Jharkhand PSC Prelim Exam 2006 Solved Paper

Suppose selling price of each $2-$ scooter $=100$
$25\%$ profit on 1 st scooter
$\therefore \mathrm{C}.\mathrm{P}$ of $1\mathrm{st}$ scooter $=\mathit{x}$
$\therefore$ Profit $=\mathit{x}\times \frac{25}{100}=\frac{\mathit{x}}{4}$
$\therefore \mathrm{S}.\mathrm{P}=\mathit{x}+\frac{\mathit{x}}{4}=\frac{5\mathit{x}}{4}$
According to question,
$\frac{5\mathit{x}}{4}=100\Rightarrow \mathit{x}=80-\mathrm{C}.\mathrm{P}$ of 1st scooter.
$25\%$ loss of 2 nd scooter
Let
$\therefore \mathrm{Loss}=\mathit{y}\times \frac{25}{100}=\frac{\mathit{y}}{4}$
$\therefore \mathrm{S}.\mathrm{P}=\mathit{y}-\frac{4}{4}=\frac{3\mathit{y}}{4}$
According to question,
$\frac{3\mathit{y}}{4}=100\Rightarrow \mathit{y}$
of $2\mathit{n}\mathit{d}$ scooter
C.P of both $=80+133.33$
$=213.33$
S.P of both $=100+100=200$
$\therefore \mathrm{loss}=213.33-200$
=13.33