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Inequality

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Inequality

by prachi18oct » Tue Mar 10, 2015 8:46 am
Q.Is x > y?
(1) x ^ 1/2 > y
(2) x^3 > y

Q.Is X > Y?
(!) x^2 >y
(2) x^ 1/2 < y

For questions like these, what are the suitable numbers that we need to test for . I take positive and negative fractions like x = 1/4, y = 1/3etc. Sometimes I get stuck due to these numbers only.
Please suggest any approach to take on these inequality questions.
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by GMATGuruNY » Tue Mar 10, 2015 8:54 am
prachi18oct wrote:Q.Is x > y?
(1) x ^ 1/2 > y
(2) x^3 > y
For this problem, I posted two approaches here:

https://www.beatthegmat.com/x-y-t280459.html
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by GMATGuruNY » Tue Mar 10, 2015 9:11 am
prachi18oct wrote:
Q.Is X > Y?

(!) x^2 > y
(2) x^ 1/2 < y
Test a value for x whose square and root can be calculated easily.
Let x=4.

The statements become:
Statement 1: 16 > y
Statement 2: 2 < y
Both statements are satisfied if y=3, in which case x>y.
Both statements are satisfied if y=5, in which case x<y.
INSUFFICIENT.

The correct answer is E.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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I unlock the best way for YOU to solve problems.

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by Matt@VeritasPrep » Wed Mar 11, 2015 1:18 am
Is x > y?
S1:: x² > y
S2:: y > √x
There are lots of ways to see that neither statement alone is sufficient. (Number picking is probably the easiest. For instance, in the first one, x could be negative and y positive, or vice versa. In the second, we could have y = 4 and √x = 1, or we could have y = 3 and √x = 2.)

Taking the two together, we have

x² > y > √x

This makes it easier to test numbers. We could have √x = 2 and y = 3, in which case x = 4 and x > y. But we could also have √x = 2 and y = 15, in which case x = 4 (as before) but now y > x.
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by Brent@GMATPrepNow » Wed Mar 11, 2015 9:21 am
Aside: Once we combine the statements, we get: √x < y < x²
If we ignore the y, we see that √x < x²

When it comes to the squares and square roots of positive numbers, there are three possible cases:
case a: If 0 < k < 1, then √k > k²
case b: If k = 1, then √k = k²
case c: If 1 < k, then √k < k²

Since the combined statements tell us that √x < x², we can conclude that 1 < x

When we test certain values of x that are greater than 1 (as others have done above), we see that the combined statements are still insufficient.

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Brent
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