SSC CGL 2024 Tier-I · 13 September 2024, Shift 2 · Quantitative Aptitude · Question 14 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.

Let a + b = 1, then the value of (1⁄a² + 1⁄b² + 2⁄ab) is:

Answer

Why

Correct — B. Put the three terms over one denominator and the numerator turns into a perfect square.

Common denominator is a²b²:
1/a² + 1/b² + 2/(ab) = (b² + a² + 2ab)/(a²b²)

The numerator is the expansion of (a + b)²:
= (a + b)²/(a²b²)

Substitute a + b = 1:
= 1²/(a²b²) = 1/(a²b²) → option (b)

The same result seen faster: the expression is (1/a + 1/b)², and 1/a + 1/b = (a + b)/ab = 1/ab.

Why the others are wrong

(a)
1/(a²b) drops a factor of b from the denominator. The common denominator of 1/a², 1/b² and 2/(ab) is a²b², and both squares survive, because only the numerator ever uses a + b = 1.
(c)
1/(ab²) drops a factor of a in the same way. Neither letter can fall to the first power: the denominator a² × b² is fixed before the condition a + b = 1 is applied at all.
(d)
1/(ab) is the square root of the answer rather than the answer. The expression equals (1/a + 1/b)², and 1/a + 1/b works out to 1/ab, so the value is 1/ab squared.

Concept

The pattern to spot is x² + y² + 2xy = (x + y)², taken with x = 1/a and y = 1/b.

Read that way, 1/a² + 1/b² + 2/(ab) is (1/a + 1/b)², and 1/a + 1/b = (a + b)/ab. The given a + b = 1 turns the bracket into 1/ab, so the whole expression is 1/(a²b²).

Notice what the condition does and does not do. It fixes the numerator only. The denominator a²b² is untouched by it, which is why every option that lowers a power of a or of b is wrong on structure, before any arithmetic.

The stem and all four options are printed as images on the response sheet. The answer stays in terms of a and b, because a + b = 1 is not enough to pin the expression to a single number — ab can be anything up to 1/4.

Key facts

x² + y² + 2xy factorises as (x + y)², and reading the identity backwards is how a hidden square is spotted.
1/a + 1/b = (a + b)/ab, which is why a condition on a + b simplifies a sum of reciprocals.
With a + b = 1, the product ab is at most 1/4, reached when a = b = 1/2.

Study next

Writing 1/a² + 1/b² as ((a + b)² − 2ab)/(ab)² and reading off values from a + b and abSymmetric expressions expressed entirely in terms of a + b and abQuestions that supply x + 1/x and ask for x² + 1/x² or x³ + 1/x³

Common traps

Substituting a + b = 1 into the denominator, which the condition says nothing about
Adding the three fractions over a²b instead of a²b²
Stopping at (a + b)²/(a²b²) without using the value that was given

A condition on a + b feeding a symmetric expression is asked at 9 Sep 2024, 16:00, Quant Q.8, where a + b = 2c is given and the four options are expressions to be tested.

Related PYQs

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