ABC26GN4802 · Volume and Surface Area

Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: 2010 · Marks: · Difficulty:

The length of the longest rod that can be placed in a room of dimensions $10 \mathrm{m} \times 10 \mathrm{m} \times 5 \mathrm{m}$ is

(a)$15 \sqrt{3}$

(b)15

(c)$10 \sqrt{2}$

(d)$5 \sqrt{3}$

Answer

Answer (as printed): B

Explanation

Required length $=\sqrt{(10)^{2}+(10)^{2}+(5)^{2}} \mathrm{m}=\sqrt{225} \mathrm{m}=15 \mathrm{m}$.