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C
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Difficulty:
95%
(hard)
Question Stats:
41% (02:14) correct
59%
(01:56)
wrong
based on 3176
sessions
History
Date
Time
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Not Attempted Yet
Is \(x > y\) ?
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
I thought this was a really tough question !
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
I thought this was a really tough question !
08 Sep 2010, 05:10
Kudos
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shrouded1
I thought this was a really tough question !
Is \(x > y\) ?
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
Is \(x > y\) ?
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
Is x>y?
(1) \(\sqrt{x}>y\) --> if \(x=4\) and \(y=1\) then the answer will be YES but if \(x=\frac{1}{4}\) and \(y=\frac{1}{3}\) then the answer will be NO. Two different answers, hence not sufficient.
Note that from this statement we can derive that \(x\geq{0}\) because an expression under the square root cannot be negative.
(2) \(x^3>y\) --> if \(x=4\) and \(y=1\) then the answer will be YES but if \(x=2\) and \(y=3\) then the answer will be NO. Two different answers, hence not sufficient.
(1)+(2) From (1) we have that \(x\geq{0}\). Now, \(\sqrt{x}\), \(x\), \(x^3\) can be positioned on a number line only in 2 ways:
1. For \(1\leq{x}\): ------\(\sqrt{x}\)----\(x\)----\(x^3\), so \(1\leq{\sqrt{x}}\leq{x}\leq{x^3}\) (the case \(\sqrt{x}=x=x^3\) is when \(x=1\)). \(y\) is somewhere in green zone (as \(y<\sqrt{x}\) and \(y<x^3\)), so if we have this case answer is always YES: \(y<x\).
2. For \(0\leq{x}<1\): \(0\)----\(x^3\)----\(x\)----\(\sqrt{x}\)----\(1\), so \(0\leq{x^3}\leq{x}\leq{\sqrt{x}}\). \(y\) is somewhere in green zone (as \(y<\sqrt{x}\) and \(y<x^3\)), so if we have this case answer is always YES: \(y<x\).
So in both cases \(y<x\). Sufficient.
Answer: C.
Hope it's clear.
29 Oct 2010, 06:56
Kudos
Bookmarks
shrouded1
I thought this was a really tough question !
Is \(x > y\) ?
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
Is \(x > y\) ?
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
One of those gorgeous questions that seem so simple at first but surprise you later...
Best way to work on these is to fall back on your drawing skills (Yes, I love diagrams!)
Statement (1): If I can say that \(x >= \sqrt{x}\) for all values of x, then I can say that x > y. The green line shows me the region where \(x >= \sqrt{x}\) but the red line shows me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
I also understand from this statement that x >= 0.
Statement (2): If I can say that \(x >= x^3\) for all values of x, then I can say that x > y. The green lines show me the region where \(x >= x^3\) but the red lines show me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
Attachment:
Ques.jpg [ 9.04 KiB | Viewed 143974 times ]
Using both together, I know x >= 0.
If 0 <= x <= 1, then we know \(x >= x^3\). Since statement (2) says that \(x^3 > y\), I can say that x > y.
If x > 1, then we know \(x > \sqrt{x}\). Since statement (1) says that \(\sqrt{x} > y\), I can deduce that x > y.
For all possible values of x, we can say x > y. Sufficient. Answer (C).
