Pipes and Cisterns Practice Test 2 with solutions

Let (x – 10) and x hrs be the time taken by two pipes to fill the reservoir, then,
$\frac{1}{\mathit{x}-10}+\frac{1}{\mathit{x}}$ = $\frac{1}{11}$
⇒ $\frac{2\mathit{x}-10}{{\mathit{x}}^2-10\mathit{x}}$ = $\frac{1}{12}$
⇒ ${\mathit{x}}^2$ - 34x + 120 = 0
⇒ (x – 30) (x – 4) = 0
∴ x = 30 (Taking 30 only).
Hence, second pipe will fill the reservoir in 30 hours.