SSC CGL 2024 Tier-I · 25 September 2024, Shift 1 · Quantitative Aptitude · Question 10 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.

The side BC of ∆ABC is produced to a point D. If AC = BC and ∠ BAC = 70°, then find the value of ∠ ACD − ∠ ABC.

(a)60°
(b)70°
(c)45°
(d)75°

Answer

Why

Correct — B. AC = BC, so the triangle is isosceles, and the angles opposite two equal sides are equal.

AC faces ∠B and BC faces ∠A, so ∠ABC = ∠BAC = 70°
∠ACB = 180° − 70° − 70° = 40°
∠ACD = 180° − 40° = 140°, since BCD is a straight line
∠ACD − ∠ABC = 140° − 70° = 70° → option (b)

Faster route: the exterior angle ∠ACD equals the sum of the two remote interior angles, 70° + 70° = 140°, so you never need ∠ACB at all.

Why the others are wrong

(a)
60° — 60° would need ∠ACD = 130° and so ∠ACB = 50°. But the two base angles are 70° each, which leaves 40° at C — nothing in the given data makes the angle at C 50°.
(c)
45° — 45° is what a right isosceles triangle would hand you. The stem fixes ∠BAC = 70°, so the equal angles are 70° each and the third is 40°, and no 45° appears anywhere in the figure.
(d)
75° — 75° needs ∠ACD = 145°, so the two remote interior angles would have to add to 145°. They are 70° and 70°, adding to 140°, and the exterior angle is fixed by them.

Concept

Two rules do all the work here.

Equal sides face equal angles. AC = BC pairs the angles opposite them — ∠ABC and ∠BAC — so one given angle hands you the other.

An exterior angle equals the sum of the two remote interior angles. Producing BC to D makes ∠ACD the exterior angle at C, so ∠ACD = ∠BAC + ∠ABC.

Together they give ∠ACD = 140° and ∠ABC = 70°, and the difference the question wants is 70°.

The lettering decides everything. AC = BC pairs the angles at B and A, not the angles at A and C.

Pair the wrong two and the working stays internally consistent while the answer is wrong, which is why this style of question rewards writing the side-angle pairing down before calculating.

Key facts

In any triangle, equal sides lie opposite equal angles, and the converse holds as well.
An exterior angle of a triangle equals the sum of the two interior angles not adjacent to it.
The three interior angles of a triangle add to 180°.
In this triangle ∠ABC = ∠BAC = 70°, ∠ACB = 40° and ∠ACD = 140°.

Study next

Exterior angle theorem, including figures where more than one side is producedAngle chases in isosceles and equilateral triangles from a single given angleAngle sum property extended to quadrilaterals and n-sided polygons

Common traps

Pairing AC = BC with the angles at A and C instead of the angles at A and B
Reading ∠ACD as the interior angle ∠ACB and computing 40° − 70°
Sketching the triangle roughly and estimating the apex angle instead of deriving it

SSC keeps this to a two-step chase: an isosceles condition, then a side produced to give an exterior angle.

Triangle side-and-angle reasoning is also asked in this paper at Quant Q.23, where two sides of an isosceles triangle are 6 cm and 12 cm and the third has to be identified.

Related PYQs

No directly related past PYQ was found.