SSC CGL 2024 Tier-I · 12 September 2024, Shift 2 · 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.

If 7 tan θ = 3, then (7 sin θ + 5 cos θ) ⁄ (7 sin θ − 2 cos θ) is:

(a)8
(b)6
(c)5
(d)7

Answer

Why

Correct — A. The stem is an image: if 7 tan θ = 3, find (7 sin θ + 5 cos θ) ⁄ (7 sin θ − 2 cos θ). Divide every term, top and bottom, by cos θ.

(7 sin θ + 5 cos θ) ⁄ (7 sin θ − 2 cos θ)
= (7 tan θ + 5) ⁄ (7 tan θ − 2)
Substitute the group 7 tan θ = 3 as it stands
= (3 + 5) ⁄ (3 − 2) = 8 ⁄ 1 = 8 → option (a)

Why the others are wrong

(b)
6 — For the value to be 6, (7 tan θ + 5) ⁄ (7 tan θ − 2) = 6 solves to 7 tan θ = 17⁄5 = 3.4. The stem fixes it at 3, so the denominator is exactly 1 and the value is 8.
(c)
5 — This is the numerator with the 3 dropped. After the division by cos θ the numerator is 7 tan θ + 5 = 3 + 5 = 8, and it is the denominator, not the numerator, that collapses to 1.
(d)
7 — 7 is the coefficient the paper puts in front of both sine terms, and it cancels nothing on its own. Substituting sin θ = 3⁄√58 and cos θ = 7⁄√58 gives 56⁄√58 ÷ 7⁄√58 = 8.

Concept

A fraction whose numerator and denominator are both first degree in sin and cos is homogeneous: divide both lines by cos θ and every term becomes a tan θ term. The angle itself is never needed.

Here the stem gives 7 tan θ = 3 rather than tan θ = 3⁄7, and the fraction is built so that 7 tan θ appears intact in both lines. Substitute the group, not the isolated value, and the arithmetic is two subtractions long.

The denominator lands on exactly 1, which is why the answer is a clean integer rather than a surd.

If your working leaves a √58 that will not cancel, you substituted sin and cos separately — not wrong, just slower, and it reaches the same 8.

Key facts

Dividing a homogeneous sin-cos ratio by cos θ turns every term into a tan θ term.
With 7 tan θ = 3, the expression reduces to (3 + 5) ⁄ (3 − 2) = 8.
tan θ = 3⁄7 corresponds to sin θ = 3⁄√58 and cos θ = 7⁄√58.

Study next

Homogeneous sin-cos expressions solved by dividing through by cosBuilding sin and cos from a given tangent with a 3-7-√58 right triangleRewriting sec and cosec expressions in terms of tan

Common traps

Reading 7 tan θ = 3 as tan θ = 3 and substituting 3 for the tangent.
Dividing through by sin θ instead of cos θ, which leaves cot terms and more work.
Cancelling the 7 sin θ that appears in both lines, which is not a common factor of either line.

The stem here is a single typeset line with no text at all, so reading it is part of the test.

Quant Q.8 of the 12 Sep 2024, 12:30 sitting is the same shape — an image expression whose keyed value is 2 cosec — so the reading skill and the identity work are tested together.

Related PYQs

No directly related past PYQ was found.