CAT 2018 Slot 1 — QA Question 5

Let there be ‘n’ tests and their average be ‘x’.

By the given conditions,

When first 10 tests are not considered:
⇒ $\frac{nx-200}{n-10}$ = x + 1

⇒ x = $\frac{nx-200}{n-10}$ - 1   ...(1)

When last 10 tests are not considered:
⇒ $\frac{nx-300}{n-10}$ = x - 1

⇒ x = $\frac{nx-300}{n-10}$ + 1   ...(2)

Equating (1) and (2)
∴ $\frac{nx-200}{n-10}$ - 1 = $\frac{nx-300}{n-10}$ + 1

∴ $\frac{nx-200}{n-10}$ - $\frac{nx-300}{n-10}$ = 2

∴ $\frac{100}{n-10}$ = 2

∴ n = 60

Hence, 60.