Hi team,
Hopefully this hasn't been asked before... re CAT #1, Quant question #2 (below for convenience)... the explanation says, "When we try to come up with x and y values that fit both conditions, we must adjust the two variables so that x is always greater than y." To that end, in the table, it states that for x = 2, y = 1 so x > y, and for x = 1/4, y = 1/128 so x > y again, but I don't understand why we can keep manipulating y to make the condition work. What I was thinking was, since there's no condition for y, if we plug in (say) y = 3 for either x = 2 or x = 1/4, then x is not greater than y, so the answer would be E bc it could be > or <.... but the answer seems to be C.
I keep getting ~half of those types of questions wrong. I'm sure your feedback will help. THANKS in advance!
_ _ _ _ _ _ _
Is x > y?
(1) sqrt(x) > y
(2) x^3 > y
CORRECT: (C) Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
(1) AND (2) SUFFICIENT: Let’s start with statement 1 and add the constraints of statement 2. From statement 1, we see that x has to be positive since we are taking the square root of x. There is no point in testing negative values for y since a positive value for x against a negative y will always yield a yes to the question. Lastly, we should consider x values between 0 and 1 and greater than 1 because proper fractions behave different than integers with regard to exponents. When we try to come up with x and y values that fit both conditions, we must adjust the two variables so that x is always greater than y.
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- rnmakrinos
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- RonPurewal
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Re: CAT 1, Quant question #2
when you consider both statements together, you can only pick values that make both statements true.
the values you've listed here don't do that:
first of all, i don't understand what you mean by "there's no condition for y".
part of the problem here is that the language of your post isn't quite clear.
the other part, to be perfectly (and bluntly) honest, is that you've posted a giant wall of text, without a single line break or paragraph break. i don't know about the other moderators, but i can't really penetrate that sort of thing.
... but, in any case, your test cases here don't work.
if you're testing both cases together, you have to make both conditions true.
i.e., you must pick numbers x and y so that √x > y, AND so that x^3 > y.
in the case of x = 2 and y = 3, the first of these is false. so that's a non-case.
in the case of x = 1/4 and y = 3, both statements are false. so that's a non-starter, too.
... so, basically, back to the drawing board there; you can't use either of these cases. remember, you have to satisfy the conditions!
the values you've listed here don't do that:
rnmakrinos Wrote:since there's no condition for y, if we plug in (say) y = 3 for either x = 2 or x = 1/4, then x is not greater than y,
first of all, i don't understand what you mean by "there's no condition for y".
part of the problem here is that the language of your post isn't quite clear.
the other part, to be perfectly (and bluntly) honest, is that you've posted a giant wall of text, without a single line break or paragraph break. i don't know about the other moderators, but i can't really penetrate that sort of thing.
... but, in any case, your test cases here don't work.
if you're testing both cases together, you have to make both conditions true.
i.e., you must pick numbers x and y so that √x > y, AND so that x^3 > y.
in the case of x = 2 and y = 3, the first of these is false. so that's a non-case.
in the case of x = 1/4 and y = 3, both statements are false. so that's a non-starter, too.
... so, basically, back to the drawing board there; you can't use either of these cases. remember, you have to satisfy the conditions!
- RonPurewal
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Re: CAT 1, Quant question #2
By the way, here's another way to prove that the answer is (c), without testing numbers. (You should also be able to test numbers; that's an essential skill.)
* You know that x > 0, because the expression √x exists. (If x were negative, you couldn't square-root it.)
Basically, there are four cases:
1/
x = 0.
In this case, √x, x^3, and x are all 0. so, statements 1 and 2 here both say 0 > y; therefore, yes, x > y.
2/
0 < x < 1.
In this case, x^3 < x < √x. (This is a thing you should know: for "fractions" between 0 and 1, the usual order of powers is all reversed.)
So, if you know that y is less than both x^3 and √x, then, since x^3 is the smallest of the three things (x^3, x, and √x), y must be less than all three of those things. So it's less than x.
3/
x = 1.
In this case, √x = x^3 = x again, so this works like case #1 above. (the values are 1 this time, rather than 0, but the logic is the same.)
4/
x > 1.
In this case, the powers are in their "normal" order:
√x < x < x^3
So, if you know that y is less than both x^3 and √x, then, since √x is the smallest of the three things (x^3, x, and √x), y must be less than all three of those things. So it's less than x.
So y is less than x in every case. Sufficient.
* You know that x > 0, because the expression √x exists. (If x were negative, you couldn't square-root it.)
Basically, there are four cases:
1/
x = 0.
In this case, √x, x^3, and x are all 0. so, statements 1 and 2 here both say 0 > y; therefore, yes, x > y.
2/
0 < x < 1.
In this case, x^3 < x < √x. (This is a thing you should know: for "fractions" between 0 and 1, the usual order of powers is all reversed.)
So, if you know that y is less than both x^3 and √x, then, since x^3 is the smallest of the three things (x^3, x, and √x), y must be less than all three of those things. So it's less than x.
3/
x = 1.
In this case, √x = x^3 = x again, so this works like case #1 above. (the values are 1 this time, rather than 0, but the logic is the same.)
4/
x > 1.
In this case, the powers are in their "normal" order:
√x < x < x^3
So, if you know that y is less than both x^3 and √x, then, since √x is the smallest of the three things (x^3, x, and √x), y must be less than all three of those things. So it's less than x.
So y is less than x in every case. Sufficient.
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