SSC CGL 2024 Tier-I · 25 September 2024, Shift 1 · General Intelligence and Reasoning · Question 25 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 option that is related to the fifth number in the same way as the second number is related to the first number and the fourth number is related to the third number. 4 : 68 :: 8 : 136 :: 9 : ? (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.)

(a)147
(b)149
(c)153
(d)152

Answer

Why

Correct — C. The second number dwarfs the first, so check the ratio before the difference.

68 ÷ 4 = 17
136 ÷ 8 = 17

Rule: the second number is the first multiplied by 17.

9 × 17 = 9 × 10 + 9 × 7 = 90 + 63 = 153 — option (c).

Why the others are wrong

(a)
147 — 147 is not a multiple of 9 — its factors are 3, 7 and 7. The relation multiplies the first number by 17, so the answer has to be a multiple of 9, and 153 is.
(b)
149 — 149 is prime, so it is a multiple of neither 9 nor 17. It is four short of 153 and survives only a rushed multiplication.
(d)
152 — 152 is one short of 9 × 17. Taking 9 × 17 as 9 × 15 + 9 × 2 gives 135 + 18 = 153, so the slip behind 152 is arithmetic rather than a wrong rule.

Concept

When the second number is far larger than the first, test the ratio before the difference.

68 ÷ 4 = 17 and 136 ÷ 8 = 17, so the multiplier is constant and two pairs settle the relation. Had the ratios disagreed, the next tests would be first × k + c, then squares and cubes.

The NOTE matters here as well: 4 : 68 must not be read as 4 giving 6 and 8. Every operation acts on the whole number.

Two pairs are supplied so the rule can be confirmed rather than guessed. Test your rule on the second pair before using it on the third — a single pair supports several rules, and 4 : 68 on its own also fits 4 cubed plus 4.

Key facts

68 = 4 × 17 and 136 = 8 × 17, which fixes the multiplier at 17.
9 × 17 = 153, the keyed value.
4 : 68 on its own also fits 4 cubed plus 4, a reading the pair 8 : 136 rules out.

Study next

Three-pair number analogies in which the third pair carries the blankChoosing between a constant multiplier and a constant differenceMultiplying by 17 as ten times the number plus seven times the number

Common traps

Working from the difference, 68 − 4 = 64, and forcing it onto 9
Fixing the rule from the first pair alone
Treating the NOTE as decoration and splitting 68 into 6 and 8

The three-pair frame recurs with different arithmetic. At 19 Sep 2024, 09:00, Reasoning Q.16 the chain is 14 : 30 :: 9 : 20 :: 4 : ? on double-and-add-two, and at 25 Sep 2024, 16:00, Reasoning Q.13 it is 45 : 56 :: 62 : 73 :: 19 : ? on plus eleven.

Related PYQs

No directly related past PYQ was found.

Select the option that is related to the fifth number in the same way as the second number is related to the first number and the fourth number is related to the third number. 4 : 68 :: 8 : 136 :: 9 : ? (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.)

(a)147
(b)149
(c)153
(d)152