SSC CGL 2024 Tier-I · 17 September 2024, Shift 2 · General Intelligence and Reasoning · Question 22 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.

Select the option in which the numbers share the same relationship as that shared by the given pair of numbers. (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.) 8 : 79 6 : 51

(a)7 : 62
(b)4 : 32
(c)9 : 96
(d)5 : 39

Answer

Why

Correct — C. Two anchor pairs are printed, so settle the rule on both of them before testing any option.

79 − 51 = 28, and 8² − 6² = 64 − 36 = 28, so the rule is built on the square.
79 − 8² = 79 − 64 = 15, and 51 − 6² = 51 − 36 = 15.

Rule: n → n² + 15.

9² = 81 → 81 + 15 = 96 → option (c).

Why the others are wrong

(a)
7 : 62 — 7² + 15 = 64, so the partner of 7 would be 64. The printed pair is 7 : 62, two short.
(b)
4 : 32 — 4² + 15 = 31, so the partner of 4 would be 31. The printed pair is 4 : 32, one over.
(d)
5 : 39 — 5² + 15 = 40, so the partner of 5 would be 40. The printed pair is 5 : 39, one short.

Concept

When a pair analogy gives you two anchors, the fastest test is whether the difference of the outputs matches the difference of the squares of the inputs.

Here 79 − 51 = 28 and 8² − 6² = 28. That single match points straight at a square rule and saves you from testing multipliers one by one.

Then read the constant off either anchor — 79 − 64 = 15 — and confirm on the other: 36 + 15 = 51. Only after both anchors agree do you touch the options.

Two pairs do not uniquely determine a rule, and this item is a clean example. The linear map n → 14n − 33 also sends 8 to 79 and 6 to 51.

But it sends 9 to 93, 7 to 65, 4 to 23 and 5 to 37, none of which is printed. The square rule is the one that lands on an option, and that is how these items are meant to be resolved.

Key facts

8 : 79 and 6 : 51 both fit n → n² + 15.
The difference of the outputs, 28, equals 8² − 6², which is what identifies the rule as a square.
9² + 15 = 96, so 9 : 96 is the matching pair.

Study next

Number-pair analogies on n² ± k and n³ ± kUsing the difference of two anchors to choose between a linear rule and a square ruleSquares to 30 and cubes to 15 held at recall speed

Common traps

Committing to a linear rule because one exists. n → 14n − 33 fits both anchors but reaches no option.
Testing one option and stopping. 4 : 32 sits one away from 4² + 15 = 31 and reads as a match at speed.
Splitting the numbers into digits, which the NOTE in the stem bans.

The stem and its NOTE are boilerplate, the two anchors sit at the end of the paragraph, and the rule changes shape from shift to shift.

Also asked 11 Sep 2024, 12:30, Reasoning Q.8, where 186 : 103 and 156 : 73 both run n − 83 and the keyed option is 95 : 12, and 11 Sep 2024, 16:00, Reasoning Q.11, where 17 : 253 and 19 : 271 run n → 9n + 100 and the keyed option is 21 : 289. Neither is a square, so test more than one shape.

Related PYQs

No directly related past PYQ was found.