SSC CGL 2024 Tier-I · 12 September 2024, Shift 1 · Quantitative Aptitude · Question 16 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 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.

If x + 1⁄x = 15, then the value of (7x² − 9x + 7) ⁄ (x² − x + 1) is:

Answer

Why

Correct — D. Divide numerator and denominator by x so the whole thing is written in x + 1⁄x.

x + 1⁄x = 15 forces x ≠ 0, so dividing through by x is legal.

Numerator ÷ x = 7x − 9 + 7⁄x = 7(x + 1⁄x) − 9
= 7(15) − 9 = 105 − 9 = 96

Denominator ÷ x = x − 1 + 1⁄x = (x + 1⁄x) − 1
= 15 − 1 = 14

Value = 96⁄14 = 48⁄7 → option (d). The value of x itself is never needed.

Why the others are wrong

(a)
−22⁄7 is wrong in both size and sign. Numerator 96 and denominator 14 are both positive, so the value cannot come out negative.
(b)
22⁄7 would need a numerator of 44, since 44⁄14 = 22⁄7. The numerator here is 7(x + 1⁄x) − 9 = 96.
(c)
−48⁄7 has the right size and the wrong sign. The slip is writing the denominator as 1 − (x + 1⁄x) = −14 instead of (x + 1⁄x) − 1 = 14.

Concept

Both quadratics here are symmetric: 7x² − 9x + 7 has the same first and last coefficient, and x² − x + 1 does too.

That symmetry is the signal to divide every term by x. Doing so converts each quadratic into an expression in the single quantity x + 1⁄x, whose value you were handed.

Numerator becomes 7(x + 1⁄x) − 9; denominator becomes (x + 1⁄x) − 1. Substituting 15 finishes it in two lines.

Solving x² − 15x + 1 = 0 for x also works, but it produces surds and wastes the gift in the stem.

The stem and all four options are printed as images, so a text-only view of this row is blank. The options read −22⁄7, 22⁄7, −48⁄7 and 48⁄7 — two sign pairs, which is why a single sign slip has a landing spot.

Key facts

x + 1⁄x = 15 rules out x = 0, which is what makes dividing through by x valid.
(7x² − 9x + 7) ÷ x = 7(x + 1⁄x) − 9.
(x² − x + 1) ÷ x = (x + 1⁄x) − 1.
96⁄14 reduces to 48⁄7.

Study next

Getting x² + 1⁄x² and x³ + 1⁄x³ from a given x + 1⁄xDividing a symmetric quadratic through by x to reduce it to one variablex − 1⁄x from x + 1⁄x, and the sign ambiguity that introduces

Common traps

Dividing the numerator by x and forgetting to divide the denominator too.
Writing the denominator as 1 − (x + 1⁄x), which flips the sign of the answer.
Solving the quadratic for x and substituting the surd, which is slow and where the arithmetic usually breaks.

SSC prints these as a ratio of two quadratics with symmetric coefficients, so the item collapses the moment you divide by x rather than solve for it.

Also asked at Quant Q.10 of this paper, where t³ + (3⁄5)³ + (9⁄5)t is a cube identity written out rather than three terms to be evaluated one by one.

Related PYQs

No directly related past PYQ was found.