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A sum of money becomes Rs. 11,880 after 4 years and Rs. 17,820 after 6 years on compound interest, if the interest is compounded annually. What is the half of the sum (in Rs.)?
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& A = P{\left( {1 + \frac{R}{{100}}} \right)^n} \cr
& 11880 = P{\left( {1 + \frac{R}{{100}}} \right)^4}.\,.\,.\,.\,.\,\left( {\text{i}} \right) \cr
& 17820 = P{\left( {1 + \frac{R}{{100}}} \right)^6}.\,.\,.\,.\,.\,\left( {{\text{ii}}} \right) \cr
& {\text{Equation }}\left( {{\text{ii}}} \right){\text{ divide by }}\left( {\text{i}} \right) \cr
& \frac{{17820}}{{11880}} = {\left( {1 + \frac{R}{{100}}} \right)^2} \cr
& \frac{3}{2} = {\left( {1 + \frac{R}{{100}}} \right)^2}.\,.\,.\,.\,.\,{\text{Put in equation}}\left( {\text{i}} \right) \cr
& 11880 = P{\left( {1 + \frac{R}{{100}}} \right)^4} \cr
& 11880 = P{\left( {1 + \frac{R}{{100}}} \right)^2}{\left( {1 + \frac{R}{{100}}} \right)^2} \cr
& 11880 = P \times \frac{3}{2} \times \frac{3}{2} \cr
& P = 11880 \times \frac{2}{3} \times \frac{2}{3} \cr
& P = 5280 \cr
& {\text{Half part of sum}} = \frac{1}{2} \times 5280 = 2640 \cr} $$
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