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# M02 Q14, what if the X and y are negative?

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M02 Q14, what if the X and y are negative? [#permalink]

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05 Mar 2009, 00:15
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$$x$$ and $$y$$ are positive integers and $$x > y$$ . What is the value of $$xy^2 + yx^2$$ ?

1. xy = 6
2. x is a prime number

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

$$xy^2+ yx^2 = xy(x+y)$$

* From S1, we know that $$xy=6$$ , but we do not know what is $$x+y$$ . Insufficient
* From S2, we know that $$x$$ is a prime number, but there is no information about $$y$$ . Insufficient

Combining the statements, we know that $$xy=6$$ . So $$x$$ is a prime number and $$x>y$$ , therefore $$y=2$$ . The problem can then be solved.
The correct answer is C.

----------------------------------------------------------
But what happens if X & Y are negative as thats a possbility? So the answer must be E
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Re: m02. Q14, what if the X and y are negative? [#permalink]

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05 Mar 2009, 06:20
Forrester300 wrote:
$$x$$ and $$y$$ are positive integers and $$x > y$$ . What is the value of $$xy^2 + yx^2$$ ?

1. xy = 6
2. x is a prime number

(C) 2008 GMAT Club - m02#14

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

$$xy^2+ yx^2 = xy(x+y)$$

* From S1, we know that $$xy=6$$ , but we do not know what is $$x+y$$ . Insufficient
* From S2, we know that $$x$$ is a prime number, but there is no information about $$y$$ . Insufficient

Combining the statements, we know that $$xy=6$$ . So $$x$$ is a prime number and $$x>y$$ , therefore $$y=2$$ . The problem can then be solved.
The correct answer is C.

----------------------------------------------------------
But what happens if X & Y are negative as thats a possbility? So the answer must be E

2 things to note:

1. x and y are +ves
2. x is a prime, which can never be -ve.

hth
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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10 Nov 2009, 22:50
The solution can be reached even if X>Y not given (because value of X is not asked, so it doesn't matter as long as we find out that X can be either 2 or 3 and depending on the value of X, Y can be one of 2 or 3).
So unless "X>Y" is meant to be a red herring, it can be removed from the Q.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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18 Mar 2010, 07:26
Forrester300 wrote:
$$x$$ and $$y$$ are positive integers and $$x > y$$ . What is the value of $$xy^2 + yx^2$$ ?

1. xy = 6
2. x is a prime number

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

$$xy^2+ yx^2 = xy(x+y)$$

* From S1, we know that $$xy=6$$ , but we do not know what is $$x+y$$ . Insufficient
* From S2, we know that $$x$$ is a prime number, but there is no information about $$y$$ . Insufficient

Combining the statements, we know that $$xy=6$$ . So $$x$$ is a prime number and $$x>y$$ , therefore $$y=2$$ . The problem can then be solved.
The correct answer is C.

----------------------------------------------------------
But what happens if X & Y are negative as thats a possbility? So the answer must be E

Its C. explanation given above.
It is given that x & y are (+)ve.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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18 Mar 2010, 10:24
Well.. Q is saying, X and Y is +ve and they are both integers and X>Y.

ST 1 is XY=6, X could be either 6 and Y=1 OR x=3 and Y=2

so its insuff

ST 2... we dont know anything about Y or any value for any of X or Y... insuff

Together x has to be 3 and Y is 2... suff

Ans C
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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19 Mar 2010, 07:29
Question is saying that both the numbers are positive integers, so there is no question of considering the case when they are negative.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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22 Mar 2011, 05:57
Question stem reworded: What's the value of XY (Y + X) when x & Y are positive integers and x > y?

1) xy = 6. We still need to determine (Y+X) (insufficient)

2) X is prime. X can be 2, 3, 5, 7, etc and we know nothing of Y (insufficient)

1 and 2) Together, xy=6. Therefore x or y has to be an even integer. Since x is prime and larger than Y, X has to be 3 and Y 2. Both statements together provide that Xy(Y+X) = 30

Ans: C
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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22 Mar 2011, 06:02
ANS IS C.

XY(X+Y)=?

ON COMBINING TWO STATEMENTS;
XY=6
AND X IS PRIME AND X>Y; ONLY 3 SATISFIES BOTH CONDITION.
SO, X=3, Y=2

ONE CAN FIND THE ANS AS 30.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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22 Mar 2011, 07:10
xy^2 + yx^2

= xy(X+y)

(1) xy = 6, so x = 6 and y = 1, or x = 3 and y = 2, (x+y) is different, so not sufficient.

(2) x is prime, but no information about y, so not sufficient.

From (1) and (2), x = 3, y = 2, so sufficient.

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Re: M02 Q14, what if the X and y are negative? [#permalink]

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22 Mar 2011, 21:43
If they state that x and y are negative integers, x cannot be a prime number.

Thus the question would become invalid.

