Question
A sum of money lent out at compound interest increases in value by 50% in 5 years. A person wants to lend three different sums x, y and z for 10, 15 and 20 years respectively at the above rate in such a way that he gets back equal sums at the end of their respective periods. The ratio x : y : z is = ?
Answer: Option C
$$\eqalign{
& P{\left( {1 + \frac{R}{{100}}} \right)^5} = 150\% \,{\text{of }}P = \frac{3}{2}P \cr
& \Rightarrow {\left( {1 + \frac{R}{{100}}} \right)^5} = \frac{3}{2} \cr} $$
$$x{\left( {1 + \frac{R}{{100}}} \right)^{10}} = y{\left( {1 + \frac{R}{{100}}} \right)^{15}} = $$ $$z{\left( {1 + \frac{R}{{100}}} \right)^{20}}$$
$$ \Rightarrow x{\left\{ {{{\left( {1 + \frac{R}{{100}}} \right)}^5}} \right\}^2} = $$ $$y{\left\{ {{{\left( {1 + \frac{R}{{100}}} \right)}^5}} \right\}^3} = $$ $$z{\left\{ {{{\left( {1 + \frac{R}{{100}}} \right)}^5}} \right\}^4}$$
$$\eqalign{
& \Rightarrow x \times {\left( {\frac{3}{2}} \right)^2} = y \times {\left( {\frac{3}{2}} \right)^3} = z \times {\left( {\frac{3}{2}} \right)^4} \cr
& \Rightarrow \frac{{9x}}{4} = \frac{{27y}}{8} = \frac{{81z}}{{16}} = k({\text{say}}) \cr
& \Rightarrow x = \frac{{4k}}{9},y = \frac{{8k}}{{27}},z = \frac{{16k}}{{81}} \cr
& \Rightarrow x:y:z = \frac{{4k}}{9}:\frac{{8k}}{{27}}:\frac{{16k}}{{81}} \cr
& \Rightarrow x:y:z = 36:24:16 \cr
& \Rightarrow x:y:z = 9:6:4 \cr} $$
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$$\eqalign{
& P{\left( {1 + \frac{R}{{100}}} \right)^5} = 150\% \,{\text{of }}P = \frac{3}{2}P \cr
& \Rightarrow {\left( {1 + \frac{R}{{100}}} \right)^5} = \frac{3}{2} \cr} $$
$$x{\left( {1 + \frac{R}{{100}}} \right)^{10}} = y{\left( {1 + \frac{R}{{100}}} \right)^{15}} = $$ $$z{\left( {1 + \frac{R}{{100}}} \right)^{20}}$$
$$ \Rightarrow x{\left\{ {{{\left( {1 + \frac{R}{{100}}} \right)}^5}} \right\}^2} = $$ $$y{\left\{ {{{\left( {1 + \frac{R}{{100}}} \right)}^5}} \right\}^3} = $$ $$z{\left\{ {{{\left( {1 + \frac{R}{{100}}} \right)}^5}} \right\}^4}$$
$$\eqalign{
& \Rightarrow x \times {\left( {\frac{3}{2}} \right)^2} = y \times {\left( {\frac{3}{2}} \right)^3} = z \times {\left( {\frac{3}{2}} \right)^4} \cr
& \Rightarrow \frac{{9x}}{4} = \frac{{27y}}{8} = \frac{{81z}}{{16}} = k({\text{say}}) \cr
& \Rightarrow x = \frac{{4k}}{9},y = \frac{{8k}}{{27}},z = \frac{{16k}}{{81}} \cr
& \Rightarrow x:y:z = \frac{{4k}}{9}:\frac{{8k}}{{27}}:\frac{{16k}}{{81}} \cr
& \Rightarrow x:y:z = 36:24:16 \cr
& \Rightarrow x:y:z = 9:6:4 \cr} $$
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