Question

A, B and C are three persons with ages 26 years, 27 years and 28 years respectively. In what ratio must they invest money at 10% p.a. compounded yearly so that each gets, same sum at the age of his retirement (i.e. at the age of 58 years).

Answer 1

Let the investments made be Rs. x, Rs. y and Rs. z respectively. Then :$x{(1+\frac{10}{100})}^{58-26}=y{(1+\frac{10}{100})}^{58-27}=z{(1+\frac{10}{100})}^{58-28}$
$\Rightarrow x\times {\left(\frac{11}{10}\right)}^{32}=y{\left(\frac{11}{10}\right)}^{31}=z{\left(\frac{11}{10}\right)}^{30}$
$\Rightarrow x\times {\left(\frac{11}{10}\right)}^2=y\times \frac{11}{10}=z$ Dividing each term by ${\left(\frac{11}{10}\right)}^{30}$
$\Rightarrow \frac{121}{100}x=\frac{11}{10}y=z\Rightarrow 121x=110y=100z=k\left(let\right)$
$\Rightarrow x=\frac{k}{121},y=\frac{k}{110}andz=\frac{k}{100}$
$\Rightarrow x:y:z=\frac{k}{121}:\frac{k}{110}:\frac{k}{100}$
$=\frac{k}{121}\times 12100:\frac{k}{110}\times 12100:\frac{k}{100}\times 12100$
$[L.C.M.of121,110and100=12100]$
$=100k:110k:121k$
$=100:110:121$