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Is x<y?

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Joined: 02 Jun 2017
Posts: 1
Is x<y?  [#permalink]

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New post 01 Oct 2017, 12:13
1
2
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

29% (02:34) correct 71% (01:46) wrong based on 48 sessions

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Is x<y?

1 \(\sqrt{x} < y\)
2 \(x^2 < y\)
Manhattan Prep Instructor
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Joined: 04 Dec 2015
Posts: 635
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170
Is x<y?  [#permalink]

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New post 01 Oct 2017, 12:42
sahil5695 wrote:
Is x<y?

1 \(\sqrt{x} < y\)
2 \(x^2 < y\)


Interesting problem! Since there are only two variables, and the math looks like it's simplified already, I'm going to start by testing cases.

Statement 1:
If \(\sqrt{x} < y\), we could have a very small x and a very large y: for instance, x = 1 and y = 100. In that case, x < y, so the answer to the question is yes.

On the other hand, we could have a situation where \(\sqrt{x}\) is smaller than y, but x itself is actually bigger than y. You could do this by picking an x that's only a little bit bigger than y. Then, when you take the square root, it gets much smaller. For instance, x = 4 and y = 3. In that case, x > y, so the answer to the question is no.

We got both 'yes' and 'no' answers to the question, so the statement is insufficient. Cross off A and D.

Statement 2:
Our first case will work here as well. x = 1 and y = 100. So, x < y, so the answer to the question is yes.

Now we need to think of a situation where \(x^2\) is smaller than y, but where x is actually bigger than y. That means x would have to be bigger than \(x^2\). That only happens when x is a fraction. Let's say that x = 1/2, so \(x^2\) = 1/4. Then we need to choose a value of y that's bigger than 1/4, but smaller than 1/2. Let's say x = 1/2 and y = 1/3. That fits the statement, and x > y, so the answer is no.

We got both 'yes' and 'no' answers to the question, so the statement is insufficient. Cross off B.

Both statements together:

The problem here is that you can't get a 'no' answer if both statements are true.

If statement 2 is true, the only way to get a 'no' answer is if x and y are fractions.

If statement 1 is true, the only way to get a 'no' answer is if x is bigger than \(\sqrt{x}\). That doesn't happen if x is a fraction.

So, there's no way to get a 'no' answer from both statements at the same time. The answer must be 'yes'. So, the right answer is C.
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Is x<y?  [#permalink]

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New post 11 Dec 2017, 11:05
Tricky question.

x < y ?

before we attack statements, we must recollect this.

if y is positive and less than 1, then \(\sqrt{y} > y > y^2 > y^3\) ......

if y is positive and greater than 1, then \(\sqrt{y} < y < y^2 < y^3\) .......

Statement 1:
\(\sqrt{x} < y\)

since sq root is always positive => y is positve, we can safely square both sides,
\(x < y^2\) => insufficient, as could be the case, \(x < y < y^2\) OR \(y < x < y^2\)

Statement 2:
\(x^2 < y\)

=> Insufficient, if x is less than 1, say x = 0.1, y = 0.02, \(x^2 = 0.01\) < 0.02 (y), but x > y
=> x is positive, since it is possible to take square root.
=> Also, y is positive, since is greater than squared value,
Taking square root on both sides of statement 2, \(|x| < \sqrt{y}\)

Statement 1 + 2:
x and y are both positive,
so from statement 1, \(x < y^2\) and from statement 2 : \(x < \sqrt{y}\)

if y < 1, then we have , \(\sqrt{y} > y > y^2 > y^3\) => \(x < y^2 < y < \sqrt{y}\), so x is less than y
if y > 1, then we have, \(\sqrt{y} < y < y^2 < y^3\) => \(x < \sqrt{y} < y < y^2\), so x is less than y

Sufficient => Answer (C)
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Is x<y? &nbs [#permalink] 11 Dec 2017, 11:05
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