A) 4620
B) 4680
C) 4480
D) 4420
Correct Answer: D
Solution :
Let the sum be Rs. P. then \[P{{\left( 1+\frac{R}{100} \right)}^{3}}=6630.....(i)\] and \[{{\left( 1+\frac{R}{100} \right)}^{6}}=9945.....(ii)\] On dividing, we get \[{{\left( 1+\frac{R}{100} \right)}^{3}}=\frac{9945}{6630}=\frac{3}{2}\] Substituting this value in (i), we get: \[P\times \frac{9945}{6630}=\frac{3}{2}=6630\] or P \[=\left( 6630\frac{2}{3} \right)=4420\] Hence, the sum is Rs.4420.
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