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?
Correct Answer : Choice D. 48
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.
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