https://www.manhattanprep.com/gre/ChallengeProblems/Archive/1289
Challenge Problem from June 10, 2013 "Average of Squares" doesn't make a whole lot of sense to me.
Can someone please clarify??
Quantity A
The average of (x – 1)^2 and (x + 3)^2
Quantity B
3
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- steve4888
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- tommywallach
- Manhattan Prep Staff
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Re: Average of Squares (Challenge Problem of June 10, 2013)
Hey Steve,
Totally! So the issue here is we need to see if there's some kind of range for what Column A can be. One thing you can try is just throwing a few values in there for x, then see what happens.
Right away, it should be obvious that it's easy to make Column A bigger than Column B (just plug in a billion for x). So the question becomes--is there any way to make Column A less than Column B?
This is where the given explanation comes in. Anything squared is either 0 or something positive, so the smallest we can make either of these two terms is 0. To make the first term (x-1) equal to zero, x would have to be 1. But if you plug in one, you end up with 16 for (x + 3), which averages to 8. That's still bigger than Column B). If we try to minimize (x + 3), we plug in -3 for x, but then the first term (x-1) ends up equalling -4, which is 16 when we square it, leading to an average of 8 again.
All of this proves that there's no way to make Column A any less than 8, so the answer is A.
Hope that helps!
-t
Totally! So the issue here is we need to see if there's some kind of range for what Column A can be. One thing you can try is just throwing a few values in there for x, then see what happens.
Right away, it should be obvious that it's easy to make Column A bigger than Column B (just plug in a billion for x). So the question becomes--is there any way to make Column A less than Column B?
This is where the given explanation comes in. Anything squared is either 0 or something positive, so the smallest we can make either of these two terms is 0. To make the first term (x-1) equal to zero, x would have to be 1. But if you plug in one, you end up with 16 for (x + 3), which averages to 8. That's still bigger than Column B). If we try to minimize (x + 3), we plug in -3 for x, but then the first term (x-1) ends up equalling -4, which is 16 when we square it, leading to an average of 8 again.
All of this proves that there's no way to make Column A any less than 8, so the answer is A.
Hope that helps!
-t
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