NIFT UG 2013 Question Paper with solutions

Let number of visitor is nNet collection = Rs. n
When fee is reduced by 20%
New entry fee = Rs. 0.8.
Again, let number of visitors increased by m.
So, net collection $=\left(\mathit{m}+\mathit{n}\right)\times 0.8$
Increase in collection $\left(\mathit{m}+\mathit{n}\right)0.8\mathit{–}\mathit{n}=0.8\mathit{m}\mathit{–}0.2\mathit{n}$
Percentage increase $\Rightarrow \frac{0.8\mathit{m}-0.2\mathit{n}}{\mathit{n}}\times 100=40$
$80\mathit{m}-20\mathit{n}=40\mathit{n}$
$80\mathit{m}=60\mathit{n}$
Percentage increase in number of visitors $\frac{\mathit{m}}{\mathit{n}}\times 100=\frac{60}{80}\times 100=75\%$