# CAT 2020 Question Paper | Quants Slot 3

###### CAT Previous Year Paper | CAT Quants Questions | Question 17

This question is from Averages. The average score of a certain number of innings is given. Our task is to find the smallest possible scores based on the given details. CAT exam is known to test on basics rather than high funda ideas. A range of CAT questions can be asked from this topic. Make sure you master the topic by solving questions from 2IIMs CAT questions bank to gear up for CAT 2022 Exam.

Question 17 : 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

1. 1
2. 3
3. 2
4. 4

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### Explanatory Answer

The average score of the last 2 innings = $$frac{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 choos eto 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.

##### The answer is, "2"

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