SSC CGL 2024 Tier-I · 13 September 2024, Shift 1 · General Intelligence and Reasoning · Question 20 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 sets. (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.) (27, 405, 3) (23, 690, 6)

(a)(14, 364, 7)
(b)(16, 448, 7)
(c)(22, 616, 7)
(d)(24, 600, 5)

Answer

Why

Correct — D. Divide the middle number by the first, then see what the quotient is made of.

(27, 405, 3): 405 ÷ 27 = 15, and 15 = 3 × 5
(23, 690, 6): 690 ÷ 23 = 30, and 30 = 6 × 5

Rule: middle = first × third × 5.

Testing option (d): 24 × 5 × 5 = 24 × 25 = 600, exactly the middle number printed → option (d).

Why the others are wrong

(a)
(14, 364, 7) — Option (a) needs 14 × 7 × 5 = 490, but the set prints 364. Its own quotient, 364 ÷ 14 = 26, is not five times the third number.
(b)
(16, 448, 7) — Option (b) needs 16 × 7 × 5 = 560, against the printed 448. Here 448 ÷ 16 = 28, well short of the required 35.
(c)
(22, 616, 7) — Option (c) needs 22 × 7 × 5 = 770, against the printed 616. Its quotient is 616 ÷ 22 = 28, not the 35 the rule demands.

Concept

A set analogy gives two worked examples and asks which fifth set obeys the same relation. With three numbers per set, the reliable first move is division, not subtraction — the middle number dwarfs the outer two, so a product is the likely link.

Compute middle ÷ first for both given sets. If those quotients come out whole, factor them against the third number: 15 against 3 and 30 against 6 both leave a constant 5, and the rule is fixed in two lines.

SSC's note bars digit-splitting, so 27 stays 27 and never becomes 2 and 7.

The constant is the whole question. All four options divide cleanly by their first number — 364 ÷ 14 = 26, 448 ÷ 16 = 28, 616 ÷ 22 = 28 and 600 ÷ 24 = 25 — so divisibility alone eliminates nothing.

Key facts

In both given sets, middle ÷ first equals five times the third number.
405 ÷ 27 = 15 = 3 × 5, and 690 ÷ 23 = 30 = 6 × 5.
600 ÷ 24 = 25 = 5 × 5, which is why option (d) fits.

Study next

Three-number set analogies where the constant is 2 or 3 instead of 5Two-number set analogies built on a constant difference

Common traps

Checking only that the middle number divides by the first, without testing the constant.
Trying differences first, which get nowhere when the middle number is an order of magnitude larger.
Splitting 27 or 23 into digits, which the question's note forbids.

SSC recycles the option strings as well as the template. The triple (24, 600, 5) is also printed at 10 Sep 2024, 12:30, Reasoning Q.12 — there as option (a), where the constant is 2 and the set is wrong.

The same template with a constant of 3 runs at 18 Sep 2024, 16:00, Reasoning Q.3, so recognising the triple is no substitute for computing its constant.

Related PYQs

No directly related past PYQ was found.