SSC CGL 2024 Tier-I · 13 September 2024, Shift 2 · Quantitative Aptitude · Question 20 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.

Simplify (25a² − 10ab − 48b²) ⁄ (5a + 6b) × (5a + 8b)

Answer

Why

Correct — B. The numerator factorises so that the printed denominator cancels outright; nothing needs multiplying out.

Split the middle term of 25a² − 10ab − 48b².
You need two numbers summing to −10 whose product is 25 × (−48) = −1200: they are +30 and −40.

25a² + 30ab − 40ab − 48b²
= 5a(5a + 6b) − 8b(5a + 6b)
= (5a + 6b)(5a − 8b)

The (5a + 6b) cancels against the denominator, leaving the printed multiplier:
(5a − 8b) × (5a + 8b)
= 25a² − 64b² → option (b)

Why the others are wrong

(a)
Option (a) shows (5a + 8b)², which would need the numerator to factor as (5a + 6b)(5a + 8b). Expand that and the middle term is +70ab with +48b² at the end — neither matches the printed −10ab − 48b².
(c)
Option (c) is (5a − 8b)². The bracket (5a − 8b) is genuinely the surviving factor, but you multiply it by the (5a + 8b) printed in the question, not by itself.
(d)
Option (d), 25a² − 36b², is (5a − 6b)(5a + 6b). It keeps 6b as the surviving term, but 6b belongs to the factor that cancels; what survives is the 8b bracket.

Concept

A quadratic in two variables factorises exactly like one in a single variable. Read 25a² − 10ab − 48b² as 25x² − 10x − 48 with x standing in for a⁄b, split the middle term, then put b back.

The whole design of the question is that the denominator, (5a + 6b), is one of the two factors. Spot that and three lines of cancelling replace a page of expansion.

The second bracket, (5a + 8b), is then paired with (5a − 8b), so the answer is forced into the difference-of-squares form x² − y².

Stem and all four options are printed as pictures, so the shape of the options is visible before any working: two are perfect squares and two are differences of squares. Deciding which family the answer belongs to settles most of the item.

Key facts

25a² − 10ab − 48b² factorises as (5a + 6b)(5a − 8b).
(5a − 8b)(5a + 8b) = 25a² − 64b², by x² − y² = (x − y)(x + y).
To split a middle term, find two numbers whose sum is the middle coefficient and whose product is the first coefficient times the last — here −10 and −1200.
(5a + 6b)(5a + 8b) expands to 25a² + 70ab + 48b², which is why option (a) cannot be the numerator.

Study next

Splitting the middle term when the quadratic carries two variablesTelling the outputs of (x ± y)² and x² − y² apart at a glanceSpotting a printed denominator as a factor before expanding anything

Common traps

Multiplying the numerator by (5a + 8b) first, which buries the cancellation.
Cancelling (5a + 6b) on sight without checking it really divides the numerator.
Getting the factors right and then writing the final product as a square.

The giveaway is the denominator: when a quadratic sits over a linear bracket, that bracket is almost always one of its factors, and the question is testing whether you factorise instead of expand.

The options are built so a sign slip inside the factorisation lands on a real option rather than on nothing.

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