.a.$ respectively. If the total amount of simple interest earned in $2$ $years$ be $Rs. 3508$,what was the amount (in $Rs.$) invested in Scheme $B$?

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