SSC CGL 2024 Tier-I · 25 September 2024, Shift 2 · Quantitative Aptitude · Question 4 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 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 quadrilateral ABCD, AB = BC, AD = DC. ∠ ABD = 68°, ∠ ADB = (2y-7)°, ∠ BDC = 33°, ∠ DBC = (3x+2)°. Then the value of 2x+3y is:

(a)104
(b)108
(c)118
(d)144

Answer

Why

Correct — A. AB = BC and AD = DC make BD the axis of symmetry, so the diagonal cuts the quadrilateral into two congruent triangles.

In △ABD and △CBD: AB = CB, AD = CD, BD common — congruent by SSS.

Matching angles at B: 3x + 2 = 68 → 3x = 66 → x = 22
Matching angles at D: 2y − 7 = 33 → 2y = 40 → y = 20

2x + 3y = 44 + 60 = 104 → option (a)

Why the others are wrong

(b)
108 — 108 would need x = 24 or y ≈ 21.3, and the congruence leaves no such room. It fixes x = 22 and y = 20 exactly, with nothing left to adjust.
(c)
118 — 118 overshoots by 14. Holding x = 22, it would need y ≈ 24.7, but ∠ADB = ∠BDC = 33° gives 2y − 7 = 33 and so y = 20.
(d)
144 — 144 would need x = 42 alongside y = 20. That contradicts 3x + 2 = 68, which is forced the moment you accept ∠ABD = ∠DBC.

Concept

A quadrilateral with two pairs of adjacent equal sides — AB = BC and AD = DC — is a kite, and the diagonal BD joining the two vertices between those pairs is its line of symmetry.

BD belongs to both △ABD and △CBD, so the triangles match by SSS and every corresponding part is equal.

That hands you both equations at once: ∠ABD = ∠DBC and ∠ADB = ∠BDC. The angles are dressed as expressions in x and y only to add a line of algebra.

The second diagonal AC never enters the working.

No figure is given. Sketch the kite yourself with BD running down the middle and mark the equal sides before starting the algebra, or the pairing of angles is easy to get backwards.

Key facts

In a kite, the diagonal joining the vertices between the equal sides is an axis of symmetry.
SSS congruence needs three pairs of equal sides and no angle information at all.
Here ∠ABD = ∠DBC = 68° gives x = 22, and ∠ADB = ∠BDC = 33° gives y = 20.

Study next

Properties of a kite, including that its diagonals meet at right anglesChoosing between SSS, SAS, ASA and RHS from the data a question givesAngle chasing in isosceles triangles where the angles are written as expressions

Common traps

Treating ABCD as a rhombus and setting all four sides equal, which the data never states
Pairing ∠ABD with ∠BDC instead of with ∠DBC, so the congruence is read across the wrong vertices
Solving for x and y correctly and then evaluating 3x + 2y instead of 2x + 3y

SSC hands you an equal-sides condition and an angle written as an expression, so the congruence is the whole question and the algebra is one line. Circle geometry in the same paper works the same way at 25 Sep 2024, 12:30, Quant Q.14, where PR and PS being diameters is what settles ∠PQR.

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