SSC CGL 2024 Tier-I · 9 September 2024, Shift 3 · Quantitative Aptitude · Question 13 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 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 aΔPQR, the bisectors of ∠ Q and ∠ R meet at point O, inside the triangle. If ∠ QOR = 107°, then the measure of ∠ P is:

(a)23°
(b)40°
(c)34°
(d)17°

Answer

Why

Correct — C. Rule: where the bisectors of two angles of a triangle meet, ∠QOR = 90° + ∠P/2.

90° + ∠P/2 = 107°
∠P/2 = 107° − 90° = 17°
∠P = 2 × 17° = 34° → option (c)

From first principles: in ΔOQR the other two angles sum to 180° − 107° = 73°, and those are the halves, so ∠Q + ∠R = 146° and ∠P = 180° − 146° = 34°.

Why the others are wrong

(a)
23° — 23° does not appear on any step. Put it back into the formula and the bisectors would meet at 90° + 11.5° = 101.5°, not the 107° the question states.
(b)
40° — 40° needs ∠QOR = 90° + 20° = 110°. Substituting your answer back into 90° + ∠P/2 is a two-second check that rules it out.
(d)
17° — 17° is ∠P/2, the value in hand the moment you subtract 90° from 107°. The formula carries a half, so the last move is to double it.

Concept

The point where the internal angle bisectors meet is the incentre. In ΔOQR the angles at Q and R are ∠Q/2 and ∠R/2, so
∠QOR = 180° − (∠Q + ∠R)/2 = 180° − (180° − ∠P)/2 = 90° + ∠P/2.

Two consequences worth carrying: the angle in the formula is the one that is not bisected, and ∠QOR is obtuse in every triangle, since ∠P/2 is always above zero.

The paper prints "In aΔPQR" with the space missing; read it as "In a ΔPQR". Nothing in the question turns on it.

Key facts

For the incentre O of ΔPQR, ∠QOR = 90° + ∠P/2.
It follows from the angle sum: ∠QOR = 180° − (∠Q + ∠R)/2.
∠QOR is obtuse in every triangle, because ∠P/2 is always greater than 0°.
Here ∠Q + ∠R = 146°, which leaves ∠P = 34°.

Study next

Incentre, circumcentre, orthocentre and centroid, and which angle formula belongs to whichThe angle between the bisectors of the exterior angles at Q and RAngle-sum and exterior-angle chains inside a single triangle

Common traps

Stopping at 17°, which is ∠P/2 and is sitting on the option list as (d).
Subtracting from 180° rather than 90° — 180° − 107° = 73° is the sum of the two halves.
Feeding ∠Q or ∠R into the formula, when the angle it takes is the one that is not bisected.

SSC runs this both directions: give ∠QOR and ask for the unbisected angle, or give that angle and ask for ∠QOR. Identifying which of the four triangle centres the stem describes is most of the work.

Related PYQs

No directly related past PYQ was found.