SSC CGL 2024 Tier-I · 26 September 2024, Shift 2 · General Intelligence and Reasoning · Question 11 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 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.

Select the set in which the numbers are related in the same way as are the numbers of the following set. 3 — 20 — 5 4 — 17 — 3 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/subtracting/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)8 — 36 — 11
(b)5 — 28 — 9
(c)9 — 77 — 8
(d)11 — 28 — 5

Answer

Why

Correct — C. Multiply the outer two numbers and see what is left over.

3 — 20 — 5: 3 × 5 = 15, and 20 – 15 = 5
4 — 17 — 3: 4 × 3 = 12, and 17 – 12 = 5

Rule: middle = (first × third) + 5

Test option (c): 9 × 8 = 72, and 72 + 5 = 77, exactly the middle number given. The set fits.

Why the others are wrong

(a)
8 — 36 — 11 — 8 — 36 — 11 gives 8 × 11 = 88, and 88 + 5 = 93, not 36. Its middle number is smaller than the product, so no add-five rule can reach it.
(b)
5 — 28 — 9 — 5 — 28 — 9 gives 5 × 9 = 45, and 45 + 5 = 50, not 28. Again the middle number sits well below the product.
(d)
11 — 28 — 5 — 11 — 28 — 5 gives 11 × 5 = 55, and 55 + 5 = 60, not 28. The form of the set is right but the arithmetic misses by 32.

Concept

A three-number set item hands you two worked examples and asks for a third obeying the same arithmetic. Two examples is the minimum needed to rule out a coincidence, so always test your rule on both.

Start with the product of the outer numbers whenever the middle number is several times larger than either outer one.

3 × 5 = 15 against a middle of 20, and 4 × 3 = 12 against a middle of 17, each leave exactly 5 over. A constant leftover across two examples is the rule.

The NOTE forbidding you to split numbers into digits removes a whole family of alternative patterns, and it is there to narrow the search rather than to complicate it.

The paper writes these sets with long dashes rather than brackets, so 3 — 20 — 5 is one set of three numbers, not a subtraction. Reasoning Q.12 in this section prints the same idea in bracketed form.

Key facts

For both given sets the middle number equals the product of the outer two plus five.
3 × 5 + 5 = 20 and 4 × 3 + 5 = 17.
Option (c) satisfies the rule: 9 × 8 + 5 = 77.

Study next

Number-pair items where the second number is built from the first aloneSet items whose rule uses a square or a cube rather than a productRules of the form (a × b) − c, where the constant is subtracted instead of added

Common traps

Settling on a rule that fits the first set and never testing it against the second
Breaking the numbers into digits, which the NOTE explicitly forbids
Trying only sums and differences when the middle number is far larger than either outer number

The same NOTE about not breaking numbers into their digits appears at Reasoning Q.16 on 9 Sep 2024 (12:30), Reasoning Q.9 on 10 Sep 2024 (09:00) and Reasoning Q.5 on 10 Sep 2024 (16:00). Derive the rule from the two given sets every time rather than recalling one.

Related PYQs

No directly related past PYQ was found.