CAT 2024 Slot 1 QA Q16 Mensuration

Question 16 of the Quantitative Ability (QA) section, CAT 2024 Slot 1, with CAT's official answer and a worked solution.

The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm , and the sum of the lengths of all its edges is 144 cm . The volume, in cubic cm , of the sphere is

A
\(750 \pi\)
B
\(1125 \pi \sqrt{2}\)
C
\(1125 \pi\)
D
\(750 \pi \sqrt{2}\)

Answer: (B) 1125 π √(2) Mensuration MCQ

Solution

Let the edges be \(l, w, h\). \(4(l + w + h) = 144 \Rightarrow l + w + h = 36\); surface area \(2(lw + wh + hl) = 846\).
\(l^2 + w^2 + h^2 = (l + w + h)^2 - 2(lw + wh + hl) = 1296 - 846 = 450\).
The box's space diagonal is the sphere's diameter: \(\sqrt{450} = 15\sqrt2\), so \(R = \frac{15\sqrt2}{2}\).
Volume \(= \frac43\pi R^3 = \frac43\pi \cdot \frac{3375 \cdot 2\sqrt2}{8} = 1125\sqrt2\,\pi\).