Question

A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is

• A. 2
• B. 3
• C. 4
• D. 1

Answer 1

The Correct Option is A

To solve for the smallest possible value of \(x\), the lowest score in the first \(n\) innings, follow these steps:

1. Total runs in \(n+2\) innings: The average score is 29, so total runs = \((n+2) \times 29 = 29n + 58\). 
2. Total runs in last 2 innings: He scored 38 and 15 runs, hence total = 38 + 15 = 53.
3. Runs scored in first \(n\) innings: Subtract runs in last 2 innings from total runs: \(29n + 58 - 53 = 29n + 5\).
4. Average score in first \(n\) innings: Given as 30, so total runs = \(30n\).
5. Equating runs for first \(n\) innings: \(29n + 5 = 30n\) implies \(30n - 29n = 5\) which gives \(n = 5\).
6. Assume scores in first \(n\) innings are \(a_1, a_2, ..., a_n\), so \(a_1 + a_2 + ... + a_5 = 150\) (since average = 30 with \(n = 5\)).
7. Since scores are less than 38 and \(\min(a_1, a_2, ..., a_5) = x\), choose scores close to 38 to minimize \(x\), say: 37, 37, 37, 37, and \(x\). Then: \(4 \times 37 + x = 150\).
8. Solving for \(x\), \(148 + x = 150\) gives \(x = 2\).

Thus, the smallest possible value of \(x\) is 2.