SSC CGL 2024 Tier-I · 18 September 2024, Shift 1 · Quantitative Aptitude · Question 23 of 99

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.

In a circle with centre O, AOC is the diameter. B is a point on the circumference of the circle such that arc AB is 1⁄5 of the arc BC. What is the degree measure of ∠BOC?

(a)120°
(b)80°
(c)30°
(d)150°

Answer

Why

Correct — D. The stem is an image. It reads: in a circle with centre O, AOC is the diameter, B is a point on the circumference such that arc AB is 1⁄5 of arc BC, and it asks for ∠BOC.

Because AOC is a diameter, arcs AB and BC together make a semicircle, so their central angles add to 180°.

Let ∠BOC = x. Arc AB is one-fifth of arc BC, so ∠AOB = x⁄5.

x⁄5 + x = 180
6x⁄5 = 180
x = 180 × 5⁄6 = 150° → option (d).

Check: ∠AOB = 150⁄5 = 30°, and 30° + 150° = 180°.

Why the others are wrong

(a)
120° — 120° passes the semicircle test and fails the ratio one. It leaves ∠AOB = 60°, and 60 : 120 is a 1 : 2 split of the arcs, not the 1 : 5 the paper states.
(b)
80° — 80° fails the semicircle test. Keeping the 1 : 5 ratio makes ∠AOB = 16°, and 16° + 80° = 96°, short of the 180° a diameter forces.
(c)
30° — 30° is ∠AOB, the smaller angle the working produces on the way. It is the fifth, not the whole — the question asks for ∠BOC.

Concept

Arc length and central angle are directly proportional in one circle, so a ratio between arcs is the same ratio between the angles they subtend at the centre.

A diameter splits the circle into two semicircular arcs of 180° each. With B on one of them, arcs AB and BC partition that semicircle.

So the whole question is one proportion: divide 180° in the ratio 1 : 5, giving 30° and 150°.

You never need the radius or any length. Ratios of arcs carry straight across to ratios of central angles, whatever the circle's size.

Key facts

Arc length = (central angle ⁄ 360°) × circumference, so arcs in one circle are proportional to their central angles.
A diameter subtends 180° at the centre, so arcs on one side of it total 180°.
Dividing 180° in the ratio 1 : 5 gives 30° and 150°.
∠BOC is a central angle, not an inscribed angle, so no halving applies here.

Study next

Dividing the angle at the centre in a given ratio of arcsCentral angle versus inscribed angle standing on the same arcAngle in a semicircle, which is always 90° at the circumference

Common traps

Reading "arc AB is 1⁄5 of arc BC" as a 1 : 5 split of the full 360° circle
Answering with ∠AOB = 30° because it is the number the ratio names first
Halving 150° to 75° by reflexively applying the inscribed-angle rule to a central angle

Circle-angle items in CGL 2024 usually hide one standard theorem behind a short stem, as with the chord produced to an external point at 23 Sep 2024, 12:30, Quant Q.16, and the arc-and-chord figure at 26 Sep 2024, 12:30, Quant Q.21.

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