gmat prep

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gmat prep

by g.shankaran » Wed Jun 01, 2011 7:48 pm
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

IMOc
Source: — Data Sufficiency |

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by Anurag@Gurome » Wed Jun 01, 2011 8:03 pm
g.shankaran wrote: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

IMOc

Solution:
If the line y = 3x +2 has to contain the point (r,s), s = 3r + 2.
First consider (1) alone.
This means that either 3r+2-s = 0 or 4r+9-s = 0.
Since nothing definite can be said, (1) alone is not sufficient.
We next consider (2) alone.
This means that either 3r + 2 - s = 0 or 4r-6-s = 0.
Again, nothing definite can be said.
So, (2) alone is not sufficient.
We next combine both the statements together and check.
On combining, we have that s = 3r + 2.
So, both statements together are sufficient to answer the question.

The correct answer is (C).
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by Frankenstein » Wed Jun 01, 2011 8:07 pm
Hi,
For (r,s) to lie on y=3x+2, we need to show that s=3r+2
From(1): (3r+2-s)(4r+9-s) = 0 => 3r+2-s=0 or 4r+9-s =0
Insufficient
From(1): (3r+2-s)(4r-6-s) = 0 => 3r+2-s=0 or 4r-6-s =0
Insufficient
Using(1)&(2): Lets assume that 3r+2-s is not equal to 0.
So 4r+9-s =0 and 4r-6-s =0 =>4r-s = -9 and 4r-s = 6. 4r-s cannot be 2 values at the same time. So, our assumption that 3r+2-s is not equal to 0 is wrong. 3r+2-s=0
Sufficient

Hence C
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by cans » Wed Jun 01, 2011 8:25 pm
a)(3r+2-s)(4r+9-s)=0
either (3r+2-s)=0 or (4r+9-s) = 0
Thus point (r,s) either lies on y=3x+2 or it lies on y=4x+9
Insufficient.
b)(3r+2-s)(4r-6-s)=0
similar to part a), point (r,s) either lies on y=3x+2 or it lies on y=4x-6
Insufficient.
a&b) (4r+9-s) and (4r-6-s) are parallel to each other (same slope = 4) and thus they don't intersect
thus for a&b to be simultaneously zero, 3r+2-s has to be 0
or (r,s) lies on y=3x+2
Sufficient
IMO C