SSC CGL 2024 Tier-I · 10 September 2024, Shift 2 · General Intelligence and Reasoning · Question 12 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 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.) (25, 300, 6) (19, 342, 9)

(a)(24, 600, 5)
(b)(14, 364, 7)
(c)(17, 476, 7)
(d)(18, 288, 8)

Answer

Why

Correct — D. Find one relation that ties all three numbers of both sample sets, not just two of them.

(25, 300, 6): 25 × 6 = 150, and 150 × 2 = 300
(19, 342, 9): 19 × 9 = 171, and 171 × 2 = 342

Rule: middle = first × third × 2.

Test the options on their outer numbers:

24 × 5 × 2 = 240
14 × 7 × 2 = 196
17 × 7 × 2 = 238
18 × 8 × 2 = 288 → option (d)

Why the others are wrong

(a)
(24, 600, 5) — 24 × 5 × 2 = 240, not 600. The printed 600 is 24 × 5 × 5, so this set uses a multiplier of 5 where the samples use 2.
(b)
(14, 364, 7) — 14 × 7 × 2 = 196. The printed 364 is not even a whole-number multiple of 14 × 7 = 98, so no constant multiplier reaches it.
(c)
(17, 476, 7) — 17 × 7 × 2 = 238, and the printed 476 is that doubled again. The set works on a multiplier of 4, twice what the samples fix.

Concept

A three-number set analogy asks for a relation that uses every number in the set. Any rule that ignores one of the three will let several options through.

Start by multiplying the outer pair and comparing the product with the middle number. If the ratio is the same in both sample sets, that ratio is the rule.

25 × 6 = 150 against a middle of 300, and 19 × 9 = 171 against 342, both give a ratio of 2.

Two of the wrong options are near misses on the same structure: option (a) multiplies by 5 and option (c) by 4. Anyone who checks only that the middle number is divisible by the outer two will find three options acceptable.

Key facts

In both sample sets the middle number is the product of the outer two, doubled.
25 × 6 = 150 and 19 × 9 = 171, against middles of 300 and 342.
For the answer, 18 × 8 = 144 and 144 × 2 = 288.

Study next

Three-number sets where the middle is the product of the outer twoSets built on n² ± n and n × (n + 1)Whole-number instructions in SSC Reasoning stems, and what they rule out

Common traps

Checking only that the middle number divides by the outer two, which options (a) and (c) also pass.
Multiplying the outer pair and stopping there, which gives 144 for the answer and makes it look wrong.
Splitting 300 into 3, 0 and 0, which the note in the stem forbids.

The same whole-number note heads a pair analogy at Q.2, a related-number item at Q.5 and an odd-one-out at Q.7 in this shift's Reasoning section (10 Sep 2024, 12:30).

Related PYQs

No directly related past PYQ was found.