In xy plane y=3x+2 contain (r,s)
1) (3r+2-s)(4r+9-s)=0
2) (4r-6-s)(3r+2-s)=0
I think each statement alone is sufficent..But it is wrong!
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Inequality (Pearson gmat practice)
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thegmatbeater
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thegmatbeater
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The point (r,s) is on the line only if s = 3r + 2. If you look at either statement, you have a product of two factors which is equal to zero. That guarantees that one (not both) of the two factors is equal to zero. So, for example, from 2):thegmatbeater wrote:In xy plane y=3x+2 contain (r,s)
1) (3r+2-s)(4r+9-s)=0
2) (4r-6-s)(3r+2-s)=0
I think each statement alone is sufficent..But it is wrong!
Either 4r-6-s = 0 --> s = 4r - 6
or: 3r+2-s = 0 --> s = 3r + 2
Only in the second case is the point certain to be on the line.
The first statement leads to a similar conclusion. So neither statement is sufficient alone. If you consider the two statements together, the point is either on the line y = 3x +2, or it is simultaneously on the lines y = 4x - 6 and y = 4x +9, which is impossible because these are distinct parallel lines.
In xy plane y=3x+2 contain (r,s)
1) (3r+2-s)(4r+9-s)=0
2) (4r-6-s)(3r+2-s)=0
I think each statement alone is sufficent..But it is wrong!
I think tht the answer should be "C"
This is because when you consider statement 1 alone, there are 2 solutions: either s = 3r + 2 or s = 4r + 9. So, statement 1 is not sufficient.
when you consider statement 2 alone, there are 2 solutions again.
either s = 4r - 6 or s = 3r - 2. so, statement 2 is not sufficient.
But, when you consider both statements, there is only one common solution which s = 3r - 2.
So, we know for sure that (r,s) is on y = 3x + 2.
Let me know if C is the correct answer?
Thanks
Paddy
1) (3r+2-s)(4r+9-s)=0
2) (4r-6-s)(3r+2-s)=0
I think each statement alone is sufficent..But it is wrong!
I think tht the answer should be "C"
This is because when you consider statement 1 alone, there are 2 solutions: either s = 3r + 2 or s = 4r + 9. So, statement 1 is not sufficient.
when you consider statement 2 alone, there are 2 solutions again.
either s = 4r - 6 or s = 3r - 2. so, statement 2 is not sufficient.
But, when you consider both statements, there is only one common solution which s = 3r - 2.
So, we know for sure that (r,s) is on y = 3x + 2.
Let me know if C is the correct answer?
Thanks
Paddy
Paddy Srinivas
This might be silly, but how does having one common solution givenpsriniva wrote: But, when you consider both statements, there is only one common solution which s = 3r - 2.
So, we know for sure that (r,s) is on y = 3x + 2.
Let me know if C is the correct answer?
Thanks
Paddy
the 2 statements guarantee that (r,s) is on y=3x + 2?
thanks,
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BlueRain
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Given the combination of statement (1) and (2), if (r,s) is NOT on the y=3x+2 line, then that means the (r,s) pair must satisfy both s=4r+9 AND s=4r-6 since (3r+2-s) cannot be zero.
Since there are no (r,s) pair that can simultaneously satisfy both s=4r+9 and s=4r-6, (r,s) must be on the y=3x+2.
Since there are no (r,s) pair that can simultaneously satisfy both s=4r+9 and s=4r-6, (r,s) must be on the y=3x+2.
