SSC CGL 2024 Tier-I · 13 September 2024, Shift 2 · Quantitative Aptitude · Question 22 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.

If , then find the value of

Answer

Why

Correct — B. Divide the numerator and the denominator by cosA. Every term then becomes tanA, which is the one quantity you were given.

(7sinA − 3cosA) ⁄ (7sinA + 3cosA)
= (7tanA − 3) ⁄ (7tanA + 3)

Put tanA = 5⁄7, so 7tanA = 5:
= (5 − 3) ⁄ (5 + 3)
= 2⁄8 = 1⁄4

Then add the 4 printed outside the fraction:
1⁄4 + 4 = 4¼ → option (b)

Why the others are wrong

(a)
7⅓ misses on both halves. The number added outside the fraction is 4, not 7, and the fraction itself reduces to 1⁄4 — nothing in the working divides by 3.
(c)
3 5⁄7 hands back the printed 5⁄7. That ratio is tanA, not the value of the fraction — substituted into (7tanA − 3)⁄(7tanA + 3) it gives 2⁄8 = 1⁄4, and the number outside is 4.
(d)
3 1⁄5 needs the fraction to be 1⁄5, which comes from a denominator of 10. The denominator is 7tanA + 3 = 8, so the fraction is 2⁄8 = 1⁄4.

Concept

The expression is homogeneous: every term upstairs and downstairs has degree one in sinA and cosA. Divide through by the highest power of cosA and the whole ratio collapses into tanA.

That is why the question gives tanA and nothing else. You never need sinA or cosA separately, and you never need to build the 5–7–√74 right triangle behind tanA = 5⁄7.

The trailing + 4 sits outside the fraction, not inside it. Read the whole expression once before you start, or the working ends correctly at 1⁄4 and the answer is still wrong.

Both the given ratio and the expression are printed as images in this item, so the + 4 is easy to skip on a first read. The numbers are chosen so 7tanA is the whole number 5, which is the signal that substitution, not triangle-building, is the intended route.

Key facts

Dividing a homogeneous sin-cos ratio by cosA turns every term into tanA.
(7sinA − 3cosA)⁄(7sinA + 3cosA) = (7tanA − 3)⁄(7tanA + 3).
With tanA = 5⁄7 the fraction is 2⁄8 = 1⁄4, so the expression equals 4¼.
tanA = 5⁄7 corresponds to a right triangle with legs 5 and 7 and hypotenuse √74.

Study next

Homogeneous trigonometric ratios and the divide-by-cosA techniqueComponendo and dividendo applied to (a sinA + b cosA)⁄(a sinA − b cosA)Rebuilding sinA and cosA from a given tanA with a right triangle

Common traps

Answering 1⁄4 and forgetting the + 4 printed outside the fraction.
Cancelling the 7 of 7sinA against the 7 in 5⁄7 before dividing by cosA.
Building the √74 triangle to get sinA and cosA — valid, but the long way round.

Given tanA, evaluate a sin-cos expression — that is the family, and here the triangle is not wanted.

17 Sep 2024, 09:00, Quant Q.22 belongs to it: tanA = 1, evaluate 4 sinA cosA. That one is a product, though, so there is nothing to divide by cosA — read tanA = 1 as A = 45°, or write the expression over sin²A + cos²A and divide through by cos²A.

When the given ratio is tidy enough for the substitution to land on whole numbers — 7tanA = 5 here — that is your cue that no triangle is needed.

Related PYQs

No directly related past PYQ was found.