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17 Jan 2020, 02:45
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A
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If x and y are unknown positive integers, is the mean of the list {6, 7, 1, 5, x, y} greater than the median of the set?
(1) x + y = 7
(2) x - y = 3
Are You Up For the Challenge: 700 Level Questions
(1) x + y = 7
(2) x - y = 3
Are You Up For the Challenge: 700 Level Questions
17 Jan 2020, 05:48
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x , y — unknown positive integers
—> set {6,7,1,5,x,y}
Is the mean of the set greater than the median of the set?
(Statement1): x+ y = 7
There are 3 cases for the values of x and y:
(Case1): 1+6 =7
—> the mean = 26/6 = 4.(3)
--> the median=( 5+6)/2= 5.5
4.(3) < 5.5 (No)
(Case2): 2+5= 7
—> the mean =4.(3).
--> The median = (5+5)/2= 5
4.(3) < 5 (No)
(Case3): 3+4 =7
—> the mean = 4.(3)
--> The median = (4+5)/2= 4.5
4.(3) < 4.5 ( No)
Sufficient
(Statement2): x—y =3
If x =5, y= 2, then
The mean= 4.(3),
the median = (5+5)/2 = 5
4.(3) < 5 (No)
If x= 20, y = 17, then
The mean = 56/6= 9.(3)
the median=( 6+7)/2 = 6.5
—> 9.(3) > 6.5 (Yes)
Insufficient
The answer is A
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—> set {6,7,1,5,x,y}
Is the mean of the set greater than the median of the set?
(Statement1): x+ y = 7
There are 3 cases for the values of x and y:
(Case1): 1+6 =7
—> the mean = 26/6 = 4.(3)
--> the median=( 5+6)/2= 5.5
4.(3) < 5.5 (No)
(Case2): 2+5= 7
—> the mean =4.(3).
--> The median = (5+5)/2= 5
4.(3) < 5 (No)
(Case3): 3+4 =7
—> the mean = 4.(3)
--> The median = (4+5)/2= 4.5
4.(3) < 4.5 ( No)
Sufficient
(Statement2): x—y =3
If x =5, y= 2, then
The mean= 4.(3),
the median = (5+5)/2 = 5
4.(3) < 5 (No)
If x= 20, y = 17, then
The mean = 56/6= 9.(3)
the median=( 6+7)/2 = 6.5
—> 9.(3) > 6.5 (Yes)
Insufficient
The answer is A
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17 Jan 2020, 06:49
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There's a problem with the wording here; in math a "set" does not have repeated elements, but presumably they intend for repetition to be possible (if it's not, Statement 1 is immediately sufficient because x and y need to be 3 and 4).
From Statement 1, we know the exact value of the sum of the six elements, so we can find the mean -- it is 26/6 = 4 + 1/3. Since x and y average to 3.5, one of x or y is less than 3.5, and that's the second-smallest value in the set, and one of them is greater than 3.5, so is at least 4. So to find the median we'll be averaging two numbers, one of which is at least 4 and one of which is at least 5, and the median will be 4.5 or greater, and the median is certainly greater than the mean. So Statement 1 is sufficient.
Using Statement 2, we know x = y + 3. If y is a huge number, the mean will clearly exceed the median. But if y = 1 (assuming repetition of values is allowed), then the mean turns out to be less than the median, so Statement 2 is not sufficient. Even if we disallow repeated values, if y = 8 the mean still turns out to be very slightly less than the median. So the answer is A.
From Statement 1, we know the exact value of the sum of the six elements, so we can find the mean -- it is 26/6 = 4 + 1/3. Since x and y average to 3.5, one of x or y is less than 3.5, and that's the second-smallest value in the set, and one of them is greater than 3.5, so is at least 4. So to find the median we'll be averaging two numbers, one of which is at least 4 and one of which is at least 5, and the median will be 4.5 or greater, and the median is certainly greater than the mean. So Statement 1 is sufficient.
Using Statement 2, we know x = y + 3. If y is a huge number, the mean will clearly exceed the median. But if y = 1 (assuming repetition of values is allowed), then the mean turns out to be less than the median, so Statement 2 is not sufficient. Even if we disallow repeated values, if y = 8 the mean still turns out to be very slightly less than the median. So the answer is A.
