XAT 2014 — QA & DI Question 21

Since the elevation increases by 1 m for every 2.6 m,

​​​​​​​

$x=\sqrt{2.6^2-1}=\sqrt{6.76-1}=\sqrt{5.76}=2.4 \mathrm{m}$

Since $\frac{48}{2.4}=20,$

The height of the deeper end of the pool is 20 m.

This can be represented as follows,

​​​​​​​

The total volume of water in the pool,

$\left(\frac{1}{2}\times 48\times 20\right)\times 20+(48\times 20\times 1)$

⇒ 48× 20 × 11

⇒ 10560 m³

Hence, option (d).