Management Aptitude Test Sep 2020 Paper

Let the price paid per trip by a passenger be $\mathit{\eta }\mathit{x}$. and auto rickshaw consumes y $\mathit{L}$ petrol.
Total expenditure in a trip $=30\times \mathit{y}=\mathit{₹}30\mathit{y}$
Total earning in a trip $=3\times \mathit{x}=$ ₹ $3\mathit{x}$
According to the question,

$\Rightarrow 3\mathit{x}-30\mathit{y}=\frac{20\times 30}{100}\mathit{y}\Rightarrow 3\mathit{x}-30\mathit{y}=6\mathit{y}$
$\Rightarrow 3\mathit{x}=36\mathit{y}\therefore \mathit{x}=12\mathit{y}$
Now, the price of petrol are reduced to ₹ 24 and passenger taken by auto-rickshaw are four.
$\therefore$ Total expenditure in a trip $=24\times \mathit{y}=$ ₹$24\mathit{y}$
Total earning in a trip $=4\times \mathit{x}=\mathit{₹}4\mathit{x}$
Profit per cent

$=\frac{4\mathit{x}-24\mathit{y}}{24\mathit{y}}\times 100\%$
$=\frac{4\times 12\mathit{y}-24\mathit{y}}{24\mathit{y}}\times 100\left[\because \mathit{x}=12\mathit{y}\right]$
$=\frac{48\mathit{y}-24\mathit{y}}{24\mathit{y}}\times 100$
$=\frac{24\mathit{y}}{24\mathit{y}}\times 100=100\%$