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Question

Refer to the following series and answer the question. (Counting to be done from left to right only.) (all numbers are single digit numbers only.) 

(Left) 9 2 3 1 4 5 6 8 5 4 3 2 1 9 3 4 5 6 7 8 1 2 9 8 3 4 5 2 6 8 2 (Right) 

How many such even digits are there, each of which is immediately preceded by a perfect square and immediately followed by an odd digit? 

(NOTE: 1 is also a perfect square.)

This question was previously asked in
RRB NTPC 2025 Under Graduate CBT 1 Question Paper PDF (20-Jun-2026) (Shift 3)
The correct answer is

More than three

The single-digit perfect squares available here are 1, 4 and 9. We scan the series for each even digit and check that the digit just before it is one of these squares and the digit just after it is odd.

Series: 9 2 3 1 4 5 6 8 5 4 3 2 1 9 3 4 5 6 7 8 1 2 9 8 3 4 5 2 6 8 2

The pattern is (perfect square)(even)(odd). Checking position by position gives the matches: 9 2 3 (9 is a square, 3 is odd) ✓; 1 4 5 (1 is a square, 5 is odd) ✓; 1 2 9 (1 is a square, 9 is odd) ✓; 9 8 3 (9 is a square, 3 is odd) ✓.

No other even digit satisfies both conditions, so exactly 4 such even digits are found.

Since 4 is greater than three, the required count is best described as "More than three".

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