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# m04 #24

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09 Oct 2008, 14:47
Is $$x = 0$$ ?

1. $$xy = x$$
2. $$x + y = y$$

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

Please explain why statement 1 is insufficient.
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09 Oct 2008, 18:26
1
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gmat4life wrote:
Is x=0?

1. xy=x

2. x+y=y

Please explain why statement 1 is insufficient.

Now xy=x means xy-x=0 or x(1-y)=0
For the equation x(1-y)=0 to hold good, either x has to be 0 or y has to be 1. It can very well be that x=0 and y=1 at the same time. We are not sure. So since it is not clear if x=0 or not, this statement is not sufficient.

Now in statement 2. x+y=y clearly shows x=0.

So it must be B

HTH
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22 Oct 2008, 06:04
Thanks for a good explanation!
+1
amitdgr wrote:
gmat4life wrote:
Is x=0?

1. xy=x

2. x+y=y

Please explain why statement 1 is insufficient.

Now xy=x means xy-x=0 or x(1-y)=0
For the equation x(1-y)=0 to hold good, either x has to be 0 or y has to be 1. It can very well be that x=0 and y=1 at the same time. We are not sure. So since it is not clear if x=0 or not, this statement is not sufficient.

Now in statement 2. x+y=y clearly shows x=0.

So it must be B

HTH

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Re: m04 #24 [#permalink]

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15 Oct 2010, 05:02
This question goes to show that when you have been looking at 700 level questions too long, you try and pick holes in something as simple as x = y - y ....

on the other hand, its good to finally get a question right!
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15 Oct 2010, 08:36
For the first option, lets assume Y = 1. So whatever the value of x, the product of XY is always X.

X * 1 = X (For any values of X). So, option 1 is ruled out..
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Re: m04 #24 [#permalink]

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16 Oct 2010, 23:07
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for statement 1 :
x*y=x
if x=0 and y=1 then 0*1=0 ; satisfies
if x=2 and y=1 then 2*1=2 ; satisfies

thus statement 1 is insufficient to say that x is 0

for statement 2:
x+y=y
if x=0 and y=any integer then 0+y = y ; satisfies
if x=2(or any int but not zero) and y=any int then x+y=y is not satisfying

thus statement 2 is sufficient to say that x=0

SO B

But what would be the answer if the second statement was : x + y = x ??
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Re: m04 #24 [#permalink]

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18 Oct 2010, 04:48
Hi Jimmy86;
if (2) where: x + y = x
--> y = 0 [no information about x after subtracting x from both sides]

combining (1) & (2) [x*y=x; x+y=x]:
x(0) = x
for all values of x, x*0 is always a zero.
So, we can say x (the result of the multiplication) is a zero.
IMO C
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18 Oct 2010, 05:20
If X+Y=X, then I go for E..

we have options,

x * y = x and

x + y = x.

Combining 1 and 2,

(x + y) * y = x,

xy + y^2 = x

y (x + y) = x and so, x will be zero only if Y is also 0 or else the condition is not satisfied. So, IMO its E.

jimmy86 wrote:
for statement 1 :
x*y=x
if x=0 and y=1 then 0*1=0 ; satisfies
if x=2 and y=1 then 2*1=2 ; satisfies

thus statement 1 is insufficient to say that x is 0

for statement 2:
x+y=y
if x=0 and y=any integer then 0+y = y ; satisfies
if x=2(or any int but not zero) and y=any int then x+y=y is not satisfying

thus statement 2 is sufficient to say that x=0

SO B

But what would be the answer if the second statement was : x + y = x ??
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18 Oct 2010, 06:06
[quote="sivai87"]If X+Y=X, then I go for E..

we have options,

x * y = x and

x + y = x.

Combining 1 and 2,

(x + y) * y = x,

xy + y^2 = x

y (x + y) = x and so, x will be zero only if Y is also 0 or else the condition is not satisfied. So, IMO its E.

