SSC CGL 2024 Tier-I · 18 September 2024, Shift 3 · General Intelligence and Reasoning · 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.

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.

Three of the following numbers are alike in a certain way and one is different. Pick the odd one out. (NOTE : Operations should be performed on the whole numbers, without breaking down the numbers into its constituent digits. E.g. 13 – Operations on 13 such as adding /deleting /multiplying etc., to 13 can be performed. Breaking down 13 into 1 and 3 and then performing mathematical operations on 1 and 3 is not allowed.)

(a)11 – 882
(b)3 – 72
(c)4 – 128
(d)7 – 392

Answer

Why

Correct — A.

Rule: the second number is 8 times the square of the first.

3 → 8 x 9 = 72
4 → 8 x 16 = 128
7 → 8 x 49 = 392

11 → 8 x 121 = 968, but the printed pair reads 882.

Three pairs obey the rule and one does not, so 11 – 882 is the odd pair — option (a).

Why the others are wrong

(b)
3 – 72 — 3 – 72 obeys the rule exactly: 8 x 3² = 72. It belongs with the majority, so it cannot be the one picked out.
(c)
4 – 128 — 4 – 128 fits as well: 8 x 4² = 128. Choosing it would leave three pairs that no longer share a single rule.
(d)
7 – 392 — 7 – 392 also fits: 8 x 7² = 392. The largest pair is not automatically the odd one, and here it is the best behaved.

Concept

'Pick the odd one out' guarantees that three of the four share one rule. Find the rule from whichever pairs are easiest to compute, then test the fourth against it.

The small numbers are the way in. 72 ÷ 9 = 8, 128 ÷ 16 = 8 and 392 ÷ 49 = 8, so each second number is eight times the square of the first.

Applying that to 11 demands 8 x 121 = 968. The paper prints 882, so that pair breaks the rule.

882 is 2 x 21², large enough to look plausible beside 968 if the rule is never actually tested.

Dividing the larger number by the square of the smaller is the fastest single test for this family, because it turns four separate checks into four divisions that either give the same constant or do not.

Key facts

8 x 3² = 72, 8 x 4² = 128 and 8 x 7² = 392.
8 x 11² = 968, not the 882 that the paper prints.
882 factorises as 2 x 21², which is why it looks like it could belong.
The bracketed note bars digit operations, so 11 must be treated as eleven.

Study next

Odd-one-out pairs built on kn² and kn³Dividing the larger number by the square or cube of the smaller as a first testOdd-one-out items that print three numbers per option instead of two

Common traps

Fixing the rule from the pair that turns out to be the odd one
Hunting for a difference rule when the pairs grow far too fast for one
Splitting 11 into its digits in spite of the note

The same shape — four pairs, three obeying one multiplier — is asked at 9 Sep 2024, 16:00, Reasoning Q.4, where 36 – 84, 66 – 154 and 72 – 168 all take seven thirds of the first number while 33 – 72 does not.

Related PYQs

No directly related past PYQ was found.