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# Is x|y| > y^2? (1) x > y (2) y > 0

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Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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11 Feb 2012, 00:54
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35% (medium)

Question Stats:

66% (01:21) correct 34% (01:15) wrong based on 598 sessions

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Is x|y| > y^2?

(1) x > y
(2) y > 0

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?
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Joined: 02 Sep 2009
Posts: 53020
Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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11 Feb 2012, 01:10
2
4
devinawilliam83 wrote:
is x|y|>y^2?
1.x>y
2.y>0

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Yes, your rephrasing is wrong. You cannot substitute $$y^2$$ with $$|y|$$, because generally they are not equal: $$\sqrt{y^2}=|y|$$. Next, even if it were "is $$x|y|>|y|$$?" you still cannot reduce it by $$|y|$$ and write "is $$x>1$$", as $$y$$ can be zero and you cannot reduce/divide by zero. $$x|y|>|y|$$ can be rephrased as $$|y|(x-1)>0$$.

Is x|y|>y^2?

(1) x>y --> if $$x>y>0$$ then obviously $$x|y|>y^2$$ but if $$0>x>y$$ then $$x|y|<0<y^2$$. Not sufficient.

(2) y>0 --> $$|y|=y$$. No info about x. Not sufficient.

(1)+(2) Since $$x>y>0$$ then $$xy>y^2$$. Sufficient.

Hope it's clear.
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26 Sep 2012, 19:56
Asked : is x|y| > y^2

Simplifying, is x|y| > |y|*|y| i.e is x > |y|

i) x > y. But this doesn't mean x > |y| unless y is positive
ii)y > 0 ie. y is positive

Hence, both statements together are sufficient. Hence C.
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Updated on: 27 Sep 2012, 03:21
piyatiwari wrote:
Asked : is x|y| > y^2

Simplifying, is x|y| > |y|*|y| i.e is x > |y|

i) x > y. But this doesn't mean x > |y| unless y is positive
ii)y > 0 ie. y is positive

Hence, both statements together are sufficient. Hence C.

Simplifying, is x|y| > |y|*|y| i.e is x > |y|
You can simplify, or more precisely, divide through by |y| if you know that y is non-zero.

The question "Is x|y| > y^2?" is equivalent to "Is y non-zero and x > |y|?"

(1) y can be 0. Also, x can be greater than |y|.
Not sufficient.
(2) Not sufficient, as nothing is known about x.

(1) and (2) together: y > 0 and x > y = |y|.
Sufficient.

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Originally posted by EvaJager on 26 Sep 2012, 23:23.
Last edited by EvaJager on 27 Sep 2012, 03:21, edited 1 time in total.
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27 Sep 2012, 02:17
I 'd like to know if my approach is correct

x | y | > y^2 ------> x < - y or x > y

1) x > y but we do not know if x is less of - y insuff

2) y> 0 we know only that y is positive but nothing about x

1) + 2) y is positive and x must be greater than y. suff.

is correct ??
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27 Sep 2012, 03:25
1
carcass wrote:
I 'd like to know if my approach is correct

x | y | > y^2 ------> x < - y or x > y

1) x > y but we do not know if x is less of - y insuff

2) y> 0 we know only that y is positive but nothing about x

1) + 2) y is positive and x must be greater than y. suff.

is correct ??

x | y | > y^2 ------> x < - y or x > y NO

First of all, y must be non-zero. Then dividing through by |y|, we obtain x > |y| > 0.
From the given inequality, $$x|y| > y^2>0$$ we deduce that x > 0, x cannot be negative.
And you cannot automatically assume that y is positive.
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Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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07 Jun 2013, 07:45
Can we take

$$Y^2$$ = $$[y][y]$$ and cancel [y] from both sides if y is not equal to zero.
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Posts: 53020
Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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07 Jun 2013, 07:59
karjan07 wrote:
Can we take

$$Y^2$$ = $$[y][y]$$ and cancel [y] from both sides if y is not equal to zero.

If $$y\neq{0}$$ and $$x|y|>y^2$$ we CAN reduce by $$|y|$$ and get: $$x>|y|$$.

Hope it's clear.
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Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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30 Jun 2013, 13:33
1
Is x|y|>y^2?

