SSC CGL 2024 Tier-I · 9 September 2024, Shift 3 · 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 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.

Let a + b = 2c, then which of the following expressions is true?

Answer

Why

Correct — D. The four options are printed as images. Option (d) reads a² + 2bc = b² + 2ac.

Test it against the given a + b = 2c.

Bring the squares to one side and the products to the other:
a² − b² = 2ac − 2bc

Factorise both sides:
(a − b)(a + b) = 2c(a − b)

Substitute a + b = 2c into the left side:
(a − b)(2c) = 2c(a − b)

The two sides are identical, so this holds for every a, b, c satisfying the condition.

Numerical check with a = 3, b = 1, c = 2 (which gives a + b = 4 = 2c): left = 9 + 4 = 13, right = 1 + 12 = 13 → option (d).

Why the others are wrong

(a)
Option (a) shows a² − ac = b² − bc. It rearranges to (a − b)(a + b − c) = 0, which under a + b = 2c becomes c(a − b) = 0 — true only if a = b. At a = 3, b = 1, c = 2 it gives 3 against −1.
(b)
Option (b) shows a² + 2ac = b² + 2bc. It reduces to (a − b)(a + b + 2c) = 0, that is 4c(a − b) = 0, so again it needs a = b. At a = 3, b = 1, c = 2 it gives 21 against 5.
(c)
Option (c) shows a² − 2bc = b² − 2ac, which collapses to the same 4c(a − b) = 0 condition and so fails for the same reason. At a = 3, b = 1, c = 2 it gives 5 against −11.

Concept

Every option here turns on one identity, the difference of squares: a² − b² = (a − b)(a + b).

Gather the squares on one side, factor (a − b) out of both sides, and the given a + b = 2c decides whether what is left is an identity or a special case.

Only the pairing that sends 2bc and 2ac across the equals sign from their own squares survives, because that is the arrangement whose right side factors to exactly 2c(a − b).

The other three reduce to (a − b) × (something non-zero) = 0, which forces a = b — a special case, not a general truth.

The options are pictures rather than text in this paper: (a) a² − ac = b² − bc, (b) a² + 2ac = b² + 2bc, (c) a² − 2bc = b² − 2ac, (d) a² + 2bc = b² + 2ac. Substituting a = 3, b = 1, c = 2 separates all four in well under a minute if the algebra does not come quickly.

Key facts

a² − b² = (a − b)(a + b) is the identity all four options turn on.
a + b = 2c says that c is the arithmetic mean of a and b.
Under a + b = 2c, the relation a² + 2bc = b² + 2ac holds for all such a, b and c.
In a "which is true" item, one counter-example is enough to eliminate an option outright.

Study next

Conditional identities solved by factoring rather than expandingDifference of squares used to simplify symmetric expressionsChoosing a legal test substitution when the algebra is slow

Common traps

Testing a = b = c = 1, which satisfies a + b = 2c and makes all four options true, proving nothing.
Cancelling (a − b) from both sides, which quietly hides the case a = b.
Expanding everything instead of factoring, which buries the (a − b) that does the work.

"Given this relation, which of the following is true" rewards a substitution over a derivation, provided the numbers you pick genuinely satisfy the condition and are not all equal to each other.

Related PYQs

No directly related past PYQ was found.