Question

A man bought two bikes for $Rs.1,85,600$. By selling one at a loss of $15$% and the other at a profit of $15$%, he found that the selling price of the first bike is more than the second bike by $Rs.10,000$. Find the $CP$ of each.

Answer 1

Let the first bike be $B_1$ and the second bike be $B_2$

Let $x$ and $y$ be the C.P of $B_1$ and $B_2$ respectively.

According to the question,$x+y=185600$ ......$\left(1\right)$

S.P of bike $B_1=$C.P of $B_1+15\%$ of C.P of $B_1$

$\Rightarrow$S.P of $B_1=x-\frac{15x}{100}=\frac{100x-15x}{100}=\frac{85x}{100}$

Also,S.P of bike $B_2=$C.P of $B_2+15\%$ of C.P of $B_2$

$\Rightarrow$S.P of $B_1=y+\frac{15y}{100}=\frac{100y+15y}{100}=\frac{115y}{100}$

Given that S.P $B_1=10000+$S.P of $B_2$

$\Rightarrow \frac{85x}{100}=10000+\frac{115y}{100}$

$\Rightarrow 85x=1000000+115y$

$\Rightarrow 85x-115y=1000000$

$\Rightarrow 17x-23y=200000$ ........$\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$ simultaneously, we get

$\left(1\right)\Rightarrow y=185600-x$

$\Rightarrow 17x-23(185600-x)=200000$

$\Rightarrow 17x+23x=200000+4268800‬=4468800‬$

$\Rightarrow 40x=4468800‬$

$\therefore x=\frac{4468800‬}{40}=111720‬$

Substituting $x=111720$ in $\left(1\right)$ we get

$y=185600-111720=73,880$‬

Hence the C.P of first bike is Rs.$111720$ and of second bike is Rs.$73880$