SSC CGL 2024 Tier-I · 23 September 2024, Shift 3 · Quantitative Aptitude · Question 7 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.

Let 3t − 1⁄(3t) = 3, then which of the following expressions has the value equal to 12?

Answer

Why

Correct — C. The stem, printed as an image, sets 3t − 1⁄(3t) = 3 and asks which listed expression equals 12.

Write x = 3t, so the condition is x − 1⁄x = 3 and 9t² + 1⁄(9t²) is exactly x² + 1⁄x².

Square both sides:
(x − 1⁄x)² = x² − 2 + 1⁄x² = 9
x² + 1⁄x² = 9 + 2 = 11

So 9t² + 1⁄(9t²) = 11, and the expression worth 12 is that quantity plus 1 → option (c).

Why the others are wrong

(a)
Option (a) is 9t² + 1⁄(9t²) − 2, which evaluates to 11 − 2 = 9. Subtracting 2 undoes the correction that the squaring introduced rather than applying it.
(b)
Option (b) is 9t² + 1⁄(9t²) + 2, which evaluates to 11 + 2 = 13. The +2 has already been used to lift 9 to 11, so adding it a second time overshoots.
(d)
Option (d) is 9t² + 1⁄(9t²) with nothing attached, and that is 11 — the squared expression itself, one short of the 12 asked for.

Concept

This is the x − 1⁄x family, and it rests on one identity: (x − 1⁄x)² = x² + 1⁄x² − 2.

Rearranged, x² + 1⁄x² = (x − 1⁄x)² + 2. The companion form (x + 1⁄x)² = x² + 1⁄x² + 2 differs only in that sign, which is why the +2 and −2 options are both on the list.

The 3t is cosmetic. Substituting x = 3t turns 9t² + 1⁄(9t²) into x² + 1⁄x², and the coefficient never has to be handled separately.

The value 12 asked for is x² + 1 + 1⁄x², which is the middle factor of x³ − 1⁄x³ = (x − 1⁄x)(x² + 1 + 1⁄x²). That gives 3 × 12 = 36 as the cube difference here, so the same 12 is reusable if a later part asks for 27t³ − 1⁄(27t³).

Key facts

(x − 1⁄x)² = x² + 1⁄x² − 2, so x² + 1⁄x² = (x − 1⁄x)² + 2.
(x + 1⁄x)² = x² + 1⁄x² + 2, the same identity with the middle sign reversed.
With 3t − 1⁄(3t) = 3, the quantity 9t² + 1⁄(9t²) equals 11.

Study next

The cube form x³ − 1⁄x³ = (x − 1⁄x)³ + 3(x − 1⁄x)Moving between x + 1⁄x and x − 1⁄x, whose squares always differ by 4Substituting x = kt to strip a coefficient off before applying a standard identity

Common traps

Squaring 3t − 1⁄(3t) = 3 and writing 9t² + 1⁄(9t²) = 9, forgetting the −2 cross term.
Squaring the 3t but leaving 1⁄(3t) unsquared, which produces 9t² + 1⁄(3t) and goes nowhere.

The condition is set on 3t rather than a bare x, so the substitution has to come first and the identity second. The same "which of the following expressions" framing on a linear condition is used at 9 Sep 2024, 16:00, Quant Q.8.

Related PYQs

No directly related past PYQ was found.

Topic: Basic algebraic identities of School Algebra & Elementary surds — all 130 SSC CGL questions on it, every solved shift.