I agree... its E.

in case statement 2 : x+y = x
x=-1 , y=0 gives -1+0 = -1 ; satisfies
x=0 , y= 0 gives 0+0 = 0 ; satisfies
we see x is not necessarily 0. Insufficient

And combining
y(x+y)=x
if x=0 ; y=0 then 0*(0+0)=0 ; satisfies
if x=1/2 ; y=1/2 then 1/2*(1/2+1/2)=1/2 ; satisfies
So x is not necessarily 0.Insufficient

Thus IMO its E

But what if the question states that x is an integer.
Then i suppose the only value that of x that satisfies y(x+y)=x is x=0. and answer would be C(not sure, feel i am missing something)
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Re: m04 #24 [#permalink]

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18 Oct 2010, 07:05
Supasubi wrote:
This question goes to show that when you have been looking at 700 level questions too long, you try and pick holes in something as simple as x = y - y ....

on the other hand, its good to finally get a question right!

Haha yes! I definitely know what you mean. You're so traumatized from every question trying to trick you, that you're reluctant to believe that the answer could be sufficient with something as easy as X=Y-Y. I stared at it for a good 30 seconds trying to pick it apart too. That's hilarious.
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18 Oct 2010, 08:30
(2) if y = x - x, then, it is sufficient to say y = "0".

plug in y (0) into (1) [x*y = x]
or x = x*y
x = x*0
RHS = x*0 = 0
for any value of a number, 0 times that number = 0
So, RHS is always "0"
LHS = x
LHS = RHS
--> x = RHS
x = 0
Sufficient

So, C IMO....except am doing something stupid somewhere.
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18 Oct 2010, 09:48
@Gmatbull..

Buddy, I dont think u went wrong... I believe I am somewhere wrong..

The first mistake i made was, in substitution...
I shud have substituted as (x+y) * y = (x+y)... instead of (X+Y)*Y= X..

But even after this I am getting (X+Y)*Y=(X+Y) ---> Y=1.. and this will finally yield that Both option are not suffiient...
And I am not sure how to proceed further...

Can anyone help me pls...

gmatbull wrote:
(2) if y = x - x, then, it is sufficient to say y = "0".

plug in y (0) into (1) [x*y = x]
or x = x*y
x = x*0
RHS = x*0 = 0
for any value of a number, 0 times that number = 0
So, RHS is always "0"
LHS = x
LHS = RHS
--> x = RHS
x = 0
Sufficient

So, C IMO....except am doing something stupid somewhere.
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Re: m04 #24 [#permalink]

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18 Oct 2010, 11:19
sivai87 wrote:
@Gmatbull..
The first mistake i made was, in substitution...
I shud have substituted as (x+y) * y = (x+y)... instead of (X+Y)*Y= X..

But even after this I am getting (X+Y)*Y=(X+Y) ---> Y=1.. and this will finally yield that Both option are not suffiient...
And I am not sure how to proceed further...

I think you seem to make the expression difficult to yourself.
First, realize that (2) has simplified the job:
from x+y = x, you know y = 0.
(x+y) * y = (x+y) ==> (x+y)0 = x+0
0 = x +0
0 = x
you still arrive at C.
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20 Oct 2010, 03:54
@ GMAT BULL

Is X = 0?

1) X * Y = X

2) X + Y = X

The statements 1 and 2 are contradicting to each other..
Option 1 says that, X * Y = X which tends to give Y = 1,

But option 2 gives, X + Y = X, and So Y = 0.
I believe such questions wont appear in GMAT and so less to worry about this question...

gmatbull wrote:
sivai87 wrote:
@Gmatbull..
The first mistake i made was, in substitution...
I shud have substituted as (x+y) * y = (x+y)... instead of (X+Y)*Y= X..

But even after this I am getting (X+Y)*Y=(X+Y) ---> Y=1.. and this will finally yield that Both option are not suffiient...
And I am not sure how to proceed further...

I think you seem to make the expression difficult to yourself.
First, realize that (2) has simplified the job:
from x+y = x, you know y = 0.
(x+y) * y = (x+y) ==> (x+y)0 = x+0
0 = x +0
0 = x
you still arrive at C.
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20 Oct 2010, 04:16
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Expert's post
sivai87 wrote:
@ GMAT BULL

Is X = 0?

