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# M12-16

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Math Expert
Joined: 02 Sep 2009
Posts: 51067

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15 Sep 2014, 23:47
1
6
00:00

Difficulty:

95% (hard)

Question Stats:

39% (00:58) correct 61% (01:00) wrong based on 135 sessions

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If $$x$$ and $$y$$ are positive integers, is $$\frac{x}{y}+ \frac{y}{x} \gt 2$$?

(1) $$x$$ does not equal $$y$$

(2) $$x$$ and $$y$$ do not share any common divisors except 1

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Joined: 02 Sep 2009
Posts: 51067

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15 Sep 2014, 23:47
1
2
Official Solution:

The question asks: is $$\frac{x}{y}+\frac{y}{x} \gt 2$$? Or: is $$\frac{x^2+y^2}{xy} \gt 2$$?

Since both unknowns are positive then we can safely multiply by $$xy$$:

Is $$x^2+y^2 \gt 2xy$$?

Is $$x^2-2xy+y^2\gt 0$$?

Is $$(x-y)^2 \gt 0$$? Now, if $$x$$ does not equal $$y$$ the answer to this question will be YES, but if $$x=y$$, then the answer will be NO, since in this case $$(x-y)^2=0$$.

(1) $$x$$ does not equal $$y$$. Directly answers the question. Sufficient.

(2) $$x$$ and $$y$$ do not share any common divisors except 1. If $$x=y=1$$ then the answer is NO, but if $$x=1$$ and $$y=2$$, then the answer is YES. Not sufficient.

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Joined: 14 Jul 2014
Posts: 93

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24 Mar 2015, 10:02
Bunuel wrote:
Official Solution:

The question asks: is $$\frac{x}{y}+\frac{y}{x} \gt 2$$? Or: is $$\frac{x^2+y^2}{xy} \gt 2$$?

Since both unknowns are positive then we can safely multiply by $$xy$$:

Is $$x^2+y^2 \gt 2xy$$?

Is $$x^2-2xy+y^2\gt 0$$?

Is $$(x-y)^2 \gt 0$$? Now, if $$x$$ does not equal $$y$$ the answer to this question will be YES, but if $$x=y$$, then the answer will be NO, since in this case $$(x-y)^2=0$$.

(1) $$x$$ does not equal $$y$$. Directly answers the question. Sufficient.

(2)$$x$$ and $$y$$ do not share any common divisors except 1. If $$x=y=1$$ then the answer is NO, but if $$x=1$$ and $$y=2$$, then the answer is YES. Not sufficient.

Hi Bunuel

Doesnt Stmnt 2 mean that x & y are PRIME ?
In that case this statement is Sufficient

Pls clarify
Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 51067

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24 Mar 2015, 11:11
buddyisraelgmat wrote:
Bunuel wrote:
Official Solution:

The question asks: is $$\frac{x}{y}+\frac{y}{x} \gt 2$$? Or: is $$\frac{x^2+y^2}{xy} \gt 2$$?

Since both unknowns are positive then we can safely multiply by $$xy$$:

Is $$x^2+y^2 \gt 2xy$$?

Is $$x^2-2xy+y^2\gt 0$$?

Is $$(x-y)^2 \gt 0$$? Now, if $$x$$ does not equal $$y$$ the answer to this question will be YES, but if $$x=y$$, then the answer will be NO, since in this case $$(x-y)^2=0$$.

(1) $$x$$ does not equal $$y$$. Directly answers the question. Sufficient.

(2)$$x$$ and $$y$$ do not share any common divisors except 1. If $$x=y=1$$ then the answer is NO, but if $$x=1$$ and $$y=2$$, then the answer is YES. Not sufficient.

Hi Bunuel

Doesnt Stmnt 2 mean that x & y are PRIME ?
In that case this statement is Sufficient

Pls clarify
Thanks

There are two examples in the solution which show that x and y are not necessarily primes.
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Joined: 27 Aug 2015
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07 Jan 2016, 13:52
I kinda get lost here.

