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X = 0 ?

by MAAJ » Mon Apr 25, 2011 10:55 am
Is x = 0?

1) 3x = 2xy
2) (x^2) = x

----------

IMO [spoiler](C)[/spoiler]

1) 3x = 2xy
3x - 2xy = 0
x (3-2y) = 0
x = 0 OR 3-2y = 0

2)(x^2) = x
x = 0 OR x = 1

3)Combining 1 and 2:
x must be 0
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by MAAJ » Mon Apr 25, 2011 11:14 am
Woops! wrong forum, should be moved to DS Section.
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by venmic » Mon Apr 25, 2011 11:09 pm
Should this answer not be A alone becuase

3x - 2xy = 0
then
x(3-2Y) = 0

then

X=0 and y = 3/2

so sufficent

how can it be C
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by jaymw » Mon Apr 25, 2011 11:37 pm
Is x = 0?

1) 3x = 2xy
2) (x^2) = x
It does not personally bother me that you posted this question in the "wrong" forum, because we're still talking about math stuff here:) However, please make sure that you include the source whenever you post a question. Also, it would be good to include the OA using the spoiler tag.

Now to the question:

Is x=0 ?

Statement 1: 3x=2xy

if y=1.5 then x could be any value, including zero.

Insufficient.

Statement 2: x²=x

This is true for x=0 and x=1.

Insufficient.

Combining the statements:

x=0 would make both statements correct.

x=1 could also make both statements correct, because:

3(1)=2(1)*y, is true if y=1.5

Insufficent.
[spoiler]
The answer is E.[/spoiler]
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by jaymw » Mon Apr 25, 2011 11:41 pm
venmic wrote:Should this answer not be A alone becuase

3x - 2xy = 0
then
x(3-2Y) = 0

then

X=0 and y = 3/2

so sufficent

how can it be C
That's actually not true. Although your rearrangement was correct, the conclusion drawn from it (X=0 and y = 3/2) was not. For a product to equal zero, it is ENOUGH when ONE of the factors is equal to zero. Thus, the proper conclusion is X=0 OR y = 3/2.
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by pankajks2010 » Tue Apr 26, 2011 12:41 am
venmic wrote:Should this answer not be A alone becuase

3x - 2xy = 0
then
x(3-2Y) = 0

then

X=0 and y = 3/2

so sufficent

how can it be C
The solution for the equation: x(3-2Y) = 0, cannot be just x=0 AND y=3/2.
It is actually, either x=0 or y=3/2 or both (x=0 and y=3/2)
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by MAAJ » Tue Apr 26, 2011 3:21 am
This is tricky :(

I blindly choose C because both 1) and 2) stated x = 0

So the lesson here is that we should consider all values given by the statements and combine them?

1) x = 0 OR y = 3/2
2) x = 0 OR x = 1

3)Combined:
x = 0 AND y = 3/2
x = 1 AND y = 3/2

What's the best method to tackle this type of questions?
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by Abhishek009 » Tue Apr 26, 2011 3:39 am
MAAJ wrote:Is x = 0?

1) 3x = 2xy
2) (x^2) = x

----------

IMO [spoiler](C)[/spoiler]

1) 3x = 2xy
3x - 2xy = 0
x (3-2y) = 0
x = 0 OR 3-2y = 0

2)(x^2) = x
x = 0 OR x = 1

3)Combining 1 and 2:
x must be 0

3x = 2xy

Can be written as

3 = 2y { Cancelling X from both sides }

Now y = 3/2

(x^2) = x

Can be written as

x = 1 { Cancelling X from both sides }

So to my knowledge none of them is sufficient to reach the conclusion x = 0.
Abhishek
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by MAAJ » Tue Apr 26, 2011 4:47 am
Abhishek009, the second statement is a quadratic equation, so you should not solve it dividing by a variable because you could miss an answer.

x² = x
x² - x + 0 = 0
(x-1)(x-0) = 0
x = 1 OR x = 0

x = 0 is a possible solution to the equation!
Abhishek009 wrote:
3x = 2xy

Can be written as

3 = 2y { Cancelling X from both sides }

Now y = 3/2

(x^2) = x

Can be written as

x = 1 { Cancelling X from both sides }

So to my knowledge none of them is sufficient to reach the conclusion x = 0.
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."
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