SSC CGL 2024 Tier-I · 25 September 2024, Shift 1 · Quantitative Aptitude · Question 18 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.

If x² + 7x + 8 = 0, then find the value of 4x ⁄ (x² − 5x + 8).

Answer

Why

Correct — A. The stem never asks you to find x — it hands you a relation to substitute.

x² + 7x + 8 = 0 → x² + 8 = −7x

Denominator: x² − 5x + 8 = (x² + 8) − 5x
= −7x − 5x = −12x

4x ⁄ (−12x) = −1⁄3 → option (a), the picture showing −1 over 3.

x = 0 is not a root (it would force 8 = 0), so cancelling x is legitimate and the value is the same for either root.

Why the others are wrong

(b)
Option (b) shows 1⁄6, a positive value. The denominator collapses to −12x, so 4x over −12x is negative whichever root you take — the x cancels against itself and the sign cannot flip.
(c)
Option (c) shows 1⁄2, which would need the denominator to come out as 8x. Substituting x² + 8 = −7x gives −12x, and 4x ⁄ (−12x) is −1⁄3.
(d)
Option (d) shows −1⁄4 — right sign, wrong multiple. −1⁄4 needs a denominator of −16x, but the substitution produces −12x.

Concept

The question is a substitution test wearing a quadratic's clothes.

Rearranged, the given equation says x² + 8 = −7x. The denominator contains that exact block, x² + 8, with a −5x sitting beside it — so one substitution turns a quadratic denominator into a single term in x.

Once numerator and denominator are both multiples of x, the x cancels and the answer is a pure number, the same for both roots.

Solving the quadratic is the slow road: its discriminant is 7² − 4 × 8 = 17, so both roots are irrational.

The constant in the denominator, 8, is the same 8 as in the given equation, and that match is what makes the substitution clear the whole denominator in one move.

Had the denominator read x² − 5x + 10, you would still write x² + 8 = −7x first and then add the extra 2 by hand.

Key facts

x² + 7x + 8 = 0 rearranges to x² + 8 = −7x, and that single substitution is the whole question.
Substituting gives x² − 5x + 8 = −12x, so the expression is 4x ⁄ (−12x) = −1⁄3.
The discriminant 7² − 4 × 1 × 8 = 17 is not a perfect square, so neither root is rational.
Neither root is 0, which is what makes cancelling x in the fraction valid.

Study next

Valuing rational expressions by substituting a given quadratic relation instead of solving itSum and product of roots, x₁ + x₂ = −b⁄a and x₁x₂ = c⁄a, used on symmetric expressionsIdentities for x + 1⁄x and x³ + 1⁄x³, which run on the same substitute-do-not-solve idea

Common traps

Reaching for the quadratic formula and then grinding √17 through the fraction
Cancelling 4x against −12x but leaving the minus sign with the numerator and answering +1⁄3
Assuming the value depends on which root you pick, and computing the fraction twice

The lever is always the same: rearrange the given equation until the square and the constant sit together, then drop that block into the expression.

The same substitute-rather-than-solve reading is wanted at Quant Q.19 of this paper, where a³ + 1⁄a³ = 2 has to be turned into a + 1⁄a.

Related PYQs

No directly related past PYQ was found.