SSC CGL 2024 Tier-I · 10 September 2024, Shift 2 · Quantitative Aptitude · Question 20 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.

If p cos A = 2q sin A and 2p cosec A − q sec A = 3, then the value of p² + 4q² is:

(a)4
(b)2
(c)3
(d)1

Answer

Why

Correct — A. The stem is an image: p cos A = 2q sin A and 2p cosec A − q sec A = 3, and it asks for p² + 4q².

The first equation fixes a ratio, not the letters:
tan A = sin A ⁄ cos A = p ⁄ 2q

Build a right triangle on that ratio — side opposite A is p, side adjacent is 2q — so the hypotenuse is k = √(p² + 4q²), which is the quantity the question wants, squared.
sin A = p⁄k, so cosec A = k⁄p
cos A = 2q⁄k, so sec A = k⁄2q

Substitute into the second equation and watch p and q cancel:
2p cosec A = 2p × k⁄p = 2k
q sec A = q × k⁄2q = k⁄2
2k − k⁄2 = 3k⁄2 = 3

k = 2, so p² + 4q² = k² = 4 → option (a).

Why the others are wrong

(b)
2 — 2 is k itself — the hypotenuse √(p² + 4q²). The question asks for p² + 4q², which is that value squared, so this stops one step early.
(c)
3 — 3 is the constant already given in the second equation. That equation equals 3k⁄2, not k², so it settles k at 2 and leaves the required value at 4.
(d)
1 — 1 would require k = 1, and then 2p cosec A − q sec A = 3k⁄2 = 3⁄2, not the 3 the stem states.

Concept

Two unknowns and one equation short, yet the answer is a number. That works because the question asks for p² + 4q², which is exactly the squared hypotenuse of the triangle the first equation describes.

p cos A = 2q sin A is a ratio statement. It pins tan A = p⁄2q without pinning p or q separately, and that is enough: once the triangle has legs p and 2q, every ratio in the problem becomes an expression in p, q and k.

The second equation then reduces to a statement about k alone, because p and q cancel against the reciprocal ratios.

Watch the cancellation, because it is what makes the item finite. The p in 2p cosec A dies against sin A = p⁄k, and the q in q sec A dies against cos A = 2q⁄k.

If either survived, the expression would still contain a free letter and no single number could be the answer.

Key facts

p cos A = 2q sin A rearranges to tan A = p⁄2q.
With legs p and 2q the hypotenuse is k = √(p² + 4q²), giving cosec A = k⁄p and sec A = k⁄2q.
2p cosec A − q sec A simplifies to 3k⁄2, so the given value 3 forces k = 2 and p² + 4q² = 4.

Study next

Building all six ratios from one given ratio using a right triangleHomogeneous trigonometric expressions, where the letters cancel and only their combination survivesReciprocal identities cosec A = 1⁄sin A and sec A = 1⁄cos A, and when they simplify faster than the direct ratios

Common traps

Answering 2, the hypotenuse, when the question asks for its square.
Dividing the first equation the wrong way round and writing tan A = 2q⁄p.
Replacing 2p cosec A with 2p sin A, which loses the reciprocal and destroys the cancellation.

Quant in this shift hands you a single ratio and asks for something built from it more than once — Q.10 gives 2 tan θ = 3 and Q.11 gives cot θ = 4⁄3.

Here the ratio is hidden one line deeper, inside p cos A = 2q sin A. The move is the same in all three: build the triangle rather than solve for the angle.

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