Two equal sums were lent at simple interest f | vedclass

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(B) Let the principal amount be $P$ and the rate of interest for the first case be $R \%$.For the first case: Time $T_1 = 4$ years,Rate $R_1 = R \%$,A

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$9$

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(B) Let the principal amount be $P$ and the rate of interest for the first case be $R \%$.For the first case: Time $T_1 = 4$ years,Rate $R_1 = R \%$,Amount $A_1 = 1088$.$A_1 = P(1 + \frac{R_1 T_1}{100}) \Rightarrow 1088 = P(1 + \frac{4R}{100})$.For the second case: Time $T_2 = 3$ years,Rate $R_2 = (R + 3) \%$,Amount $A_2 = 1088$.$A_2 = P(1 + \frac{R_2 T_2}{100}) \Rightarrow 1088 = P(1 + \frac{3(R + 3)}{100})$.Since both amounts are equal,we equate the expressions:$P(1 + \frac{4R}{100}) = P(1 + \frac{3(R + 3)}{100})$.Dividing both sides by $P$ (assuming $P 
eq 0$):$1 + \frac{4R}{100} = 1 + \frac{3R + 9}{100}$.Subtracting $1$ from both sides:$\frac{4R}{100} = \frac{3R + 9}{100}$.Multiplying by $100$:$4R = 3R + 9$.Solving for $R$:$R = 9$.Thus,the rate of interest for the former case is $9 \%$.

Two equal sums were lent at simple interest for $4$ years and for $3$ years respectively. The rate of interest in the latter case was $3 \%$ higher than that of the former case,but the amount in each was $Rs. 1088$. Find the rate of interest (in $\%$) of the former case?

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