Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: 2005 · Marks: · Difficulty:
A rectangular tank measuring $5 \mathrm{m} \times 4.5 \mathrm{m} \times 2.1 \mathrm{m}$ is dug in the centre of the field measuring 13.5 m by 2.5 m. The earth dug out is evenly spread over the remaining portion of the field. How much is the level of the field raised?
(a)4 m
(b)4.1 m
(c)4.2 m
(d)4.3 m
Answer
Answer (as printed): C
Explanation
Volume of earth dug out $=(5 \times 4.5 \times 2.1) \mathrm{m}^{3}=47.25 \mathrm{m}^{3}$. Area over which earth is spread $=(13.5 \times 2.5-5 \times 4.5)$ $\mathrm{m}^{2}=(33.75-22.5) \mathrm{m}^{2}=11.25 \mathrm{m}^{2}$. $$\text { ∴ } \text { Rise in level }=\frac{\text { Volume }}{\text { Area }}=\left(\frac{47.25}{11.25}\right) \mathrm{m}=4.2 \mathrm{m} .$$