SSC CHSL 5 Mar 2018 Shift 3 Paper

Let cost price = Rs. 100x
Profit % = 25%
⇒ Selling price = $100\mathit{x}+\frac{25}{100}\times 100\mathit{x}=\mathit{R}\mathit{s}.125\mathit{x}$
If he reduces its selling price by Rs 240, then new Selling price = Rs. $\left(125\mathit{x}-240\right)$
Now, loss % = $\frac{\left(\mathit{C}.\mathit{P}.-\mathit{S}.\mathit{P}.\right)}{\mathit{C}.\mathit{P}.}\times 100$
⇒ $\frac{100\mathit{x}-\left(125\mathit{x}-240\right)}{100\mathit{x}}\times 100=5$
⇒ $240-25\mathit{x}=5\mathit{x}$
⇒ $5\mathit{x}+25\mathit{x}=30\mathit{x}=240$
⇒ $\mathit{x}=\frac{240}{30}=8$
∴ Initial selling price = $125\times 8=\mathit{R}\mathit{s}.1000$