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If x and y are unknown positive integers, is the mean of the

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by AAPL » Wed Nov 14, 2018 4:59 am

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If x and y are unknown positive integers, is the mean of the set {6, 7, 1, 5, x, y} greater than the median of the set?

1) x + y = 7.
2) x - y = 3.

OA A.
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by GMATGuruNY » Wed Nov 14, 2018 5:50 am
AAPL wrote:Manhattan Prep

If x and y are unknown positive integers, is the mean of the set {6, 7, 1, 5, x, y} greater than the median of the set?

1) x + y = 7.
2) x - y = 3.
Known values, in ascending order:
1, 5, 6, 7

Statement 1: x+y = 7
Mean = sum/quantity = (1+5+6+7+x+y)/6 = (1+5+6+7+7)/6 = 26/6 = 13/3 = 4.33.
Text EXTREMES.

Case 1: x and y are far from each other
If x=1 and y=6, we get:
1, 1. 5, 6, 6, 7
Median = (5+6)/2 = 11/2 = 5.5

Case 2: x and y are near each other
If x=3 and y=4, we get:
1, 3, 4, 5, 6, 7
Median = (4+5)/2 = 9/2 = 4.5

In both cases, the mean is LESS than the median, so the answer to the question stem is NO.
SUFFICIENT.

Statement 2: x-y = 3
Again, test EXTREMES.

Case 1: x and y are as small as possible.
If x=4 and y=1, we get:
1, 1, 4, 5, 6, 7
Mean = (1+1+4+5+6+7)/6 = 24/6 = 4
Median = (4+5)/2 = 9/2 = 4.5.
In this case, the mean is LESS than the median, so the answer to the question stem is NO.

Case 2: x and y are large
If x=100 and y=97, we get:
1, 5, 6, 7, 97, 100
Mean = (1+5+6+7+97+100)/6 = 216/6 = 108/3 = 36.
Median = (6+7)/2 = 13/2 = 6.5.
In this case, the mean is GREATER than the median, so the answer to the question stem is YES.

Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.

The correct answer is A.
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by Jay@ManhattanReview » Wed Nov 14, 2018 9:24 pm
AAPL wrote:Manhattan Prep

If x and y are unknown positive integers, is the mean of the set {6, 7, 1, 5, x, y} greater than the median of the set?

1) x + y = 7.
2) x - y = 3.

OA A.
Let's take each statement one by one.

1) x + y = 7.

Mean = (1 + 5 + 6 + 7 + x + y)/6 = (19 + x + y)/6 = (19 + 7)/6 = 4.33

To find out median, we must first arrange terms in an ascending order.

The terms, excluding x and y are: 1, 5, 6, 7.

Median, excluding x and y: 5.5

If one of the terms (x and y) is less than equal to 5 and the other term is greater than or equal 6, i.e. one is added to the left of 5.5 and the other is added to its right, the median would not change.

Given x + y = 7, the pairs of x and y are: (1, 6); (2, 5); and (3, 4)

Needless to state that if the pair (1, 6) is added, the median would not change. Or, Median = 5.5 . Mean = 4.33. The answer is No.

It will be interesting to see how would median change when we add (2, 5) and (3, 4).

Since for both the pairs, the maximum values (5 and 4) are less than the median value (5.5), the new median would be less than 5.5. But will the median be greater than 4.33? Let's see.

Again, since for the pair (2, 5), the maximum value (5) is greater than mean (4.33), the new median, would though be less than 5.33, it would be greater than mean 4.33. The answer is No.

So, we are left with the last pair (3, 4).

The set would be 1, 3, 4, 5, 6, 7 and its median would be (4 + 5)/2 = 4.5 > Mean 4.33. The answer is still No.

Unique answer. Sufficient.

2) x - y = 3.

If x and y both are very big, then both of them added to the right of median, 5.5,

Mean = Mean = (19 + x + y)/6 = a very big number

The set would be 1, 5, 6, 7, x, y and its median would be (6 + 7)/2 = 6.5 < Very big Mean. The answer is Yes.

However, given x - y = 3, we can have x and y as: (4, 1); (5, 2); etc.

Let's see the pair (4, 1).

Mean = Mean = (19 + x + y)/6 = (19 + 4 + 1)/6 = 4

The set would be 1, 1, 4, 5, 6, 7 and its median would be (4 + 5)/2 = 4.5 > Mean 4. The answer is No.

No unique answer. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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