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CAT 2020 Slot 3 — QA Question 3

Change in AverageEasy

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

Correct answer: 2

  • A

    3

  • B

    4

  • C

    1

  • 2

Solution

Easy

Use total runs (average ×\times innings) to find nn, then the first nn innings' total. To make one score as small as possible, push the other scores as high as the constraint "less than 3838" allows.

1

Find nn from the two averages. Total over n+2n+2 innings =29(n+2)=29(n+2). This equals the first nn innings' total 30n30n plus the last two scores 3838 and 1515.

30n+38+15=29(n+2)⇒ 30n+53=29n+58⇒ n=5\begin{aligned} &30n+38+15=29(n+2)\\ &\Rightarrow\ 30n+53=29n+58\\ &\Rightarrow\ n=5 \end{aligned}
2

Total of the first n=5n=5 innings. Their average is 3030.

Total=5×30⇒ Total=150\begin{aligned} &\text{Total}=5\times30\\ &\Rightarrow\ \text{Total}=150 \end{aligned}
3

Minimise one score. Each of these 55 scores is less than 3838, so the maximum any one can be is 3737. To make the lowest score xx smallest, set the other 44 scores to 3737.

x+4×37=150(from step 2)⇒ x+148=150⇒ x=2\begin{aligned} &x+4\times37=150 \quad\text{(from step 2)}\\ &\Rightarrow\ x+148=150\\ &\Rightarrow\ x=2 \end{aligned}
x=2x=2

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CAT 2020 Slot 3 QA Q3: A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 r — Solution | TheCATExam