SSC CGL 2024 Tier-I · 23 September 2024, Shift 2 · Quantitative Aptitude · Question 13 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 a + b + c = 0, then the value of (a² + b² + 2ab) is equal to:

Answer

Why

Correct — B. The stem is printed as an image: if a + b + c = 0, find the value of (a² + b² + 2ab).

The bracket is a perfect square: a² + b² + 2ab = (a + b)²

From the condition, a + b = −c

Square both sides: (a + b)² = (−c)² = c² → option (b).

Why the others are wrong

(a)
−c is the value of a + b itself, not of its square. The bracket is one squaring step beyond that.
(c)
c would require (a + b)² = c, which holds only for special values such as c = 0 or c = 1, not for every triple summing to zero.
(d)
−c² keeps the minus sign outside the square. The sign is inside it: (−c)² = +c², and a real square is never negative.

Concept

Two standard moves solve the whole item.

First, recognise the identity: a² + 2ab + b² is the expansion of (a + b)², so the three-term bracket is really one squared quantity.

Second, use the linear condition: a + b + c = 0 lets any two variables be replaced by the negative of the third, so a + b = −c, b + c = −a and c + a = −b.

Substituting the second into the first collapses a two-variable expression into c².

Option (b) and option (d) differ only by a sign, which is where the item is actually decided.

Key facts

a² + b² + 2ab is (a + b)² written out.
If a + b + c = 0 then a + b = −c, b + c = −a and c + a = −b.
The same condition gives a³ + b³ + c³ = 3abc.
(−c)² = c², so squaring erases a minus sign.

Study next

a³ + b³ + c³ − 3abc factorised as (a + b + c)(a² + b² + c² − ab − bc − ca)Expressing a² + b² + c² through ab + bc + ca when the sum of the variables is zeroSubstituting a linear condition to remove a variable before expanding

Common traps

Expanding the condition instead of the bracket and losing the 2ab term.
Carrying the minus sign outside the square and answering −c².
Misreading the bracket as (a + b + c)² and answering 0.

An identity item hands you a linear condition and an expression that collapses under it, so the work is recognition rather than algebra. The trigonometric identity item at Quant Q.18 of this paper rewards the same substitute-the-condition move.

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