Daily Olympiad: Quantitative Aptitude - Mensuration [20260511]

Challenge yourself with today's CAT practice! This test covers 'Mensuration' for Quantitative Aptitude (CAT - Graduate). Level: Hard | Duration: 45 mins.

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1. A solid cube of side 8 cm is melted and recast into smaller cubes of side 2 cm each. The ratio of the total surface area of all small cubes to the surface area of the original cube is:

Solution
Correct: A
Original cube surface area = 6(8^2) = 384 cm². Number of small cubes = 8³/2³ = 64. Total surface area of small cubes = 64×6×(2²) = 1536 cm². Ratio = 1536/384 = 4:1. The correct answer is 4:1.

2. A right circular cone of radius 6 cm is placed over a hemisphere of radius 6 cm. The slant height of the cone is equal to its height. The total surface area of the composite figure is:

Solution
Correct: A
Let slant height = l = height = h. For cone: l² = r² + h² → l = 6√2. Total surface area = hemisphere (2πr²) + cone (πrl) = 2π×6² + π×6×6√2 = 72π + 36√2π. However, the overlapping circular base is excluded. Correct calculation is hemisphere (2πr²) + cone (πrl) = 72π + 36√2π ≈ 216π. Correct answer: 216π.

3. The vertices of a triangle are A(2,3), B(–4,5), and C(2,–1). If the triangle is reflected over the y-axis, what is the area of the resulting triangle?

Solution
Correct: A
Reflection does not change area. Use determinant: Area = ½ |(2(5–(–1)) + (–4)(–1–3) + 2(3–5))| = ½ |(12 +16 + (–4))| = ½×24 = 12. Reflected triangle area is same, 12 sq. units.

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