Is (x−y)2>0? Now, if x does not equal y the answer to this question will be YES, but if x=y, then the answer will be NO, since in this case (x−y)2=0.

(1) x does not equal y. Directly answers the question. Sufficient.

(2)x and y do not share any common divisors except 1. If x=y=1 then the answer is NO, but if x=1 and y=2, then the answer is YES. Not sufficient.

And how does statement not say x and y are not prime?
Math Expert
Joined: 02 Sep 2009
Posts: 51067

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08 Jan 2016, 00:34
kkahuja91 wrote:
I kinda get lost here.

Is (x−y)2>0? Now, if x does not equal y the answer to this question will be YES, but if x=y, then the answer will be NO, since in this case (x−y)2=0.

(1) x does not equal y. Directly answers the question. Sufficient.

(2)x and y do not share any common divisors except 1. If x=y=1 then the answer is NO, but if x=1 and y=2, then the answer is YES. Not sufficient.

And how does statement not say x and y are not prime?

Your question is not clear. Only thing we want to know to answer the question is whether x = y. Why do we need to know whether w and y re primes?
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Math Expert
Joined: 02 Aug 2009
Posts: 7096

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08 Jan 2016, 00:56
kkahuja91 wrote:
I kinda get lost here.

Is (x−y)2>0? Now, if x does not equal y the answer to this question will be YES, but if x=y, then the answer will be NO, since in this case (x−y)2=0.

(1) x does not equal y. Directly answers the question. Sufficient.

(2)x and y do not share any common divisors except 1. If x=y=1 then the answer is NO, but if x=1 and y=2, then the answer is YES. Not sufficient.

And how does statement not say x and y are not prime?

Hi,
firstly the statement two does not say that x and y are prime, even if it included the info that x and y are not same..
It would have just meant that the two are prime to each other, say 3,8 or 5,7 etc

Now,if you intend to ask the way statement 2 would have been written to say that X and y are not prime...
Two scenarios..
X and Y are prime..
1) X and y are two different positive integers greater than 1, having only two factors( or having only one factor other than 1)

X and Y are not prime..
1) X and Y are two integers having more than One common factor..

I could make this out as your query..
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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

GMAT online Tutor

Intern
Joined: 29 Mar 2015
Posts: 11

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21 Aug 2016, 03:38
The definition of a prime number is: a number that doesn't have a factor apart from 1 and itself.
The "and itself" part is very important as it shows that all prime numbers have EXACTLY 2 factors while 1 just has 1 factor and thus 1 is not a prime number.

Hope it helps.
Manager
Joined: 19 Jul 2016
Posts: 50

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25 Dec 2016, 08:12
kindly explain how it could be A
from S1
CASE 1::
if X=12 and Y =13 then,x/y+y/x is not >2
CASE 2::
X=1,Y=2 then X/Y+Y/X is >2
NOT SUFFICIENT
Math Expert
Joined: 02 Sep 2009
Posts: 51067

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25 Dec 2016, 08:16
1
gupta87 wrote:
kindly explain how it could be A
from S1
CASE 1::
if X=12 and Y =13 then,x/y+y/x is not >2
CASE 2::
X=1,Y=2 then X/Y+Y/X is >2
NOT SUFFICIENT

12/13 + 13/12 = 313/156 > 2
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Joined: 26 Sep 2017
Posts: 3

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05 Oct 2017, 17:37
Quick question:

Statement 2: would imply that the numbers are consecutive, BUT only if the initial question stem characterized X and Y as two distinct positive integers, correct?

I picked D for this question, upon review I understand why it is not the correct answer. I would like to make sure my understanding of statement two is correct.

Thank you very much.
Math Expert
Joined: 02 Sep 2009
Posts: 51067

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05 Oct 2017, 20:10
georgetags10 wrote:
Quick question:

Statement 2: would imply that the numbers are consecutive, BUT only if the initial question stem characterized X and Y as two distinct positive integers, correct?