~Nilay
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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26 Mar 2012, 09:50
The correct answer is C.

if notice that that statement one gives you (x)(y)=6

then you can see that the only factors for 6 is (1)(6) and (2)(3)
for X to be greater than Y. you will need x to be either 6 or 3

Now the second statement tells you that x is prime leaving us with the possibility of 3.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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26 Mar 2012, 12:18
Easy one from the GMATCLUB which rarely happens. Such questions really motivate you if face such questions after doing many wrong.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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27 Mar 2013, 05:13
x and y are positive integers and x>y . What is the value of xy2 + yx2 ?

1. xy = 6
2. x is a prime number

For Case 1:
xy = 6 and x>y, has two solution x,y = {3,2 and 6,1}
for 3,2 the value is 12 + 18 = 30
for 6,1 the value is 6 + 36 = 42

For Case 2:
x is prime=> x can be 2,3,5
and y can have any value so the equation has infinite values.

Combining case 1 and 2:
The only possible solution for x,y is 3,2 and the equation solves to 30.

Hence C.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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27 Mar 2013, 05:29
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Can anyone tell the aproximate level of this question?

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Re: M02 Q14, what if the X and y are negative? [#permalink]

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27 Mar 2013, 05:45
Solution C

1) xy = 6

Considering x>y, below two scenarios can come up
a) x = 6
y = 1
Putting the above x & Y values in the equuation xy2+ yx2 = 6+36 = 42

b) x = 3
y = 2
Putting the above x & Y values in the equuation xy2+ yx2 = (3*4)+(2*9) = 40

Two solutions so 1 itself is not sufficient.

2) x is a prime number

x can be anything 1,3,5, 7.. so as y. not sufficient

Taking 1 & 2 together
XY = 6, X is a prime number & X >Y we get unique values for X & Y

i.e. X=3 & Y = 2, this will get you the unique solutin of the equation xy2+ yx2
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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27 Mar 2013, 06:36
chose A ... x+y is not known but has a fixed value .... therefore sufficient . Can anybody please explain why this cannot be correct?
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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27 Mar 2013, 06:50
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TheNona wrote:
chose A ... x+y is not known but has a fixed value .... therefore sufficient . Can anybody please explain why this cannot be correct?

When a DS question asks about the value of some variable (or an expression with variable(a)), then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable (expression).

x and y are positive integers and x > y . What is the value of xy^2 + yx^2 ?

What is the value of $$xy^2 + yx^2=xy(y+x)$$?

(1) xy = 6 --> $$xy(y+x)=6(y+x)$$. Now, since x and y are positive integers and x > y, then from xy = 6 it follows that x=6 and y=1 OR x=3 and y=2, thus 6(y+x)=42 or 6(y+x)=30. Two different values, thus this statement is NOT sufficient.

(2) x is a prime number. Clearly insufficient.

(1)+(2) Since from (2) we know that x is a prime number, then x=3=prime and y=2, thus 6(y+x)=42. Sufficient.

Hope it's clear.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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27 Mar 2013, 06:54
Recobita wrote:
Can anyone tell the aproximate level of this question?

I'd say the difficulty level is ~600.
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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27 Mar 2013, 07:15
Bunuel wrote:
TheNona wrote:
chose A ... x+y is not known but has a fixed value .... therefore sufficient . Can anybody please explain why this cannot be correct?

When a DS question asks about the value of some variable (or an expression with variable(a)), then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable (expression).

x and y are positive integers and x > y . What is the value of xy^2 + yx^2 ?

What is the value of $$xy^2 + yx^2=xy(y+x)$$?

(1) xy = 6 --> $$xy(y+x)=6(y+x)$$. Now, since x and y are positive integers and x > y, then from xy = 6 it follows that x=6 and y=1 OR x=3 and y=2, thus 6(y+x)=42 or 6(y+x)=30. Two different values, thus this statement is NOT sufficient.

(2) x is a prime number. Clearly insufficient.

(1)+(2) Since from (2) we know that x is a prime number, then x=3=prime and y=2, thus 6(y+x)=42. Sufficient.

Hope it's clear.

yessssssssssssssssss ... I did not consider the option of 6& 1... only 2&3 . Thanks Bunuel
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Re: M02 Q14, what if the X and y are negative? [#permalink]

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27 Mar 2013, 11:38
saratchandra wrote:
Solution C

1) xy = 6

Considering x>y, below two scenarios can come up
a) x = 6
y = 1
Putting the above x & Y values in the equuation xy2+ yx2 = 6+36 = 42

b) x = 3
y = 2
Putting the above x & Y values in the equuation xy2+ yx2 = (3*4)+(2*9) = 40

Two solutions so 1 itself is not sufficient.

2) x is a prime number

x can be anything 1,3,5, 7.. so as y. not sufficient

Taking 1 & 2 together
XY = 6, X is a prime number & X >Y we get unique values for X & Y

i.e. X=3 & Y = 2, this will get you the unique solutin of the equation xy2+ yx2

So far the best explanation-Thanks Dude.
Re: M02 Q14, what if the X and y are negative?   [#permalink] 27 Mar 2013, 11:38

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# M02 Q14, what if the X and y are negative?

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