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m10 Q02

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m10 Q02 [#permalink] New post 22 Apr 2012, 23:57
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Question Stats:

66% (02:34) correct 33% (02:26) wrong based on 0 sessions
If 2X + 3Y = 12 , what is X ?

1) X = 6 - (\frac{3}{2})Y
2) Y = 4 - (\frac{3}{2})X

Why is statement 1 not sufficient?

Can I not find the value of Y and substitute in 2X + 3Y = 12. I did this, please tell me what mistake I am making

2X = 12-3Y
From Statement 1 X = 6 - (\frac{3}{2})Y
Which can be simplified to X = (\frac{9}{2})Y
Therefore, Y = (\frac{2}{9})X

Now if i can substitute this Y in 2X = 12-3Y
2X = 12-\frac{6}{9}X
2X = 12-\frac{2}{3}X
2X+\frac{2}{3}X = 12
\frac{8X}{3} = 12
8X = 36
X = \frac{36}{8}
[Reveal] Spoiler: OA

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Re: m10 Q02 [#permalink] New post 23 Apr 2012, 00:35
ENAFEX wrote:
If 2X + 3Y = 12 , what is X ?

1) X = 6 - (\frac{3}{2})Y
2) Y = 4 - (\frac{3}{2})X

Why is statement 1 not sufficient?

Can I not find the value of Y and substitute in 2X + 3Y = 12. I did this, please tell me what mistake I am making

2X = 12-3Y
From Statement 1 X = 6 - (\frac{3}{2})Y
Which can be simplified to X = (\frac{9}{2})Y
Therefore, Y = (\frac{2}{9})X

Now if i can substitute this Y in 2X = 12-3Y
2X = 12-\frac{6}{9}X
2X = 12-\frac{2}{3}X
2X+\frac{2}{3}X = 12
\frac{8X}{3} = 12
8X = 36
X = \frac{36}{8}


The red part is not correct: x = 6 - \frac{3}{2}y --> x=\frac{12-3y}{2}, not x=\frac{9}{2}y

If 2x + 3y = 12, what is x ?

(1) x = 6 - (\frac{3}{2})y --> multiply both parts by 2: 2x=12-3y --> rearrange: 2x + 3y = 12, so we have the same equation as in the stem. Not sufficient.

(2) y = 4 - (\frac{3}{2})x --> 2y=8-3x --> 3x+2y=8. We have two distinct linear equations, hence we can get the single numerical value of x. Sufficient.

Answer: B.
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Manager
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Status: And the Prep starts again...
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Re: m10 Q02 [#permalink] New post 23 Apr 2012, 03:22
That was silly!!

Sorry to waste your time. :D GGRRR
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Re: m10 Q02   [#permalink] 23 Apr 2012, 03:22
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