SSC CGL 2025 Tier-I · 15 September 2025, Shift 2 · Quantitative Aptitude · Question 16 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 11 September 2026.

Where this paper and its answers come from

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 right circular cone has radius of 7 cm and a height of 24 cm. A sphere is placed inside the cone such that it touches the base and the slanted surface of the cone. Find the radius of this inscribed sphere.

(a)5.25 cm
(b)4.86 cm
(c)3.82 cm
(d)6.24 cm

Answer

Why

Correct — A.
Cut the cone through its axis: an isosceles triangle, base 14, height 24.
Slant side = √(7² + 24²) = √625 = 25
Area = ½ × 14 × 24 = 168

Semi-perimeter s = (25 + 25 + 14) ÷ 2 = 32
The sphere shows up as this triangle's incircle: r = area ÷ s
r = 168 ÷ 32 = 5.25 cm → option (a)

Why the others are wrong

(b)
4.86 cm — 4.86 cm is too small to reach the slant side. Its centre would sit 4.86 cm up the axis, about 5.36 cm from the slant line, leaving a gap.
(c)
3.82 cm — 3.82 cm leaves a wider gap: with its centre 3.82 cm up the axis, the slant surface is about 5.65 cm away, so the sphere touches only the base.
(d)
6.24 cm — 6.24 cm is too large. With its centre 6.24 cm up the axis, the slant line is only about 4.97 cm away, so the sphere would cut through the cone's side.

Concept

A sphere touching the base and the curved surface of a cone appears, in the cross-section through the axis, as a circle touching all three sides of an isosceles triangle: its incircle.

For any triangle, inradius = area ÷ semi-perimeter. For a cone of radius R, height h and slant l this becomes r = Rh ÷ (R + l) = 7 × 24 ÷ 32.

Check by distance: put the base on the x-axis and the apex at (0, 24), so the slant line is 24x + 7y = 168.
The centre (0, r) must be r away from it: (168 − 7r) ÷ 25 = r, so 32r = 168 and r = 5.25.

Key facts

7, 24, 25 is a Pythagorean triple, so the slant height is 25 cm.
Inradius of any triangle = area ÷ semi-perimeter.
Sphere inscribed in a cone: r = Rh ÷ (R + l), here 168 ÷ 32 = 5.25 cm.

Study next

Inradius and circumradius of a right triangleSlant height, curved surface area and volume of a right circular coneA sphere inscribed in a cube or a cylinder

Common traps

Putting the full base 14 into the slant root: √(14² + 24²) is not the slant height
Dividing the area by the whole perimeter, 64, which halves the answer to 2.625 cm

19 Sep 2025, 09:00, Quant Q.24 asks for an incircle directly: a right triangle with legs 5 and 12 has area 30 and semi-perimeter 15, so its inscribed circle has radius 2.

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 51 SSC CGL questions on it, every solved shift.