SSC CGL 2025 Tier-I · 24 September 2025, Shift 2 · Quantitative Aptitude · Question 4 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 13 September 2026.

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SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

A spherical shell has an outer diameter of 20 cm and is 1 cm thick. What is the volume of the material in the shell? (Use π≈3.14).

(a)1134.7 cm³
(b)1256.6 cm³
(c)1340.4 cm³
(d)1046.7 cm³

Answer

Why

Correct — A. Material = outer sphere − hollow core = 4⁄3 π (R³ − r³).

Outer radius: R = 20 ÷ 2 = 10 cm
Inner radius: r = 10 − 1 = 9 cm
Difference of cubes: 1000 − 729 = 271

Volume: 4⁄3 × 3.14 × 271 = 3403.76 ÷ 3 ≈ 1134.59 cm³
Nearest option: 1134.7 cm³ → option (a)

Why the others are wrong

(b)
1256.6 cm³ — 1256.6 ≈ 400π, the outer surface area 4π × 10², times the 1 cm thickness. The inner surface is smaller than the outer one, so this shortcut overestimates the material.
(c)
1340.4 cm³ — 1340.4 cm³ would need R³ − r³ ≈ 320. With R = 10 and r = 9 the difference of cubes is 271, so this is about 206 cm³ too large.
(d)
1046.7 cm³ — 1046.7 cm³ equals 4⁄3 × 3.14 × 250, so it uses 250 where 10³ − 9³ = 271 belongs. It falls about 88 cm³ short.

Concept

The material in a hollow sphere is the difference of two sphere volumes: 4⁄3 πR³ for the outer ball minus 4⁄3 πr³ for the empty core.

The thickness links the radii: r = R − thickness. The question gives a diameter, so halve it before cubing: R = 10, not 20.

Subtract the cubes first, then multiply by 4⁄3 π once.

With π = 3.14 the exact value is 1134.59 cm³, which rounds to 1134.6, not the printed 1134.7. The printed figure is still the nearest option by more than 85 cm³, so choose it.

Key facts

Volume of a sphere = 4⁄3 πr³.
Volume of a spherical shell = 4⁄3 π(R³ − r³), with r = R − thickness.
10³ − 9³ = 1000 − 729 = 271.

Study next

Hemispherical shells: 2⁄3 π(R³ − r³)Melting and recasting, where the volume of metal stays the same

Common traps

Cubing the diameters: 20³ − 18³ = 2168, eight times the true 271.
Cubing the thickness. The shell needs R³ − r³, not (R − r)³ = 1.

The same difference of cubes decides 18 Sep 2025, 16:00, Quant Q.12, a hollow hemispherical shell with inner radius 7 cm and outer radius 14 cm.

At 17 Sep 2025, 09:00, Quant Q.10 a hollow sphere of outer radius 10 cm is melted into 50 solid spheres, and the inner radius must be found.

Related PYQs

No directly related past PYQ was found.

Topic: Right Prism, Right Circular Cone, Right Circular Cylinder, Sphere, Hemispheres, Rectangular Parallelepiped, Regular Right Pyramid with triangular or square base — all 99 SSC CGL questions on it, every solved shift.