SSC CGL 2024 Tier-I · 26 September 2024, Shift 2 · Quantitative Aptitude · Question 21 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 a circle with center O, an arc ABC subtends an angle of at the centre of the circle. The chord AB is produced to a point P. Then, the measure of is:

Answer

Why

Correct — C. Two pieces of the stem survive only as images: the arc ABC subtends 138° at O, and the angle wanted is ∠CBP.

Arc ABC = 138°, so the rest of the circle — arc CA, the one without B — is 360° − 138° = 222°
∠ABC is the inscribed angle standing on that 222° arc, so ∠ABC = 222°⁄2 = 111°

AB is produced to P, so ∠CBP is the supplement of ∠ABC:
∠CBP = 180° − 111° = 69° → option (c)

Why the others are wrong

(a)
42° is 180° − 138°, which treats the 138° as though it were an angle at B. It is a central angle, so it has to be halved before any supplement is taken.
(b)
111° is ∠ABC itself, the angle inside the circle. The question asks for ∠CBP, the angle on the far side of B once AB has been produced to P.
(d)
108° does not come out of this figure. A 138° arc yields only two angles at B — 111° inside and 69° outside — and neither halving 138° nor subtracting it gives 108°.

Concept

An inscribed angle is half the central angle standing on the same arc, so an angle at the circumference is half the arc it faces.

∠ABC sits at B and faces arc AC, the arc that does not contain B: 360° − 138° = 222°, hence 111°.

The shortcut worth carrying is the cyclic-quadrilateral one — an exterior angle equals the interior opposite angle. Put D anywhere on the major arc; then ∠CBP = ∠ADC = 138°⁄2 = 69°, in a single line.

Both the 138° and the '∠CBP' are missing from the row's text and live only in its images, and all four options are pictures too, reading 42°, 111°, 69° and 108°.

Key facts

An inscribed angle is half the central angle standing on the same arc.
An arc of 138° leaves 222° for the rest of the circle, so the inscribed angle facing it is 111°.
In a cyclic quadrilateral the exterior angle equals the interior opposite angle, which gives ∠CBP = 69° directly.
Half of arc ABC itself, 138°⁄2 = 69°, is the same answer by the shorter route.

Study next

Cyclic quadrilaterals: opposite angles supplementary, exterior equal to interior oppositeAngles in the same segment, and the angle in a semicircleThe alternate segment theorem, where a tangent replaces the produced chord

Common traps

Halving the wrong arc — taking 138°⁄2 for ∠ABC instead of 222°⁄2.
Reading the given 138° as ∠ABC and answering its supplement, 42°.
Stopping at ∠ABC = 111° because the eye settles on the angle inside the circle.

One angle is given and a second is wanted a produced line away, so the closing step here is a supplement.

The same 'side produced to a point' move runs on a triangle at 26 Sep 2024, 12:30, Quant Q.3 and again at 25 Sep 2024, 09:00, Quant Q.10.

Related PYQs

No directly related past PYQ was found.