SSC CGL 2025 Tier-I · 21 September 2025, Shift 3 · General Intelligence and Reasoning · Question 22 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

We doubt SSC's key on this question.

a² + ab + b², b³ − a³ and 3ab + 1 each fit both examples and give 5 $ 6 = 91, option (b); the keyed 89 needs the leftovers after a² + b² (12, 20) to keep rising by 8, a step two examples cannot establish.

The answer below is still the option SSC's key marks, and we solve to that option.

If 3 $ 4 = 37 and 4 $ 5 = 61, what is 5 $ 6?

(a)85
(b)91
(c)96
(d)89

Answer

Why

Correct — D. The key marks 89. The working below reaches it by extending a step seen in only two examples: an extrapolation, not a standard rule. The note on the key explains the doubt.

Compare each result with a² + b²:
3 $ 4: 3² + 4² = 25, leftover 37 − 25 = 12
4 $ 5: 4² + 5² = 41, leftover 61 − 41 = 20

The leftover rose by 8; extend it: 20 + 8 = 28
5 $ 6: 5² + 6² = 61
61 + 28 = 89 → option (d).

Why the others are wrong

(a)
85 — 85 is 61 + 24, the jump from 37 to 61 repeated as if 5 $ 6 were the next term of a series. After 5² + 6² = 61 it needs a leftover of 24; the key's +8 step gives 28.
(b)
91 — 91 needs a leftover of 30 after 5² + 6² = 61. The key's reading steps the leftovers 12, 20 by +8 to 28, not 30. SSC's key marks 89; see the note on the key.
(c)
96 — 96 needs a leftover of 35 after 5² + 6² = 61; the key's +8 step from 12, 20 gives 28.

Concept

An operator defined by examples is solved by finding one rule that fits every example, then applying it to the new pair. A useful first move is to compare each result with a² + b² and look at what is left over.

Two examples leave the rule open. Here the leftovers are 12 and 20, and two numbers do not fix how a pattern continues. The key's 89 follows if the leftover keeps rising by 8.

SSC's key, read from the colour-marked cell on the response sheet, sits on (d), 89. The stem on that sheet prints 37 and 61 in both English and Hindi, so the doubt in the note on the key is not a copying error.

Key facts

Compare an operator's results with a² + b² and study what is left over.
Here 3² + 4² = 25 and 4² + 5² = 41 leave 12 and 20.

Study next

Operators defined by three or more examplesNumber analogies built on a² + b² or b³ − a³Missing number in a figure from row or column rules

Common traps

Adding the last jump again (61 + 24 = 85), as if 5 $ 6 were the next term of a series.
Taking the first rule that fits both examples as settled without checking whether another rule fits too.

14 Sep 2025, 12:30, Reasoning Q.18 defines $ by two examples: 9 $ 3 = 18 and 8 $ 2 = 12 fit (a − b) × b, keyed 6 $ 4 = 8.

17 Sep 2025, 16:00, Reasoning Q.5 gives three examples of @ (3 @ 2 = 7, 6 @ 3 = 21, 8 @ 2 = 22), which fit a × b + (a − b), keyed 5 @ 4 = 21.

Related PYQs

No directly related past PYQ was found.

Topic: Symbolic/Number Analogy — all 169 SSC CGL questions on it, every solved shift.