SSC CGL 2024 Tier-I · 26 September 2024, Shift 1 · Quantitative Aptitude · Question 8 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 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 3(16³ − 6³) ⁄ (16² + 6² + Q) = 30, then find the value of Q.

(a)112
(b)98
(c)96
(d)108

Answer

Why

Correct — C. Factorise the numerator with a³ − b³ = (a − b)(a² + ab + b²), taking a = 16 and b = 6.

16³ − 6³ = 10 × (256 + 96 + 36) = 10 × 388 = 3880
Numerator = 3 × 3880 = 11640

The equation becomes 11640 ⁄ (256 + 36 + Q) = 30, so 292 + Q = 11640 ⁄ 30 = 388
Q = 388 − 292 = 96 → option (c).

Faster route: the fraction only collapses to 3(a − b) = 30 if the denominator is the complete a² + ab + b², so Q must be ab = 16 × 6 = 96.

Why the others are wrong

(a)
112 — 112 puts the denominator at 404 and the fraction at 11640 ⁄ 404, about 28.81. Only a denominator of 388 returns 30.
(b)
98 — 98 is the near miss, giving a denominator of 390 and a value near 29.85. The middle term of the identity is the product 16 × 6, which is 96.
(d)
108 — 108 makes the denominator a tidy-looking 400 and the fraction 29.1. The identity fixes the denominator at 16² + 16 × 6 + 6² = 388, not at a round number.

Concept

The whole question is the identity a³ − b³ = (a − b)(a² + ab + b²). The denominator has been printed as 16² + 6² + Q, which is that second factor with its middle term hidden, so Q has to be the product ab.

Seen that way the fraction reduces to 3(a − b) with no cubes computed at all: 3 × 10 = 30, matching the value the question states.

Reading the equation this way turns a four-digit arithmetic problem into a one-line substitution, which is the point of the item.

The stated value 30 is not decoration. It equals 3 × (16 − 6), which is what the fraction must come to once Q is right, so it doubles as a check: if your Q makes the left side anything other than 30, it is wrong.

Key facts

a³ − b³ = (a − b)(a² + ab + b²), with a plus sign on the middle term.
a³ + b³ = (a + b)(a² − ab + b²), where that middle term turns negative.
16² + 16 × 6 + 6² = 256 + 96 + 36 = 388, and 16³ − 6³ = 4096 − 216 = 3880 = 10 × 388.
The fraction 3(a³ − b³) ⁄ (a² + ab + b²) equals 3(a − b) for any a and b.

Study next

The identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)Getting x³ ± 1⁄x³ from a given x ± 1⁄xCancelling a difference of squares against a difference of cubes in one fraction

Common traps

Writing the middle term as 2ab, borrowed from the expansion of (a − b)².
Making the middle term negative, which belongs to the sum-of-cubes factorisation.
Computing 16³ = 4096 and 6³ = 216 the long way and running out of time.

The identity is set as a fraction that only cancels once you factorise, so the arithmetic stays small if you spot it.

A sibling identity carries 10 Sep 2024, 09:00, Quant Q.22, where the sum of three numbers and the sum of their squares are enough to fix the value of a³ + b³ + c³ − 3abc.

Related PYQs

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