1) X * Y = X

2) X + Y = X

The statements 1 and 2 are contradicting to each other..
Option 1 says that, X * Y = X which tends to give Y = 1,

But option 2 gives, X + Y = X, and So Y = 0.
I believe such questions wont appear in GMAT and so less to worry about this question...

Not so. First of all statement (2) is $$x+y=y$$ and not $$x+y=x$$.

Next, (1) $$xy=x$$ --> $$x(y-1)=0$$ --> either $$x=0$$ (and $$y$$ can take any value) OR $$y=1$$ (and $$x$$ can take any value). So we can not say that $$y=1$$ from this statement.

If the question were:

Is $$x = 0$$?

(1) $$xy = x$$
(2) $$x+y=x$$ (Note that this statement is changed)

Then the answer would be C:

(1) $$xy=x$$ --> $$x(y-1)=0$$ --> either $$x=0$$ (and $$y$$ can take any value) OR $$y=1$$ (and $$x$$ can take any value). Not sufficient.
(2) $$x+y=x$$ --> $$y=0$$. Not sufficient to answer whether $$x=0$$.

(1)+(2) As from (2) $$y=0\neq{1}$$ then according to (1) $$x=0$$. Sufficient.

So for this revised question answer would be C.

Hope it helps.
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Re: m04 #24 [#permalink]

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19 Oct 2011, 08:14
statement 1 XY=X means y=1 but X can take any value not sufficient
statement 2 x + y = Y means X = y-y or X=0 sufficient
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19 Oct 2011, 11:51
Bunuel wrote:
sivai87 wrote:
@ GMAT BULL

Is X = 0?

1) X * Y = X

2) X + Y = X

The statements 1 and 2 are contradicting to each other..
Option 1 says that, X * Y = X which tends to give Y = 1,

But option 2 gives, X + Y = X, and So Y = 0.
I believe such questions wont appear in GMAT and so less to worry about this question...

Not so. First of all statement (2) is $$x+y=y$$ and not $$x+y=x$$.

Next, (1) $$xy=x$$ --> $$x(y-1)=0$$ --> either $$x=0$$ (and $$y$$ can take any value) OR $$y=1$$ (and $$x$$ can take any value). So we can not say that $$y=1$$ from this statement.

If the question were:

Is $$x = 0$$?

(1) $$xy = x$$
(2) $$x+y=x$$ (Note that this statement is changed)

Then the answer would be C:

(1) $$xy=x$$ --> $$x(y-1)=0$$ --> either $$x=0$$ (and $$y$$ can take any value) OR $$y=1$$ (and $$x$$ can take any value). Not sufficient.
(2) $$x+y=x$$ --> $$y=0$$. Not sufficient to answer whether $$x=0$$.

(1)+(2) As from (2) $$y=0\neq{1}$$ then according to (1) $$x=0$$. Sufficient.

So for this revised question answer would be C.

Hope it helps.

I seem to recall that you can't add/subtract variables from both sides of the equation unless you know the sign of the variables. But the question doesn't tell you the sign of x or y so I don't think the problem can be approached this way. I think the only way to answer the question is to plug in numbers for x and y then check to see if the equation holds.
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19 Oct 2011, 12:32
St 1:
xy = x. Either x=0 and y can take any value OR x can be any value and y=1. So insufficient.

St2
x+y = Y => X=0, Y can be anything. Hence Sufficient.

OA is B.
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http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
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Re: m04 #24 [#permalink]

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20 Nov 2011, 06:44
B.

1 tells us that y=1 and 2 tells us that x=0.
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Re: m04 #24 [#permalink]

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07 Oct 2012, 11:47
Quote:
Is $$x = 0$$ ?

1. $$xy = x$$
2. $$x + y = y$$

For Option1, which of the below method need to be followed?

Method1 ---- $$xy = x$$ ----> $$xy-x = 0$$ ----> $$x(y-1) = 0$$ ----> $$x = 0, y = 1$$ ---- I thought this method is always true in determining a value
Method2 ---- $$xy = x$$ - Substitute values $$x=0$$ & $$y$$can be any value? or $$x$$ can be any value and $$y=1?$$
Re: m04 #24   [#permalink] 07 Oct 2012, 11:47

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