CAT 2020 Slot 3 — QA Question 3
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
Answer & solution
- A
3
- B
4
- C
1
2
Average score in n innings is 30 and in n + 2 innings is 29. Score in last 2 innings is 38 and 15
∴ 30n + 38 + 15 = 29(n + 2)
⇒ n = 5
∴ Total runs score in first n innings = 5 × 30 = 150
Also, he scored less than 38 runs in each of these 5 innings.
To calculate the lowest possible score in an innings we try to maximize the score of other 4 innings. Maximum runs score in each of the 4 innings can be 37.
∴ x + 4 × 37 = 150
⇒ x = 2
Hence, option (d).