SSC CGL 2025 Tier-I · 15 September 2025, Shift 1 · Quantitative Aptitude · Question 24 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.

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.

There are two parallel chords measuring 16 cm and 12 cm, both situated on the same side of the center of a circle. The space between the two chords is 2 cm. What is the radius of the circle?

(a)8 cm
(b)10 cm
(c)12 cm
(d)15 cm

Answer

Why

Correct — B. The perpendicular from the centre bisects each chord, so half-chord, distance and radius form a right triangle.

Halve the chords: 16⁄2 = 8 cm and 12⁄2 = 6 cm
Place them: the longer chord is nearer the centre, at x, the shorter at x + 2
Equate r²: 8² + x² = 6² + (x + 2)²

Expand: 64 + x² = 36 + x² + 4x + 4
Solve: 64 = 40 + 4x, so x = 6 cm
Radius: r² = 8² + 6² = 100, r = 10 cm → option (b)

Why the others are wrong

(a)
8 cm — 8 cm is half of the 16 cm chord. A radius of 8 would make that chord a diameter, and the 12 cm chord would then sit √(64 − 36) ≈ 5.3 cm from it, not 2 cm.
(c)
12 cm — 12 cm puts the chords √(144 − 64) ≈ 8.94 cm and √(144 − 36) ≈ 10.39 cm from the centre, about 1.45 cm apart instead of 2 cm.
(d)
15 cm — 15 cm puts the chords √161 ≈ 12.69 cm and √189 ≈ 13.75 cm from the centre, about 1.06 cm apart instead of 2 cm.

Concept

The perpendicular from the centre to a chord bisects it, so r² = (half-chord)² + (distance)² for every chord.

Longer chords lie nearer the centre. With both chords on the same side, the gap between them is the difference of their distances. On opposite sides it is the sum.

Set the two expressions for r² equal and the x² terms cancel, leaving a linear equation in the unknown distance.

The side matters here. On opposite sides the distances would add to 2, and 64 + x² = 36 + (2 − x)² gives x = −6, which is impossible. The same-side arrangement the question states is the one that works.

Key facts

The perpendicular from the centre to a chord bisects the chord.
r² = (half-chord)² + (distance from centre)².
Of two chords in the same circle, the longer one is nearer the centre.

Study next

Parallel chords on opposite sides of the centre, where the distances addLength of a chord at a given distance from the centreDistance between two parallel chords from the radius and the chord lengths

Common traps

Placing the shorter chord nearer the centre. The 12 cm chord is the one farther away, at x + 2.
Using the full chord lengths 16 and 12 in the Pythagoras step instead of the halves 8 and 6.

The same-side setup decides 9 Sep 2024, 09:00, Quant Q.4: chords 10 and 24 cm, 7 cm apart, give radius 13 and diameter 26 cm.

17 Sep 2024, 12:30, Quant Q.2 (chords 24 and 18 cm, 21 cm apart) works only with the chords on opposite sides: distances 9 and 12 add to 21, giving radius 15 cm.

Related PYQs

No directly related past PYQ was found.

Topic: Circle and its chords, tangents, angles subtended by chords of a circle, common tangents to two or more circles — all 140 SSC CGL questions on it, every solved shift.