I picked D for this question, upon review I understand why it is not the correct answer. I would like to make sure my understanding of statement two is correct.

Thank you very much.

No. Even then, x and y have not to be consecutive integers, just co-prime. For example, 1 and any other integer, any two distinct primes (for example 2 and 7), ...
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Joined: 04 Jan 2018
Posts: 8
Location: India
Concentration: Finance, General Management
GMAT 1: 750 Q49 V44

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07 May 2018, 04:57
Bunuel wrote:
Official Solution:

The question asks: is $$\frac{x}{y}+\frac{y}{x} \gt 2$$? Or: is $$\frac{x^2+y^2}{xy} \gt 2$$?

Since both unknowns are positive then we can safely multiply by $$xy$$:

Is $$x^2+y^2 \gt 2xy$$?

Is $$x^2-2xy+y^2\gt 0$$?

Is $$(x-y)^2 \gt 0$$? Now, if $$x$$ does not equal $$y$$ the answer to this question will be YES, but if $$x=y$$, then the answer will be NO, since in this case $$(x-y)^2=0$$.

(1) $$x$$ does not equal $$y$$. Directly answers the question. Sufficient.

(2) $$x$$ and $$y$$ do not share any common divisors except 1. If $$x=y=1$$ then the answer is NO, but if $$x=1$$ and $$y=2$$, then the answer is YES. Not sufficient.

I did the square root of (x-y)^2>0 to get x>y. Can we not do that in this case?
Math Expert
Joined: 02 Sep 2009
Posts: 51067

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07 May 2018, 05:01
rhnbansal wrote:
Bunuel wrote:
Official Solution:

The question asks: is $$\frac{x}{y}+\frac{y}{x} \gt 2$$? Or: is $$\frac{x^2+y^2}{xy} \gt 2$$?

Since both unknowns are positive then we can safely multiply by $$xy$$:

Is $$x^2+y^2 \gt 2xy$$?

Is $$x^2-2xy+y^2\gt 0$$?

Is $$(x-y)^2 \gt 0$$? Now, if $$x$$ does not equal $$y$$ the answer to this question will be YES, but if $$x=y$$, then the answer will be NO, since in this case $$(x-y)^2=0$$.

(1) $$x$$ does not equal $$y$$. Directly answers the question. Sufficient.

(2) $$x$$ and $$y$$ do not share any common divisors except 1. If $$x=y=1$$ then the answer is NO, but if $$x=1$$ and $$y=2$$, then the answer is YES. Not sufficient.

I did the square root of (x-y)^2>0 to get x>y. Can we not do that in this case?

The point is that $$\sqrt{x^2}=|x|$$. So, if you take the square root from $$(x-y)^2 \gt 0$$ you'll get $$|x-y| \gt 0$$, which is true for all values of x and y but the case when x = y (the same conclusion as we got in the solution).
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Joined: 04 Jan 2018
Posts: 8
Location: India
Concentration: Finance, General Management
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07 May 2018, 05:15
Bunuel wrote:
rhnbansal wrote:
Bunuel wrote:
Official Solution:

The question asks: is $$\frac{x}{y}+\frac{y}{x} \gt 2$$? Or: is $$\frac{x^2+y^2}{xy} \gt 2$$?

Since both unknowns are positive then we can safely multiply by $$xy$$:

Is $$x^2+y^2 \gt 2xy$$?

Is $$x^2-2xy+y^2\gt 0$$?

Is $$(x-y)^2 \gt 0$$? Now, if $$x$$ does not equal $$y$$ the answer to this question will be YES, but if $$x=y$$, then the answer will be NO, since in this case $$(x-y)^2=0$$.

(1) $$x$$ does not equal $$y$$. Directly answers the question. Sufficient.

(2) $$x$$ and $$y$$ do not share any common divisors except 1. If $$x=y=1$$ then the answer is NO, but if $$x=1$$ and $$y=2$$, then the answer is YES. Not sufficient.

