SSC CGL 2024 Tier-I · 26 September 2024, Shift 3 · Quantitative Aptitude · Question 15 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 triangle XYZ, A is a point on YZ such that XA = YA. If ∠XYA = 50° and ∠AXZ = 19°, what is the degree measure of ∠XZA?

Answer

Why

Correct — A. XA = YA makes triangle XYA isosceles, so the angles facing those equal sides are equal.

∠AXY = ∠AYX = 50°
∠XAY = 180° − 50° − 50° = 80°

Y, A and Z lie on one line, so ∠XAZ = 180° − 80° = 100°

In triangle XAZ: ∠XZA = 180° − 100° − 19° = 61° → option (a).

Why the others are wrong

(b)
∠XYZ is 50° and ∠YXZ is 50° + 19° = 69°, both already fixed. With 53° at Z the three angles total 172°, not 180°.
(c)
Same failure, larger: 50° + 69° + 49° = 168°. Nothing in the figure is free once the isosceles pair is used.
(d)
50° + 69° + 41° = 160°, twenty degrees short of a triangle. Inside triangle XAZ the 100° at A and 19° at X leave exactly 61°.

Concept

Two facts carry this: equal sides face equal angles, and a triangle's angles sum to 180°.

XA = YA sits inside triangle XYA, which forces ∠AXY = ∠AYX = 50°. Because A lies on YZ, ∠XAY and ∠XAZ form a linear pair and add to 180°, carrying the 80° at A across as the 100° you need next.

That is the standard shape of an angle-chase: solve the triangle you can, move one angle across the shared vertex, solve the other.

The stored stem image is cut off — it ends mid-word, at "what is the degree measu", so the named angle is not visible in the capture.

The keyed 61° is ∠XZY, the angle at Z, which is precisely what the chain above produces. ∠YXZ would be 69° and ∠XAZ 100°, and neither is offered.

Key facts

In a triangle, sides of equal length face angles of equal measure.
Angles on a straight line at a point add to 180°, which turns ∠XAY = 80° into ∠XAZ = 100°.
An exterior angle equals the sum of the two remote interior angles, so ∠XAZ = 50° + 50° = 100° in a single step.

Study next

Isosceles triangle base-angle theorem and its converseExterior angle theorem as a shortcut across a shared vertexAngle chasing in figures where a cevian splits a triangle in two

Common traps

Reading XA = YA as XA = XY, which pairs the wrong angles and changes everything downstream.
Using the 19° inside triangle XYA, where it does not belong.
Carrying 80° into triangle XAZ instead of its supplement, 100°.

SSC's angle-chase items pair one equal-sides condition with one stray angle and expect two triangles to be solved in sequence. A cevian setup is also used on 17 Sep 2024, 16:00, Quant Q.16, where D is the mid-point of BC and the two perpendiculars dropped from it are equal.

Related PYQs

No directly related past PYQ was found.