Question

###### If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?

A. 50

B. 52

C. 49

D. 48

E. 44

###### Correct Answer :** Choice D. 48**

###### Explanation

The average of 5 positive integers is 40. i.e., the sum of these integers = 5*40 = 200

Let the least of these 5 numbers be x.

Then the largest of these 5 numbers will be x + 10.

If we have to maximize the largest of these numbers, we have to minimize all the other numbers.

That is 4 of these numbers are all at the least value possible = x.

So, x + x + x + x + x + 10 = 200

Or x = 38.

So, the largest of these 5 integers is 48.

## Queries, answers, comments welcome