SSC CGL 2024 Tier-I · 10 September 2024, Shift 1 · Quantitative Aptitude · Question 18 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 10 September 2026.

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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 + 1⁄a = 12, then find the value of a² + 1⁄a².

(a)146
(b)140
(c)142
(d)144

Answer

Why

Correct — C. The stem is printed as an image: if a + 1⁄a = 12, find a² + 1⁄a². Square the given relation and watch the cross term.

(a + 1⁄a)² = a² + 2·a·(1⁄a) + 1⁄a²
= a² + 1⁄a² + 2, because a·(1⁄a) = 1

Left side = 12² = 144
So a² + 1⁄a² = 144 − 2 = 142 → option (c).

Why the others are wrong

(a)
146 — 146 adds the 2 instead of removing it. Squaring gives 144 = a² + 1⁄a² + 2, so the cross term has to come off the 144, not go on to it.
(b)
140 — 140 takes off 4 rather than 2. The middle term of the expansion is 2·a·(1⁄a), and a·(1⁄a) is 1, so the constant to remove is exactly 2.
(d)
144 — 144 is only 12², the square of the left-hand side. It is the full expansion a² + 1⁄a² + 2, so it still carries the cross term that must be stripped out.

Concept

This is the squaring identity for a reciprocal pair: (a + 1⁄a)² = a² + 1⁄a² + 2.

It works because a and 1⁄a multiply to 1, so the cross term 2·a·(1⁄a) collapses to the constant 2 and no unknown survives it.

The mirror form is (a − 1⁄a)² = a² + 1⁄a² − 2. So a² + 1⁄a² sits 2 below one square and 2 above the other, and the two squares differ by 4.

You could solve for a: a² − 12a + 1 = 0 gives a = 6 ± √35.

That is real but irrational, and squaring it back is slow and error-prone. The identity route never touches the surd, which is the point of asking the question this way.

Key facts

(a + 1⁄a)² = a² + 1⁄a² + 2, so a² + 1⁄a² = (a + 1⁄a)² − 2.
(a − 1⁄a)² = a² + 1⁄a² − 2, so the two squares always differ by exactly 4.
a³ + 1⁄a³ = (a + 1⁄a)³ − 3(a + 1⁄a), the cube form built from the same given.
a + 1⁄a = 12 means a² − 12a + 1 = 0, so a = 6 ± √35.

Study next

a³ + 1⁄a³ from a + 1⁄a using the cube identitya⁴ + 1⁄a⁴ by squaring a² + 1⁄a² a second timeFinding a − 1⁄a from a + 1⁄a, and arguing the sign of the root

Common traps

Adding the 2 instead of subtracting it, which turns 142 into 146.
Doubling the constant to 4 because the expansion appears to hold two cross terms.
Solving a² − 12a + 1 = 0 for a = 6 ± √35 and grinding the surd through the square.

The given is a reciprocal pair and the target is a power of that pair, so one squaring step serves the square, and one cubing step serves the cube.

Recognise the shape from the a and 1⁄a pairing rather than from the letter used.

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