Problem Solving

nafishasan wrote: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

The average of 5 positive integers = 40 implies that the sum of 5 positive integers = 40 * 5 = 200
Let the smallest integer = s and the largest integer = l , then l - s = 10 or l = s + 10
We have to maximize s + 10, and for s + 10 to be maximum, all other terms should be minimum, and the smallest integer is s. So other terms should be equal to s.
Hence, s + s + s + s + (s + 10) = 200
5s = 190
s = 38
s + 10 = 48

The correct answer is D.

nafishasan wrote: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

We are given that we have a sum of 40 x 5 = 200 and a range of 10.

Let's strategically check our answer choices:

If the largest integer is 50, then the 4 other numbers could be 40. However, 50 + 40 x 4 = 210, which is greater than 200. Thus, we see 50 (and also 52) cannot be correct.

Let's check 49:

49 + 4 x 39 = 49 + 156 = 205, which is also greater than 200.

Let's check 48:

48 + 38 x 4 = 48 + 152 = 200, so 48 is the largest number in the set.
Alternate Solution:

Let us denote the greatest element in this set by M and the smallest element in this set by m.

To maximize the greatest integer in the set, we should keep the remaining integers as small as possible. For that purpose, let's assume all the elements in the set besides M are equal to m. Then, according to the information given in the question, we have

(4m + M)/5 = 40

4m + M = 200

and

M - m = 10

Let's multiply the last equation by -1 and add to the preceding equation:

5m = 190

m = 38

Substituting m = 38 in M - m = 10, we see that the greatest value of M is 48.

Answer: D