SSC CGL 2024 Tier-I · 25 September 2024, Shift 3 · Quantitative Aptitude · Question 17 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 10 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.

In the figure shown above, quadrilateral PQRS has its vertices on the circumference of the circle with centre O. If m ∠ QOS = 20x° and m ∠ QRS = 26x°, then what is the value of ‘x’?

(a)4
(b)3
(c)6
(d)5

Answer

Why

Correct — D. In the figure S, P, Q and R lie on the circle in that order and O is the centre. The 20x° marker is the angle QOS on the side of O facing R, and 26x° is the angle QRS at the vertex R.

∠QRS stands at the circumference on the arc QS that goes through P, so its central angle is the reflex one, 360° − 20x°.

Angle at centre = 2 × angle at circumference:
360 − 20x = 2 × 26x
360 = 52x + 20x = 72x
x = 5 → option (d).

Check: 20x = 100° and 26x = 130°, so ∠QPS = ½ × 100° = 50° and the opposite pair 50° + 130° = 180°, as a cyclic quadrilateral requires.

Why the others are wrong

(a)
4 — 4 is where you drift by treating 20x° as the vertex angle opposite R and writing 20x + 26x = 180. That gives x ≈ 3.9 and 4 is the nearest option, but 20x° is measured at the centre O, not at P.
(b)
3 — Substitute x = 3: ∠QOS = 60°, so the angle at the circumference on the far arc is ½ × 300° = 150°, while 26x would be only 78°. The two readings of the same arc disagree.
(c)
6 — Substitute x = 6: ∠QOS = 120°, so the circumference angle is ½ × 240° = 120°, but 26x is 156°. Only x = 5 makes the two agree.

Concept

Two circle facts do the work: the angle at the centre is twice the angle at the circumference on the same arc, and the opposite angles of a cyclic quadrilateral sum to 180°.

The trap is which arc. ∠QRS has its vertex at R, so it stands on the arc QS that does not contain R — the one through P. The central angle for that arc is the reflex angle at O, 360° − 20x°, not the marked 20x°.

Either route closes: use the reflex angle directly, or halve 20x° to get ∠QPS and pair it with ∠QRS against 180°. Both give 72x = 360.

The figure is marked NOT TO SCALE, so the drawn sizes carry no information — only the two labels and the order of the four points around the circle do.

Key facts

An arc subtends at the centre twice the angle it subtends anywhere on the remaining circle.
Opposite angles of a cyclic quadrilateral add to 180°.
With x = 5 the figure reads ∠QOS = 100°, ∠QRS = 130° and ∠QPS = 50°.

Study next

Choosing the reflex or the non-reflex central angle from where the vertex sitsCyclic quadrilaterals and the 180° pairing of opposite anglesAngles in the same segment, which are equal to one another

Common traps

Using the marked 20x° as the central angle for ∠QRS, which collapses to 26x = 10x.
Adding the central angle to the vertex angle and setting the sum to 180°.
Trusting the drawing after it has told you it is not to scale.

SSC sets the centre-versus-circumference rule plainly — 80° at the centre, find the circumference angles, at 24 Sep 2024, 12:30, Quant Q.24 — and inside an inscribed quadrilateral at 19 Sep 2024, 12:30, Quant Q.2. Here the two dressings are combined and both angles are given in terms of x.

Related PYQs

No directly related past PYQ was found.