I did the square root of (x-y)^2>0 to get x>y. Can we not do that in this case?

The point is that $$\sqrt{x^2}=|x|$$. So, if you take the square root from $$(x-y)^2 \gt 0$$ you'll get $$|x-y| \gt 0$$, which is true for all values of x and y but the case when x = y (the same conclusion as we got in the solution).

Understood. I assumed that just because they both are positive integers, we could do the square root and get x>y. Totally missed out on the fact that it could be zero as well.

Thank you so much!
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Joined: 25 Dec 2017
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28 Nov 2018, 10:06
while simplifying we have got
is (x-y)>0?
is x>y?

1- says x does not equal y
how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall.
Math Expert
Joined: 02 Sep 2009
Posts: 51067

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28 Nov 2018, 20:20
1
shivam5511 wrote:
while simplifying we have got
is (x-y)>0?
is x>y?

1- says x does not equal y
how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall.

Please read carefully. After simplifying we got "is (x - y)^2 > 0"
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29 Nov 2018, 06:52
Bunuel wrote:
shivam5511 wrote:
while simplifying we have got
is (x-y)>0?
is x>y?

1- says x does not equal y
how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall.

Please read carefully. After simplifying we got "is (x - y)^2 > 0"

Exactly i am asking the same thing.
(x-y)^2>0
we can also right it as-
(x-y) (x+y) >0
so x and y is positive integer
then, we arrived at
is (x-y)>0?
again, is x>y?

statement 1 - x does not equal y ( This statement tell us x not equal to y. But nothing to say about x is greater than y or smaller than y).

Please tell me where i fall. while simplifying is there any probability of x=y?
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Posts: 16974
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GMAT 1: 750 Q49 V42

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29 Nov 2018, 07:00
2
shivam5511 wrote:
Bunuel wrote:
shivam5511 wrote:
while simplifying we have got
is (x-y)>0?
is x>y?

1- says x does not equal y
how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall.

Please read carefully. After simplifying we got "is (x - y)^2 > 0"

Exactly i am asking the same thing.
(x-y)^2>0
we can also right it as-
(x-y) (x+y) >0
so x and y is positive integer
then, we arrived at
is (x-y)>0?
again, is x>y?

statement 1 - x does not equal y ( This statement tell us x not equal to y. But nothing to say about x is greater than y or smaller than y).

Please tell me where i fall. while simplifying is there any probability of x=y?

Here is on in a million trick for you - PLUG NUMBERS to check your math and your answers. It can also help you solve the question.
Let's say X=10 and Y=3
1. (x-y)^2 = 49
2. (x-y)(x+y)=91
3. x^2-2xy+y^2 = 100-60+9 = 49
4. (x-y)(x-y) = 49

You spot your mistake? It is the fourth grade formula. This is a very basic mistake - it shows you have a weak base - you need to go way back to fix it.
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Joined: 02 Sep 2009
Posts: 51067

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29 Nov 2018, 08:57
shivam5511 wrote:
Bunuel wrote:
shivam5511 wrote:
while simplifying we have got
is (x-y)>0?
is x>y?

1- says x does not equal y
how this can be correct? whether this statement haven't say x is greater or y is smaller. Please tell me where i fall.

Please read carefully. After simplifying we got "is (x - y)^2 > 0"

Exactly i am asking the same thing.
(x-y)^2>0
we can also right it as-
(x-y) (x+y) >0

so x and y is positive integer
then, we arrived at
is (x-y)>0?
again, is x>y?

statement 1 - x does not equal y ( This statement tell us x not equal to y. But nothing to say about x is greater than y or smaller than y).

Please tell me where i fall. while simplifying is there any probability of x=y?

Red part is not correct.

$$(x-y)^2 = x^2-2xy+y^2$$ NOT $$(x-y) (x+y)$$, which in turn equals to $$x^2 - y^2$$.
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Re: M12-16 &nbs [#permalink] 29 Nov 2018, 08:57

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# M12-16

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