a data from gmat forum

[diebeatsthegmat]

Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.
my answer is B dunno why its E

[Geva@EconomistGMAT]

Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.
my answer is B dunno why its E

EDIT: Spoke too soon.

[Ian Stewart]

Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.
my answer is B dunno why its E

The answer ought to be B - who is saying that it's E?

If the range of X is 2, then one value in the set must be greater than 2, since all the values are positive. If you square that value, it must increase by more than 2. Then the only way for the set of squares (set Y) to have a lower sum than the set X is if the other values *decrease* by a total of more than 2. The fraction which decreases most when you square it is 1/2 (it decreases by 1/4), but even if all 8 of our remaining values were equal to 1/2, they would only decrease by a total of 8(1/4) = 2 after we square them. That's not enough to compensate for the increase in the largest value, which we know must be greater than 2. So the sum of the values in Y must be greater than the sum of values in X, and thus the average of the values in Y is greater. The answer is B (Statement 1 is not sufficient, since the set X could contain five values equal to 1.01, and four values equal to 0.5 - then the sum of the squares would be smaller than the sum of the original values).

In any case, the math involved here is not GMAT-like, so I wouldn't worry about the question much. For example, the fact that 1/2 decreases, when we square it, more than any other fraction is easy to prove with calculus, but not so easy to prove from first principles, and it's not the kind of fact you'd ever be expected to know for a real GMAT question.

[diebeatsthegmat]

Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.
my answer is B dunno why its E

The answer ought to be B - who is saying that it's E?

If the range of X is 2, then one value in the set must be greater than 2, since all the values are positive. If you square that value, it must increase by more than 2. Then the only way for the set of squares (set Y) to have a lower sum than the set X is if the other values *decrease* by a total of more than 2. The fraction which decreases most when you square it is 1/2 (it decreases by 1/4), but even if all 8 of our remaining values were equal to 1/2, they would only decrease by a total of 8(1/4) = 2 after we square them. That's not enough to compensate for the increase in the largest value, which we know must be greater than 2. So the sum of the values in Y must be greater than the sum of values in X, and thus the average of the values in Y is greater. The answer is B (Statement 1 is not sufficient, since the set X could contain five values equal to 1.01, and four values equal to 0.5 - then the sum of the squares would be smaller than the sum of the original values).

In any case, the math involved here is not GMAT-like, so I wouldn't worry about the question much. For example, the fact that 1/2 decreases, when we square it, more than any other fraction is easy to prove with calculus, but not so easy to prove from first principles, and it's not the kind of fact you'd ever be expected to know for a real GMAT question.

Thank you a lot, Geva...

[TimeforGMAT]

Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.
my answer is B dunno why its E

The answer ought to be B - who is saying that it's E?

If the range of X is 2, then one value in the set must be greater than 2, since all the values are positive. If you square that value, it must increase by more than 2. Then the only way for the set of squares (set Y) to have a lower sum than the set X is if the other values *decrease* by a total of more than 2. The fraction which decreases most when you square it is 1/2 (it decreases by 1/4), but even if all 8 of our remaining values were equal to 1/2, they would only decrease by a total of 8(1/4) = 2 after we square them. That's not enough to compensate for the increase in the largest value, which we know must be greater than 2. So the sum of the values in Y must be greater than the sum of values in X, and thus the average of the values in Y is greater. The answer is B (Statement 1 is not sufficient, since the set X could contain five values equal to 1.01, and four values equal to 0.5 - then the sum of the squares would be smaller than the sum of the original values).

In any case, the math involved here is not GMAT-like, so I wouldn't worry about the question much. For example, the fact that 1/2 decreases, when we square it, more than any other fraction is easy to prove with calculus, but not so easy to prove from first principles, and it's not the kind of fact you'd ever be expected to know for a real GMAT question.

Hi,

I am little confused on your explanation. Why is that 1/2 decreases the most when it is squared? 0.0001, let's say, will decrease by even a greater extent when squared. Right? Can you please clarify?

Thanks

[Ian Stewart]

Hi,

I am little confused on your explanation. Why is that 1/2 decreases the most when it is squared? 0.0001, let's say, will decrease by even a greater extent when squared. Right? Can you please clarify?

No, when you square, say, 1/100, you will get 1/10,000. That hasn't decreased by all that much - the decrease is only 1/100 - 1/10,000 = 99/10,000 = 0.0099. When you square 1/2, the decrease is much larger - the decrease is 0.25.

The standard way to prove that x - x^2 is greatest when x = 1/2 is using calculus (if you know that subject, you take a first derivative and set it to zero to find the maximum). That's way beyond the scope of the GMAT, which is why I said in my post that the question was not a realistic GMAT question.