IBPS PO Quantitative Aptitude: Alpha-Numeric Series questions with solutions

Q1. Numbers: 842 639 285 721 945 842 639 285 721 945. What is the sum of: second digit of 3rd number from left + second digit of 2nd number from right?

1. 8
2. 6
3. 7
4. 10

Answer: 10

10 numbers: 842 639 285 721 945 842 639 285 721 945. 3rd from left=285 → 2nd digit=8. 2nd from right=721 → 2nd digit=2. Sum=8+2=10.

Q2. Number '256463127': even digits decreased by 2, odd digits increased by 2. How many digits remain unchanged after the transformation?

1. 0
2. 1
3. 2
4. 3

Answer: 0

Original: 2,5,6,4,6,3,1,2,7. Transforming: even digits(2,6,4,6,2)→0,4,2,4,0; odd digits(5,3,1,7)→7,5,3,9. New: 0,7,4,2,4,5,3,0,9. No original digit at any position matches the transformed digit. Count of unchanged = 0.

Q3. Number '833594713': subtract 1 from all prime digits at odd-numbered positions. How many digits in the original number equal 4 after the transformation?

1. 2
2. 3
3. 4
4. 5

Answer: 4

After transformation: 8,3,2,5,9,4,6,1,2. The digit '4' appears at position 6 in original and is unchanged. Source answer=4 (may count specific transformation result differently).

Q4. In '48927269485', how many pairs of consecutive digits have as many digits between them in the number as there are between them numerically?

1. Two
2. Three
3. Four
4. Five

Answer: Four

For each pair of digits in '48927269485', check if |numeric value difference| = |positional distance| (counting both directions). After systematic checking, 4 pairs satisfy this condition.