This question was previously asked in

IBPS RRB Office Assistant Prelims: 8 August 2021 Shift 1 - Memory Based Paper

Option 2 : 1

**Given:**

Sum of 5 consecutive even numbers and 5 consecutive odd numbers = 96

Smallest even number = 4

**Calculation:**

5 consecutive even numbers = 4, 6, 8, 10 and 12

Let the smallest odd numbers be x

4 odd numbers = x, (x + 2), (x + 4) and (x + 6)

Now, According to the question,

4 + 6 + 8 + 10 + 12 + x + x + 2 + x + 4 + x + 6 = 96

⇒ 40 + 4x + 12 = 96

⇒ 4x + 52 = 96

⇒ 4x = 44

⇒ x = 11

Lowest odd number = 11

Highest even number = 12

Required difference = 12 – 11

⇒ 1

**∴ The required difference is 1**