GRE® Quant Practice | Statistics & Averages #3

The GRE maths sample question given below is a problem solving question in statistics. It tests the concept of maximum possible range of a set of numbers if their average (arithmetic mean) is known.

Question 3 :If the average (arithmetic mean) of 5 positive integers is 11, what is the maximum possible difference between the largest and the smallest of these 5 numbers?

1. 50
2. 44
3. 39
4. 4
5. 20

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Video Explanation

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Explanatory Answer | GRE Averages Practice Question 3

Given Data

Let a, b, c, d, and e be the five positive integers.
The average of these five positive integers is 11.
i.e., \\frac{\text{a + b + c + d + e}}{5}) = 11 → a + b + c + d + e = 55 ......(1)

When will the Range be Maximum?

Let e be the largest number and a be the smallest number.
For the difference to be maximum 'e' has to be maximized.
If e has to be maximized, then the remaining four integers have to be minimized.
It is given that all five are positive integers. So, the minimum value possible for any of these integers is 1.
Please note that these integers need not be distinct. The question does not specify that.
So, all 4 numbers could take the same value. i.e., a = b = c = d = 1 is the minimum possible value for a, b, c, and d.

From equation 1 → a + b + c + d + e = 55
1 + 1 + 1 + 1 + e = 55
Or e = 51
Maximum possible range = maximum possible difference between the largest and smallest number = 51 – 1 = 50

Choice A is the correct answer