(1) x>y

Though we know that x>y, we don't know anything about the signs of x and y. For example:

x=2, y=1
x|y|>y^2
2|1|>1^2
2>1 Valid

x=2, y=-3
x|y|>y^2
2|-3|>-3^2
6>9 Invalid

INSUFFICIENT

(2) y>0
This tells us nothing about x. For example:

x=10, y=1
x|y|>y^2
10|1|>1^2
10>1 Valid

x=1, y=10
x|y|>y^2
1|10|>10^2
10>100 Invalid

INSUFFICIENT

1+2) x>y, y>0 ===> x>y>0
If y>0 then x is greater than zero (and y)
x=10, y=9
x|y|>y^2
10|9|>9^2
90>81

Think of it like this: The RHS is the smaller number times the smaller number. The LHS is the smaller number times a number larger than the smaller number.
SUFFICIENT

(C)
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Posts: 421
Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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22 Jul 2013, 09:16
Is x|y|>y^2?

(1) x>y
x may be greater than y but we still cannot be sure if x|y|>y^2.
For example:

x>y
2>1
2|1| > 1^2
2>1 Valid

x>y
-1>-2
-1|-2| > -2^2
-2 > 4 Invalid
INSUFFICIENT

(2) y>0
This tells us nothing about x.
INSUFFICIENT

1+2) x>y and y>0. Therefore, x>y>0. We saw in #1 that when y is greater than zero, the inequality holds true. When it is less than zero, the inequality does not hold true. Just to be sure:
x>y>0
2>1>0
x|y|>y^2
2|1|>1^2
2>1

Also, take note that x|y| is y*a number larger than y. This will always be greater than y*y (y^2) so long as x is positive.

(C)
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Re: Is x*|y| > y^2?  [#permalink]

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17 May 2014, 19:31
1
russ9 wrote:
Is x·|y| > y2?

(1) x > y
(2) y > 0

I'll paste my question in the second post. Good Luck!

We have x|y|> y2

Now, to solve this, we need to understand, what do we actually want to know.

y2 = |y|2

x|y|>|y|2
x|y| - |y|2 > 0
or (x-|y|)|y|>0

Now, we know that |y| is always greater than 0, Thus, our inequality yield x-|y|>0

So Is x|y|>y2 translates to Is x> |y|

Now, lets come to the statements,

Statement 1 says, x>y; which is not sufficient. In case y is negative, we can't say anything about x being greater than absolute value of y

Statement 2 say, y>0, which in itself is insufficient as we don't know anything about x here.

But when we combine these two statements, we would get the answer. As we know when y > 0; |y| = y and statement 1 says x > y, and if y > 0 we can write x > |y|

Hope it helps!!!

Kudos if it helped!!!
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Posts: 53020
Re: Is x*|y| > y^2?  [#permalink]

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18 May 2014, 00:27
mittalg wrote:
russ9 wrote:
Is x·|y| > y2?

(1) x > y
(2) y > 0

I'll paste my question in the second post. Good Luck!

We have x|y|> y2

Now, to solve this, we need to understand, what do we actually want to know.

y2 = |y|2

x|y|>|y|2
x|y| - |y|2 > 0
or (x-|y|)|y|>0

Now, we know that |y| is always greater than 0, Thus, our inequality yield x-|y|>0

So Is x|y|>y2 translates to Is x> |y|

Now, lets come to the statements,

Statement 1 says, x>y; which is not sufficient. In case y is negative, we can't say anything about x being greater than absolute value of y

Statement 2 say, y>0, which in itself is insufficient as we don't know anything about x here.

But when we combine these two statements, we would get the answer. As we know when y > 0; |y| = y and statement 1 says x > y, and if y > 0 we can write x > |y|

Hope it helps!!!

Kudos if it helped!!!

Everything is correct except that |y| is more than or equal to 0. So, x*|y| > y^2 after reducing by |y| can be translated: is $$x > |y|>{0}$$?
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Re: Is x*|y| > y^2?  [#permalink]

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05 Oct 2014, 15:19
Hi Experts,

Is the following solution correct?
Question: Is x·|y| > y^2?
True when:
Case 1:
y>0 and x > y
Case 2:
y < 0 and x < -y

(1) NS; we don't know whether y > 0
(2) NS; we don't know anything about x

(1) + (2)
Case 1, so sufficient
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Re: Is x*|y| > y^2?  [#permalink]

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05 Oct 2014, 23:40
TooLong150 wrote:
Hi Experts,

Is the following solution correct?
Question: Is x·|y| > y^2?
True when:
Case 1:
y>0 and x > y
Case 2:
y < 0 and x < -y

(1) NS; we don't know whether y > 0
(2) NS; we don't know anything about x

(1) + (2)
Case 1, so sufficient

For $$x*|y| > y^2$$ to be true either $$0 < y < x$$ or $$-x < y < 0$$ (notice that in both cases x must be positive).
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Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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06 Feb 2016, 10:21
devinawilliam83 wrote:
Is x|y|>y^2?

