SSC CGL 2024 Tier-I · 9 September 2024, Shift 1 · Quantitative Aptitude · Question 4 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

Let O be the centre of the circle and AB and CD are two parallel chords on the same side of the radius. OP is perpendicular to AB and OQ is perpendicular to CD. If AB = 10 cm, CD = 24 cm and PQ = 7 cm, then the diameter (in cm) of the circle is equal to:

(a)24
(b)13
(c)12
(d)26

Answer

Why

Correct — D. A perpendicular from the centre bisects its chord, so P and Q are the midpoints of AB and CD.

AP = 10 ⁄ 2 = 5 cm and CQ = 24 ⁄ 2 = 12 cm.

With r for the radius:
OP² = r² − 5² = r² − 25
OQ² = r² − 12² = r² − 144

Both chords sit on the same side of the centre, so OP − OQ = PQ = 7.

OP² − OQ² = (r² − 25) − (r² − 144) = 119
(OP − OQ)(OP + OQ) = 119 → 7 × (OP + OQ) = 119
OP + OQ = 17

Solving OP − OQ = 7 with OP + OQ = 17 gives OP = 12 and OQ = 5.

r² = 12² + 5² = 169 → r = 13
Diameter = 2r = 26 → option (d).

Why the others are wrong

(a)
24 — 24 cm is the length of chord CD, not the diameter. A radius of 12 would make OQ = 0, putting CD through the centre and leaving PQ ≈ 10.9 cm instead of 7.
(b)
13 — 13 cm is the radius the working reaches, and the question asks for the diameter. Double it.
(c)
12 — No chord can be longer than the diameter, and CD is 24 cm — so a 12 cm diameter is impossible before any calculation begins.

Concept

Two facts do all the work in a parallel-chords question.

First, the perpendicular from the centre to a chord bisects it, so half-chord, distance-from-centre and radius form a right triangle.

Second, the arrangement decides the sign: chords on the same side of the centre give PQ = OP − OQ, chords on opposite sides give PQ = OP + OQ.

Choosing that sign correctly is the entire difficulty of the item.

The paper says the chords lie 'on the same side of the radius'. What makes the arithmetic work is that they lie on the same side of the centre, which is what OP − OQ = 7 encodes. On opposite sides the equation OP + OQ = 7 has no solution, since a 24 cm chord already forces OP to exceed 10.9 cm.

Key facts

A perpendicular dropped from the centre of a circle to a chord bisects that chord.
A chord of length 2a in a circle of radius r lies √(r² − a²) from the centre.
In a circle of radius 13 cm, chords of 10 cm and 24 cm sit 12 cm and 5 cm from the centre.

Study next

Parallel chords on opposite sides of the centre, where the two distances addEqual chords are equidistant from the centre, and the converse of that resultLength of the common chord of two intersecting circles

Common traps

Answering 13, the radius, when the question asks for the diameter
Adding the distances from the centre when the chords are on the same side

The numbers are chosen so the surds vanish — the 5-12-13 triple carries this one. Circle geometry returns in this shift at Quant Q.17, where two circles of radii 22 cm and 10 cm share a direct common tangent.

Related PYQs

No directly related past PYQ was found.