SSC CGL 2024 Tier-I · 18 September 2024, Shift 2 · Quantitative Aptitude · Question 2 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. [(7.3)³ − (4.7)³] ⁄ [(7.3)² + 7.3 × 4.7 + (4.7)²]

(a)3.2
(b)2.6
(c)11
(d)12

Answer

Why

Correct — B. The denominator is the second bracket of the difference-of-cubes identity: a³ − b³ = (a − b)(a² + ab + b²).

With a = 7.3 and b = 4.7, the printed denominator is exactly a² + ab + b².

(a³ − b³) ⁄ (a² + ab + b²) = a − b
= 7.3 − 4.7 = 2.6 → option (b).

Why the others are wrong

(a)
3.2 — 3.2 is not a − b for these numbers. 7.3 − 4.7 = 2.6, and once the bracket cancels there is nothing else left to adjust.
(c)
11 — 11 is neither the difference nor the sum: 7.3 − 4.7 = 2.6 and 7.3 + 4.7 = 12. No step of the identity produces 11.
(d)
12 — 12 is a + b, which belongs to a³ + b³ = (a + b)(a² − ab + b²). That form needs a minus ab in the middle, and the paper prints + 7.3 × 4.7.

Concept

Two factorisations carry most of SSC's simplify-the-fraction algebra:

a³ − b³ = (a − b)(a² + ab + b²)
a³ + b³ = (a + b)(a² − ab + b²)

The sign of the middle term of the quadratic bracket tells you which one you are holding: + ab goes with the difference of cubes, − ab with the sum.

Spot the pattern and the decimals never get cubed — the quadratic bracket cancels and only a − b survives.

Nothing here depends on the numbers being decimals. The same cancellation runs identically for 73 and 47, or for any pair of values.

Key facts

a³ − b³ = (a − b)(a² + ab + b²), so dividing a³ − b³ by a² + ab + b² leaves a − b.
a³ + b³ = (a + b)(a² − ab + b²), where the middle term carries a minus sign.
With a = 7.3 and b = 4.7 the whole fraction reduces to 7.3 − 4.7 = 2.6.

Study next

The a³ ± b³ factorisations and which way the middle sign runsa² − b² = (a + b)(a − b) applied to decimal pairsa³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Common traps

Cubing 7.3 and 4.7 by hand and losing a decimal place under time pressure
Reading the + 7.3 × 4.7 in the denominator as the sum-of-cubes pattern and answering 12

SSC reuses this exact cancellation with fresh decimals. At 11 Sep 2024, 12:30, Quant Q.16 the fraction is [(4.2)³ − (1.2)³] ⁄ [(4.2)² + 5.04 + (1.2)²], where 5.04 is 4.2 × 1.2 in disguise and the value is 4.2 − 1.2.

At 10 Sep 2024, 12:30, Quant Q.6 the second number is hidden as a decimal: 0.001 is 0.1³, 0.058 is 0.58 × 0.1 and 0.01 is 0.1², so the answer is 0.58 − 0.1.

The stem text of this row is empty — the word Simplify and the whole fraction sit inside the image. That same layout carries the arithmetic simplifications at 17 Sep 2024, 09:00, Quant Q.13 and at 26 Sep 2024, 16:00, Quant Q.16.

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