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CAT 2018 Slot 1 — QA Question 5

Change in AverageEasy

A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is

Answer & solution

Correct answer: 60

Answer: 60

Solution

Easy

Let there be nn tests with overall average xx, so total =nx= nx. Dropping the first 10 (total 200200, since their average is 2020) raises the average by 1; dropping the last 10 (total 300300, average 3030) lowers it by 1. Two equations in n,xn,x — subtract to eliminate xx.

1

Drop the first 10 tests. Their total is 10×20=20010\times20 = 200; the remaining average is x+1x+1.

nx−200n−10=x+1...(1)\begin{aligned} &\frac{nx - 200}{n - 10} = x + 1 \quad\text{...(1)} \end{aligned}
2

Drop the last 10 tests. Their total is 10×30=30010\times30 = 300; the remaining average is x−1x-1.

nx−300n−10=x−1...(2)\begin{aligned} &\frac{nx - 300}{n - 10} = x - 1 \quad\text{...(2)} \end{aligned}
3

Subtract to eliminate xx. Compute (1) −- (2).

(nx−200)−(nx−300)n−10=(x+1)−(x−1)⇒ 100n−10=2[(1)−(2)]⇒ n−10=50\begin{aligned} &\frac{(nx-200) - (nx-300)}{n-10} = (x+1)-(x-1)\\ &\Rightarrow\ \frac{100}{n-10} = 2 \quad\text{[(1)}-\text{(2)]}\\ &\Rightarrow\ n - 10 = 50 \end{aligned}
4

Solve for nn.

n=60\begin{aligned} &n = 60 \end{aligned}
60 tests\boxed{60 \text{ tests}}

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CAT 2018 Slot 1 QA Q5: A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests a — Solution | TheCATExam