Daily Olympiad: Quantitative Aptitude - Mensuration [20260512]

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

🏆 Free — No Login Required
← Back to All Olympiads

1. A solid cone with base radius 6 cm and height 8 cm is melted into a right circular cylinder. If the radius of the cylinder is half of the cone’s radius, what is its height?

Solution
Correct: A
Volume of the cone = (1/3)πr²h = (1/3)π(6²)(8) = 96π cm³. Let height of cylinder be H. Radius of cylinder = 3 cm (half of 6 cm). Volume of cylinder = π(3)²H = 9πH cm³. Equating volumes: 9πH = 96π ⇒ H = 96/9 = 32/3. Correct answer is closest option A.

2. A right circular cylinder is inscribed in a sphere of radius R. What is the maximum volume of such a cylinder?

Solution
Correct: A
Let the cylinder’s radius be r and height be h. From geometry: r² + (h/2)² = R². Volume = πr²h. Express r² = R² - h²/4. Substitute into volume: V = π(R²h - h³/4). Maximize V w.r.t h. Set dV/dh = 0 ⇒ π(R² - 3h²/4) = 0 ⇒ h = (2R√3)/3. Substituting, max volume = π(R²*(2R√3/3) - (2R√3/3)³/4) = 4πR³/3√3.

3. A cube of side 3 cm is cut into smaller cubes of side 1 cm. How many such small cubes will have exactly two faces painted if the original cube is painted on all faces?

Solution
Correct: A
On each face of the cube, the cubes with exactly two faces painted lie on edges but not at corners. Since the original cube is 3 cm, each edge has 1 such cube. A cube has 12 edges. Total = 12 × 1 = 12.

4. A regular tetrahedron has edge length 4 cm. What is its total surface area?

Solution
Correct: A
A regular tetrahedron has 4 equilateral triangular faces. Area of one face = (√3/4)(edge)² = (√3/4)(16) = 4√3 cm². Total = 4 × 4√3 = 16√3 cm².

5. A wire bent into a circle of radius 14 cm is rebent into a square. What is the area of the square?

Solution
Correct: D
Circumference of circle = 2πr = 2×22/7×14 = 88 cm. Side of square = 88/4 = 22 cm. Area = 22² = 484 cm².

6. A spherical iron ball is melted and recast into 8 smaller spheres of equal size. What is the ratio of the surface area of the original sphere to that of one of the smaller spheres?

Solution
Correct: C
Volume of original sphere = (4/3)πR³. Volume of each small sphere = (4/3)π(r³)/8 ⇒ r = R/2. Surface area ratio = 4πR² / 4πr² = R²/r² = (R)²/( (R/2)² ) = 4/1.