Question:medium

Identify the missing number:
1 and 3, 4 and 6, 7 and 9, ___ and 12

Updated On: Jul 15, 2026
  • 10
  • 11
  • 12
  • 13
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The Correct Option is A

Approach Solution - 1

Step 1: Write the whole series as one flat sequence: 1, 3, 4, 6, 7, 9, __, 12, and look at the difference between each consecutive term instead of treating them as separate pairs.

Step 2: The differences so far are +2 (1 to 3), +1 (3 to 4), +2 (4 to 6), +1 (6 to 7), +2 (7 to 9). The pattern alternates strictly between +2 and +1.

Step 3: Following the alternating pattern, the next difference after +2 (7 to 9) must be +1, so the missing term = 9 + 1 = 10.

Step 4: Check this against the final gap: the difference from the missing term to 12 should then be +2, and 10 + 2 = 12, which matches exactly, confirming the answer.
\[ \boxed{10} \]
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Approach Solution -2

Instead of treating the numbers as separate pairs, line up every number given so far as one list: 1, 3, 4, 6, 7, 9, ?, 12, and check each number's remainder when divided by 3. This reveals a clean split: 1, 4, 7 each leave a remainder of 1 when divided by 3, while 3, 6, 9, 12 each leave a remainder of 0.

  1. 10: Dividing 10 by 3 gives a remainder of 1 (since 10 = 3 x 3 + 1), matching the remainder-1 group that already contains 1, 4, and 7. It slots into that group as the next term after 7, keeping the remainder pattern intact.
  2. 11: Dividing 11 by 3 gives a remainder of 2, a remainder that appears nowhere else in the sequence. Introducing a brand-new remainder class breaks the two-group structure the rest of the numbers follow.
  3. 12: 12 already appears as the last given term itself, and it belongs to the remainder-0 group along with 3, 6, and 9. Repeating it as the missing term as well would mean two terms sharing the same value and the same remainder role, which the sequence never does elsewhere.
  4. 13: Dividing 13 by 3 gives a remainder of 1, which does belong to the correct group, but placing 13 there breaks the even spacing within that group, since 1, 4, 7 climb by exactly 3 each time, and 13 is 6 more than 7, not 3 more.

Only 10 keeps both the remainder-1 grouping and the even spacing within that grouping consistent with the rest of the sequence.

Therefore, the correct answer is 10.

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