A batsman's average score for a certain number of innings is 44. If he scores 132 runs in the next innings, his average becomes 48. How many innings did he play earlier?

Correct — A. Let n be the number of innings played earlier.

Earlier total = average × count = 44n
New total = 44n + 132, over n + 1 innings
New average: 44n + 132 = 48(n + 1) = 48n + 48
Solve: 132 − 48 = 48n − 44n, so 84 = 4n and n = 21 → option (a)

Check: 44 × 21 = 924, 924 + 132 = 1,056, and 1,056 ÷ 22 = 48.

Why the others are wrong

• (b)
  22 — 22 is the count after the new innings, n + 1. The question asks how many he played earlier, which is 22 − 1 = 21.
• (c)
  23 — Test it: 44 × 23 = 1,012, plus 132 is 1,144, and 1,144 ÷ 24 ≈ 47.67, not 48.
• (d)
  24 — Test it: 44 × 24 = 1,056, plus 132 is 1,188, and 1,188 ÷ 25 = 47.52, not 48.

A new score shifts an average by sharing its excess among all the innings. The 132 is 84 above the new average of 48.

That 84 is exactly what lifts each of the n earlier innings from 44 to 48, a rise of 4 each. So 4n = 84 and n = 21, the same equation as the algebra, read as a balance.

Key facts

• New score = new average + (earlier innings × rise in average): 132 = 48 + 4n.
• Total = average × count, so the earlier total is 44n and the new total is 48(n + 1).
• After the new innings the count is n + 1 = 22.

Common traps

• Answering n + 1 = 22, the count after the new innings, when the question asks for the innings played earlier.
• Dividing 132 − 44 = 88 by the rise of 4: that also gives 22, because the excess over the old average is spread over n + 1 innings.

No directly related past PYQ was found.