SSC CGL 2024 Tier-I · 13 September 2024, Shift 3 · Quantitative Aptitude · Question 4 of 98

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.

Simplify (1.5 × 1.5 + 2.5 × 2.5 + 3.5 × 3.5 + 2 × 1.5 × 2.5 + 2 × 2.5 × 3.5 + 2 × 1.5 × 3.5) ⁄ (1.5 + 2.5 + 3.5).

(a)9.5
(b)7.5
(c)6.5
(d)8.5

Answer

Why

Correct — B. The numerator is a squared sum written out the long way. Put a = 1.5, b = 2.5, c = 3.5.

Identity: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

All six terms are present — the squares 1.5×1.5, 2.5×2.5, 3.5×3.5, and the doubled products 2×1.5×2.5, 2×2.5×3.5, 2×1.5×3.5.

Sum inside the square: 1.5 + 2.5 + 3.5 = 7.5
Numerator = 7.5² = 56.25
56.25 ⁄ 7.5 = 7.5 → option (b)

Shortcut: the whole thing is (a + b + c)² ⁄ (a + b + c), which collapses to a + b + c — you never have to compute 56.25 at all.

Why the others are wrong

(a)
9.5 — 9.5 is two more than 1.5 + 2.5 + 3.5. Once the numerator is read as (a + b + c)², the fraction equals that sum exactly, with nothing left over to add.
(c)
6.5 — 6.5 is one short. Check the addition itself: 1.5 + 2.5 = 4, and 4 + 3.5 = 7.5, which is both the denominator and the answer.
(d)
8.5 — 8.5 is one too many. The bracket inside the square and the denominator are the same three numbers, so the fraction can only come out as their total, 7.5.

Concept

When an expression lists three squares and three doubled cross-products of the same three numbers, it is (a + b + c)² written out. Recognising that turns a page of decimal multiplication into one addition.

Here the denominator is the very sum being squared, so the fraction reduces to (a + b + c)² ⁄ (a + b + c) = a + b + c. The decimals are never multiplied at all.

The check worth doing before you cancel is the count: three squares and three doubled products, six terms, each pair appearing once.

The question is set entirely in decimals so that brute force is slow and error-prone. Counting the six terms first costs about ten seconds and removes the arithmetic completely.

Key facts

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca — three squares and three doubled products.
1.5 + 2.5 + 3.5 = 7.5, and 7.5² = 56.25.
An expression of the form (a + b + c)² ⁄ (a + b + c) equals a + b + c whenever that sum is not zero.

Study next

Rearranging the same identity as a² + b² + c² = (a + b + c)² − 2(ab + bc + ca)Spotting (a − b)² and (a + b)(a − b) inside long expressionsa³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Common traps

Multiplying all six decimal terms out and losing a decimal place under time pressure.
Assuming the identity after checking only the three squares, when a cross term may be missing.
Cancelling the sum before confirming that every one of the six terms is actually there.

SSC files two different skills under the bare instruction Simplify. This one wants an algebraic identity spotted.

The other kind is a BODMAS chain, set at Quant Q.22 of this shift (16 ÷ 2 − 7 × 15 ÷ 3 + 5 × 5 + 4 × 3 − 10) and at 25 Sep 2024, 12:30, Quant Q.23, where nested brackets and 'of' carry the difficulty instead.

Related PYQs

No directly related past PYQ was found.