(1) x>y
(2) y>0

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Can we rephrase the question as IxI>IyI??

reason.. if we square both side xIyI>y^2
we get x^2Y^2>Y^4

X^2>Y^2 or IXI>IYI...????
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Posts: 53020
Re: Is x|y| > y^2? (1) x > y (2) y > 0  [#permalink]

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06 Feb 2016, 10:27
rohit8865 wrote:
devinawilliam83 wrote:
Is x|y|>y^2?

(1) x>y
(2) y>0

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Can we rephrase the question as IxI>IyI??

reason.. if we square both side xIyI>y^2
we get x^2Y^2>Y^4

X^2>Y^2 or IXI>IYI...????

No, we cannot do this. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Check the following post for more: How to manipulate inequalities (adding, subtracting, squaring etc.).

Hope it helps.
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Re: Is x·|y| > y^2?  [#permalink]

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21 May 2016, 05:47
2
nishatfarhat87 wrote:
Is x·|y| > y^2?

(1) x > y
(2) y > 0

Hi Bunuel,

Hi,
Bunuel too will be able to give a approach, but i will give you one--

$$x·|y| > y^2.....$$
what does this mean...
$$y^2-x|y|<0.............$$
since y^2 and |y| will be positive, we have to answer if x>|y|......OR x>0 and x>y...

lets see the statements-

$$(1) x > y$$
say y is -ive and x is also -ive...... ans will be NO
if x is +ive, ans is YES
Insuff

$$(2) y > 0$$
nothing about relation between x and y..
Insuff...

Combined-
x>y and y>0, so x>0..
ans is YES
Suff
C
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html

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Re: Is x·|y| > y^2?  [#permalink]

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21 May 2016, 05:59
chetan2u wrote:
nishatfarhat87 wrote:
Is x·|y| > y^2?

(1) x > y
(2) y > 0

Hi Bunuel,

Hi,
Bunuel too will be able to give a approach, but i will give you one--

$$x·|y| > y^2.....$$
what does this mean...
$$y^2-x|y|<0.............$$
since y^2 and |y| will be positive, we have to answer if x>|y|......OR x>0 and x>y...

lets see the statements-

$$(1) x > y$$
say y is -ive and x is also -ive...... ans will be NO
if x is +ive, ans is YES
Insuff

$$(2) y > 0$$
nothing about relation between x and y..
Insuff...

Combined-
x>y and y>0, so x>0..
ans is YES
Suff
C

Hi Chetan,

Thanks for your response. How did you get this: since y^2 and |y| will be positive, we have to answer if x>|y|......OR x>0 and x>y..
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Posts: 7334
Re: Is x·|y| > y^2?  [#permalink]

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22 May 2016, 00:28
nishatfarhat87 wrote:
Hi Chetan,

Thanks for your response. How did you get this: since y^2 and |y| will be positive, we have to answer if x>|y|......OR x>0 and x>y..

Hi,

Quote:
$$x·|y| > y^2.....$$
what does this mean...
$$y^2-x|y|<0.............$$
since y^2 and |y| will be positive, we have to answer if x>|y|......OR x>0 and x>y...

Now when will x·|y| > y^2....
RHS= Y^2, which is always 0 or +ive....
LHS has two terms x and |y|, |y| is again 0 or +..
so if x is -ive , x|y| will be -ive and it will not be > + value... so x has to be +ive ..
and the numeric value has to be >y for x*y to be > than y*y
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html

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Posts: 53020

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22 May 2016, 02:39
nishatfarhat87 wrote:
Is x·|y| > y^2?

(1) x > y
(2) y > 0

Hi Bunuel,

Is x·|y| > y^2?

(1) x > y. If y=0, then no matter what x is x*|y| = y^2 = 0, and we'll have a NO answer to the question but if y=1 and x=2, then x*|y| > y^2, and we'll have an YES answer to the question. Not sufficient.

(2) y > 0. This implies that |y| = y, thus the question becomes is x·y > y^2. Reduce by y (we can safely do this since we know that y is positive): is x > y. We don't know that. Not sufficient.

(1)+(2) From (2) the question became whether x > y and (1) confirms that. Sufficient.

Hope it's clear.
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Is x*|y| > y^2?   [#permalink] 22 May 2016, 02:39

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