SSC CGL 16 Aug 2015 Shift 2 Paper

Let the radius of swimming pool be r metre.
Breadth of shaded part = 4 metre
∴ OB = (r + 4) metre
According to the question,
π × $\mathit{O}{\mathit{B}}^2$ - π × $\mathit{O}{\mathit{A}}^2$ = $\frac{11}{25}$ π × $\mathit{O}{\mathit{A}}^2$
⇒ ${\left(\mathit{r}+4\right)}^2-{\mathit{r}}^2$ = $\frac{11}{25}{\mathit{r}}^2$
⇒ ${\mathit{r}}^2$ + 8r + 16 - ${\mathit{r}}^2$ = $\frac{11}{25}{\mathit{r}}^2$
⇒ 8r + 16 = $\frac{11}{25}{\mathit{r}}^2$
⇒ 200r + 400 = $11{\mathit{r}}^2$
⇒ 11 ${\mathit{r}}^2$ – 200r – 400 = 0
⇒ 11 ${\mathit{r}}^2$ – 220r + 20r – 400 = 0
⇒ 11r (r – 20) + 20 (r – 20) = 0
⇒ (r – 20) (11r + 20) = 0
⇒ r = 20 metre because
r ≠ - $\frac{20}{11}$ metre