SSC CGL 2024 Tier-I · 24 September 2024, Shift 1 · Quantitative Aptitude · Question 13 of 98

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.

PQ is the diameter of a circle with centre A. M is a point on the circumference of the circle. If m∠MAP = 110°, then find the value of m∠MQA.

(a)70°
(b)55°
(c)35°
(d)65°

Answer

Why

Correct — B. PQ is a diameter through the centre A, so ∠MAP and ∠MAQ make a straight line:
∠MAQ = 180° − 110° = 70°

AM and AQ are both radii, so △AMQ is isosceles and ∠AMQ = ∠MQA.

2 × ∠MQA = 180° − 70° = 110°
∠MQA = 55° → option (b).

Why the others are wrong

(a)
70° — 70° is ∠MAQ, the angle at the centre. It is the vertex angle of △AMQ, and the two equal base angles share what is left of 180°.
(c)
35° — 35° is half of the central 70°, which gives ∠MPQ, the angle arc MQ makes at P on the circumference. The stem asks for the angle at Q.
(d)
65° — 65° follows from a central angle of 50°, that is from ∠MAP = 130°. The figure-free stem prints 110°.

Concept

Two facts do all the work. Every radius has the same length, so a triangle made of two radii and a chord is isosceles.

And an inscribed angle is half the central angle standing on the same arc.

That second fact gives a one-line route here: ∠MQP stands on arc MP, whose central angle is the given ∠MAP = 110°, so ∠MQP = 55° directly.

The isosceles route is worth knowing anyway, because it survives when the angle you are given is not the one standing on the arc you want.

No figure is printed with this question, so you draw your own. M may sit on either side of PQ; the answer does not change, because the two positions are mirror images.

Key facts

Any two radii of a circle are equal, so the triangle they form with a chord is isosceles.
Angles on one side of a straight line sum to 180°, so ∠MAP + ∠MAQ = 180° when PQ is a diameter.
An inscribed angle is half the central angle on the same arc, which gives ∠MQP = 110° ⁄ 2 = 55°.
The angle in a semicircle is a right angle, so ∠PMQ = 90° and the third angle of △PMQ checks out at 55°.

Study next

Inscribed angle theorem and the arcs each angle stands onAngle in a semicircle is 90 degreesCyclic quadrilaterals, where opposite angles are supplementary

Common traps

Reporting the central angle 70° as the answer
Forgetting that A is the centre and treating ∠MAP as an angle on the circumference
Mislabelling which of ∠MAP and ∠MAQ is the given 110° in your own sketch

The half-the-central-angle rule is asked directly at 24 Sep 2024, 12:30, Quant Q.24, where arc PQ subtends 80° at the centre and the angles it subtends at two points on the major arc are both 40°.

Circle geometry also appears earlier in this paper at Quant Q.10, on the chord-distance relation.

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 67 SSC CGL questions on it, every solved shift.