SSC CGL 8 Aug 2017 Shift 1 Paper

Let foe price of foe pen= ₹ x
Pens that can be purchased for $\mathrm{₹}100=\frac{100}{\mathit{x}}$
and foe new price of foe pen $=\mathit{x}\times \frac{80}{100}=\frac{4\mathit{x}}{5}$
Pens that can be purchased for ₹100 with reducted cost $=\frac{100\times 5}{4\mathit{x}}=\frac{125}{\mathit{x}}$
According to foe question,
$\frac{125}{\mathit{x}}-\frac{100}{\mathit{x}}=10$
⇒ $\frac{25}{\mathit{x}}=10\therefore \mathit{x}=2.5$
Hence, foe new price of the pen $=\frac{4}{5}\times 2.5=\mathrm{₹}2$