SSC CGL 2024 Tier-I · 18 September 2024, Shift 3 · General Intelligence and Reasoning · Question 17 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.

40 is related to 53 following certain logic. Following the same logic, 8 is related to 13. To which of the following numbers is 68 related, following the same logic? (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)84
(b)85
(c)83
(d)73

Answer

Why

Correct — B.

Rule: each number is a perfect square plus 4, and its partner is the next square plus 4.

40 = 6² + 4 → 7² + 4 = 53
8 = 2² + 4 → 3² + 4 = 13

68 = 8² + 4 → 9² + 4 = 81 + 4 = 85 → option (b)

Subtract 4 from all four given numbers and the pattern is bare: 36 → 49, and 4 → 9.

Why the others are wrong

(a)
84 — 84 − 4 = 80, which is not a perfect square, so 84 cannot be any number's partner under this rule. It is 68 + 16, one short of the true step of 2n + 1 = 17.
(c)
83 — 83 − 4 = 79, again not a square. Adding a flat 15 to 68 ignores that the step grows with the number: it was 5 for 8 and 13 for 40.
(d)
73 — 73 − 4 = 69, not a square. 73 is 68 + 5, which borrows the step from the small pair 8 to 13 instead of computing 2n + 1 for n = 8.

Concept

The bracketed note forbids splitting a number into digits, which is SSC telling you the rule is arithmetic on the whole number.

Two worked pairs give four numbers to test: 40, 53, 8 and 13. Trying a small constant first is the cheapest move, and subtracting 4 turns all four into 36, 49, 4 and 9 — consecutive squares within each pair.

That fixes the rule as n² + 4 going to (n + 1)² + 4. Since 68 = 8² + 4, its partner is 9² + 4.

The step size is therefore 2n + 1, which is why the two worked pairs move by different amounts.

The distractors are all close to 85 and none of them is four more than a square, so the square test alone eliminates three options without any further work.

Key facts

40 = 6² + 4, 53 = 7² + 4, 8 = 2² + 4 and 13 = 3² + 4.
68 = 8² + 4, so its partner is 9² + 4 = 85.
The gap within a pair is 2n + 1: 5 for 8, 13 for 40 and 17 for 68.
The note bars digit operations, which rules out reading 68 as a 6 and an 8.

Study next

Number analogies solved by subtracting a constant to expose squares or cubesRecognising the n² ± k and n³ ± k families at speedWhy the 'do not break the number into digits' note narrows the search

Common traps

Carrying the difference from one worked pair straight to the third number
Testing differences only, and stopping when they turn out unequal
Splitting 68 into 6 and 8 in spite of the note

The same square-plus-a-constant family appears at 10 Sep 2024, 16:00, Reasoning Q.17, where 62 → 79 and 98 → 119 are 8² − 2 → 9² − 2 and 10² − 2 → 11² − 2. A plainer ratio version is at 10 Sep 2024, 09:00, Reasoning Q.6, with 10 → 15 and 42 → 63.

Related PYQs

No directly related past PYQ was found.