Is x > y ? (1) x = 4y^2 (2) x > 1

Check below links.

DS questions on inequalities: search.php?search_id=tag&tag_id=184
PS questions on inequalities: search.php?search_id=tag&tag_id=64

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Hi Bunuel,

i got 2 approaches yielding different results. Can you please take a look?

1. Approach

x = 4y^2 ; 1 < x

=> 1 < 4y^2 => 1/4 < y^2

2. Approach

x = 4y^2 ; 1 < x

=> x/x = (4y^2)/x
1 = (4y^2)/x

but if 1 < x then (4y^2)/x < x => 4y^2 < 1 => y^2 < 1/4


Where is the flaw ?

Thanks in advance

\(\frac{(4y^2)}{x} < x\) --> \(4y^2 < x^2\), not \(4y^2 < 1\).

easy to eliminate ADB

answer is C

Stat 1: x = 4y^2
Stat 2: x>1

1+2: since x > 1( x^2 is always greater that x) and y^2 is also positive. It follows that lxl = 4ly^2l
hence, x> y

To solve these types of inequality questions, it's useful to see the ranges where x>y, x<y, and where x=y.

1) To start off, we know that \(x = 4y^2\). If we wanna see where x=y, set both variables equal to each other.

x = 4x^2
1 = 4x (dividing by y means we have to consider the case when x = 0. Notice in this case that x = y.
x = 1/4.

So our critical values here are y = 0 and x = 1/4.

<-------(0)--------(1/4)-------->

If x<0 (for instance, -1, then y = 4), x<y
if 0<x<1/4 (for instance, 1/8, then y =1/16) x>y
If x>1/4 (for instance, x = 1, y= 4) x<y

So, for x to be greater than y, x has be greater than 0 and less than 1/4. This is not enough information on its own. Insufficient.

2) is obviously not enough information on its own. Insufficient.

However, if we know that x>1, we know that it's outside this range and therefore that x<y.

We need both pieces of information. Therefore answer: C

Ans C

Please check the solution in the picture attached

For any inequality we must the check the bound (-1,0) and (0,1)
THEY BEHAVE FUNNY WHEN SQUARED....
WRITE THAT DOWN...

(1)

when y=\(\frac{1}{2}\),then x=4\((\frac{1}{2})^2\)=1,so x>Y

When y=\(\frac{1}{4}\),then x=4\((\frac{1}{4})^2\)=\(\frac{1}{4}\),x=y,Not Sufficient

(2) No information about y ,Clearly Not sufficient

(1)+(2) when x>1 ,x is always greater than y in terms of the equation from Statement (1),Sufficient

Correct Answer C

Is x > y ?

(1) x = 4y^2 ---> It represents the equation of parabola symmetric towards Ist and 4th quadrant , x can be +ve or zero
if any +ve value , y> x , for x=0 , x=y=0; ( INSUFF)

(2) x > 1 --- standalone not sufficient

by combining we know that x is value greater than > 1 , and it is always smaller than y

(C) is the answer

Is x > y ?

(1) x = 4y^2
(2) x > 1

(1) x = 4y^2

if y =0 => x =0 No
if y =1 & x= 4 yes
if y= 1/2 x = 1 yes

(2) x > 1

both combined yes

The highlighted part is incorrect. The variables X & Y are interchanged in the post above. Please find below the corrected form.

y = 4y^2
1 = 4y (dividing by y means we have to consider the case when y = 0. Notice in this case that x = y.
y = 1/4.

So our critical values here are y = 0 and y = 1/4.

<-------(0)--------(1/4)-------->

If y<0 (for instance,y= -1, then x= 4), x>y
if 0<y<1/4 (for instance, y = 1/8, then x =1/16) x<y... PS: So when 0<x<1/4, x<y... (Statement 2 tells us that x>1, so, x>y.. Both statement sufficient)
If y>1/4 (for instance, y = 1, x= 4)x>y
So, for x to be greater than y, y has to be greater than 1/4 or less than 0. This is not enough information on its own. Insufficient.

Is \(x > y\)?

(1) \(x = 4y^2\)

= \(\frac{x}{4 }= y^2\)

= \(\frac{\sqrt{x} }{2} = y\)

x > \(\frac{\sqrt{x} }{2}?\)

We don't know if x = 0. Insufficient.


(2) \(x > 1\)

Clearly insufficient.

(1&2) If x > 1 then:

let x = 2

2 > \(\frac{\sqrt{2} }{2}?\)

True for all scenarios. SUFFICIENT.

Bunuel chetan2u can we also solve it by putting 4y^2>1, which gives us -1/2<y<1/2 and since x>1 then x>y?

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