Question

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?

• A. 21
• B. 22
• C. 23
• D. 24

Answer 1

Answer:

A. 21

Let the number of earlier innings be (n).

Average score earlier = 44

So, total runs earlier:

$44n$

After scoring 132 in the next innings:

• Total innings (= n+1)

• New average (= 48)

Therefore,

44n + 132 = 48(n+1)
44n + 132 = 48n + 48
132 - 48 = 48n - 44n
84 = 4n
n = 21

$\boxed{21}$