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(C) Let the principal sum be $P$. The rate of interest $R = 5 \%$ per annum and time $T = 3$ years.Simple Interest $(SI)$ is given by: $SI = \frac{P \

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$24000$

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(C) Let the principal sum be $P$. The rate of interest $R = 5 \%$ per annum and time $T = 3$ years.Simple Interest $(SI)$ is given by: $SI = \frac{P 	imes R 	imes T}{100} = \frac{P 	imes 5 	imes 3}{100} = \frac{15P}{100} = \frac{3P}{20}$.Compound Interest $(CI)$ is given by: $CI = P \left[ (1 + \frac{R}{100})^T - 1 ight] = P \left[ (1 + \frac{5}{100})^3 - 1 ight] = P \left[ (\frac{21}{20})^3 - 1 ight] = P \left[ \frac{9261}{8000} - 1 ight] = P \left[ \frac{1261}{8000} ight]$.According to the problem,$CI - SI = 183$.Substituting the values: $\frac{1261P}{8000} - \frac{3P}{20} = 183$.To subtract,make the denominators equal: $\frac{1261P}{8000} - \frac{1200P}{8000} = 183$.$\frac{61P}{8000} = 183$.$P = \frac{183 	imes 8000}{61} = 3 	imes 8000 = 24000$.Thus,the sum is $Rs. 24000$.

If the compound interest on a certain sum for $3$ years at $5 \%$ p.a. exceeds the simple interest on the same sum for the same time and at the same rate by $Rs. 183$,find the sum (in $Rs.$).

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