For the positive integers x, x + 2, x + 4, x + 7, and

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For the positive integers x, x + 2, x + 4, x + 7, and x + 12, the mean is how much greater than the median?

A. 0
B. 1
C. 2
D. 4
E. 7

[spoiler]OA=B[/spoiler].

How can I calculate this difference without knowing the value of x? What is the way to solve it? Thanks in advanced.

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Answer

by Vincen » Sat Apr 14, 2018 5:52 am
Hello Gmat_mission.

It's not necessary to know the value of x.

- The median of a set with an odd number of elements is the middle element, hence the median of the given set is x+4.

- Now the mean is $$\frac{x+\left(x+2\right)+\left(x+4\right)+\left(x+7\right)+\left(x+12\right)}{5}=\frac{5x+25}{5}=x+5.$$ Now, the difference between the average and the median is $$x+5-\left(x+4\right)=1.$$ And this is how we can calculate the difference without knowing the value of x. The OA is B.

I hope this answer can help you.

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by Brent@GMATPrepNow » Sat Apr 14, 2018 6:43 am
Gmat_mission wrote:For the positive integers x, x + 2, x + 4, x + 7, and x + 12, the mean is how much greater than the median?

A. 0
B. 1
C. 2
D. 4
E. 7
MEAN
mean = [(x) + (x + 2) + (x + 4) + (x + 7) + (x + 12)/5
= (5x + 25)/5
= x + 5

MEDIAN
The 5 values are already arranged in ASCENDING order {x, x + 2, x + 4, x + 7, x + 12}
So, the median = the middle value = x + 4

The mean is how much greater than the median?
In other words x + 5 is how much greater than the x + 4?
Answer = (x + 5) - (x + 4)
= 1

Answer: B

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by GMATGuruNY » Sun Apr 15, 2018 3:23 am
Gmat_mission wrote:For the positive integers x, x + 2, x + 4, x + 7, and x + 12, the mean is how much greater than the median?

A. 0
B. 1
C. 2
D. 4
E. 7
Let x=1, yielding the following set of values:
x = 1
x+2 = 3
x+4 = 5
x+7 = 8
x+12 = 13.

Mean = (1+3+5+8+13)/5 = 30/5 = 6.
Median = 5.
Mean - median = 6-5 = 1.

The correct answer is B.

Since only one answer choice can be correct, there is no need to test any other values for x.
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