SSC CGL 2024 Tier-I · 9 September 2024, Shift 1 · Quantitative Aptitude · Question 23 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 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.

Which of the following can be the value of k, if (88 ÷ 8 × k − 3 × 3) ⁄ (6² − 7 × 5 + k²) = 1?

(a)2,7
(b)4,7
(c)3,10
(d)1,10

Answer

Why

Correct — D. The stem is an image: Which of the following can be the value of k, if (88 ÷ 8 × k − 3 × 3)/(6² − 7 × 5 + k²) = 1?

Clear each line by BODMAS first.
Numerator: 88 ÷ 8 = 11, so 88 ÷ 8 × k = 11k, and 3 × 3 = 9 → 11k − 9
Denominator: 6² = 36 and 7 × 5 = 35, so 36 − 35 = 1 → 1 + k²

A fraction equals 1 only when the two lines are equal:
11k − 9 = 1 + k²
k² − 11k + 10 = 0
(k − 1)(k − 10) = 0

k = 1 or 10 → option (d).

Check k = 1: (11 − 9)/(1 + 1) = 2/2 = 1.

Why the others are wrong

(a)
2,7 — Neither value fits. k = 2 gives (22 − 9)/(1 + 4) = 13/5, and k = 7 gives 34/25. The fraction has to come out at exactly 1.
(b)
4,7 — 4 + 7 = 11 does match the sum of the roots, but the product must be 10 and 4 × 7 = 28. Substituting k = 4 gives 35/17, not 1.
(c)
3,10 — Only 10 is a root. k = 3 gives (33 − 9)/(1 + 9) = 2.4, so the pair fails even though its second member is right.

Concept

Two ideas are stacked here, and the order matters.

First BODMAS: division and multiplication run left to right, so 88 ÷ 8 × k is (88 ÷ 8) × k = 11k, not 88 ÷ (8k). Getting that wrong changes the equation entirely.

Second, a fraction equals 1 exactly when numerator and denominator are equal and non-zero. Setting 11k − 9 = 1 + k² turns the whole thing into the quadratic k² − 11k + 10 = 0.

It factorises on sight because 1 and 10 multiply to 10 and add to 11.

The stem says "can be the value of k", which is why an option carrying both roots is acceptable. The denominator 1 + k² is never zero for real k, so neither root has to be discarded.

Key facts

In BODMAS, × and ÷ have equal precedence and are applied left to right
88 ÷ 8 × k evaluates to 11k
6² − 7 × 5 = 36 − 35 = 1, so the denominator is 1 + k²
k² − 11k + 10 = (k − 1)(k − 10), giving roots 1 and 10

Study next

BODMAS with a left-to-right chain of × and ÷Factorising a quadratic by splitting the middle termSum and product of roots: for k² − bk + c, the roots add to b and multiply to c

Common traps

Reading 88 ÷ 8 × k as 88 ÷ (8k) and ending with a different quadratic
Computing 6² − 7 × 5 as (6² − 7) × 5
Accepting an option because one of its two numbers is a root

A which-value-of-k item usually hides an equation behind an arithmetic string, and the string is the part candidates rush.

This shift reuses the same phrasing at 09 Sep 2024, 09:00, Quant Q.11, though there k is a single digit and the test is divisibility by 6 rather than a quadratic.

Related PYQs

No directly related past PYQ was found.