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# DS Question from GMATPrep

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Manager
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10 May 2006, 15:25
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Hi,

I can't figure out this GMATPrep question that I got wrong. If someone could help, I'd greatly appreciate it. The answer was C (standard DS answer choices).

Thanks,
Marcus

In the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?

(1) (3r+2-s)(4r+9-s)=0
(2) (4r-6-s)(3r+2-s)=0
Director
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10 May 2006, 15:49
In the xy-plane, does the line with equation y=3x+2 contain the point (r,s)?

(1) (3r+2-s)(4r+9-s)=0
(2) (4r-6-s)(3r+2-s)=0

From (1):
3r+2-s = 0
=> s = 3r + 2 (its on the line y=3x+2)

or 4r+9-s = 0
=> s = 4r + 9 (its not on line y=3x+2)

Hence not sufficient.

From (2):
s = 4r - 6 (not on the line)

or s = 3r + 2 (it is on the line)

Hence not sufficient.

If both (1) and (2) are taken into account:
3r+2-s=0 is true and hence they say the answer is C. But I'm not sure how to prove that 3r+2-s=0 is true and not the other 2 equations ... if someone can show that then it would be great.
Manager
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10 May 2006, 21:59
from 1

3r+2-s=0 insuff

from 2

4r -6-s=0 insuff

combining 1 and 2 ,solving

r=8 s=26

(8,26) ---> is on line y=3x+2

therefore ans =C
Intern
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10 May 2006, 22:28
From 1

3r + 2 - s = 0 OR 4r + 9 -s = 0 (Not Suff)

From 2

3r + 2 - s = 0 OR 4r - 6 - s = 0 (Not Suff)

Now combing both together we could have either

4r + 9 - s = 0 AND 4r - 6 - s = 0
OR
3r + 2 - s = 0

(reason is obvious, but let me know if it has to be explained)

if 4r + 9 - s = 0 and 4r - 6 - s = 0 then

4r + 9 - s = 4r - 6 - s
=>9 = 6 an impossibility hence
3r+ 2 - s = 0

Therefore (C) is correct
Intern
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10 May 2006, 22:42
Quote:
Now combing both together we could have either

4r + 9 - s = 0 AND 4r - 6 - s = 0
OR
3r + 2 - s = 0

(reason is obvious, but let me know if it has to be explained)

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If A equals success, then the formula is: A = X + Y + Z, X is work. Y is play. Z is keep your mouth shut.
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Senior Manager
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11 May 2006, 00:50
take a look at this discussion!

http://www.gmatclub.com/phpbb/viewtopic.php?t=28246
Intern
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13 May 2006, 19:20
mendiratta_1812 wrote:
Quote:
Now combing both together we could have either

4r + 9 - s = 0 AND 4r - 6 - s = 0
OR
3r + 2 - s = 0

(reason is obvious, but let me know if it has to be explained)

Well if the product or 2 numbers is 0 the number1 = 0 OR number2 =0, right ?
Here we have
XY = 0 AND
XZ = 0 if we take Y = 4r + 9 -s and Z = 4r - 6 - s and 3r + 2 - s)

Since X is common in both equations then either X = 0 i.e both expressions become 0 OR X != 0 and Y = 0 as well as Z = 0 for both expressions to be zero. The third case is X, Y, Z are all 0 which is a subset of Y = 0 and Z = 0.
Director
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11 Jun 2006, 16:47
saha wrote:
From 1

3r + 2 - s = 0 OR 4r + 9 -s = 0 (Not Suff)

From 2

3r + 2 - s = 0 OR 4r - 6 - s = 0 (Not Suff)

Now combing both together we could have either

4r + 9 - s = 0 AND 4r - 6 - s = 0
OR
3r + 2 - s = 0

(reason is obvious, but let me know if it has to be explained)

if 4r + 9 - s = 0 and 4r - 6 - s = 0 then

4r + 9 - s = 4r - 6 - s
=>9 = 6 an impossibility hence
3r+ 2 - s = 0

Therefore (C) is correct

I believe saha showed it correctly, OA is C
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