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​

Answer 1

Answer:

Sum of all the scores In first n innings = 29(n+2) – 38 -15 = 29n +5

30n = 29n + 5

n = 5

so total score in first 5 innings = 5*30 = 150

to minimize x , the score of other 5 innings must be maximum which is 37 in each

so x + 4*37 = 150

x = 2

Step-by-step explanation:

Answer 2

The average score of the last 2 innings = (38+15)/2 = 26.5

The average of the first n innings = 30

The average of all the n+2 innings = 29

29(n + 2) = n(30) + 2(26.5)

n = 5 (You may choose to use the allegation method too)

The average of the first 5 innings = 30 and in none of the innings did the batsman score 38 or above.

Let's assume the scores of the 5 innings of the batsman in ascending order to be 30, 30, 30, 30, 30. (This complies with the avberage being 30)

The maximum score that he can score is 37 runs, Let's distribute the score of the first innings among the rest 4, as 7 per innings so that the score of the first innings is minimized.

Now the scores will be 2, 37, 37, 37, 37

Now we can't distribute the score of the first innings anymore, because, that makes the score of atleast one of the remainin 4 overs 38 or greater.

Hence the least score of any inning is 2.

Answer

2