earnest10 wrote:
xy coordinate line
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Source: Beat The GMAT — Data Sufficiency |
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VivianKerr
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Here's my thinking:
y = 3x + 2 contain (r,s)?
- yes/no question
- sufficiency determined if we know the value of BOTH r AND s.
(1) (3r + 2 - s)(4r + 9 - s) = 0
One equation with both r and s in it. We cannot solve for the variables individually.
(2) (4r - 6 - s)(3r + 2 - s) = 0
Same issue - one equation with two variables.
If we combine the statements, we have 2 equations with the same 2 variables. Should be suff.
y = 3x + 2 contain (r,s)?
- yes/no question
- sufficiency determined if we know the value of BOTH r AND s.
(1) (3r + 2 - s)(4r + 9 - s) = 0
One equation with both r and s in it. We cannot solve for the variables individually.
(2) (4r - 6 - s)(3r + 2 - s) = 0
Same issue - one equation with two variables.
If we combine the statements, we have 2 equations with the same 2 variables. Should be suff.
Vivian Kerr
GMAT Rockstar, Tutor
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Thank you for all the "thanks" and "follows"!
GMAT Rockstar, Tutor
https://www.GMATrockstar.com
https://www.yelp.com/biz/gmat-rockstar-los-angeles
Former Kaplan and Grockit instructor, freelance GMAT content creator, now offering affordable, effective, Skype-tutoring for the GMAT at $150/hr. Contact: [email protected]
Thank you for all the "thanks" and "follows"!
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Night reader
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Hello my friend, function is like equation
/I know you smile too
st(1) BECOMES ridiculously simple ( 3r+2-s)(4r+9-s) = 0 is Not Sufficient, because we have also (4r+9-s) can be 0
st(2) heh, (4r-6-s)(3r+2-s) =0; you know everything here --> (3r+2-s) =0 ? Not Sufficient
may be Combined ? st(1&2): both ( 3r+2-s)(4r+9-s) = 0 AND (4r-6-s)(3r+2-s) =0 can be zero if one of common factors is 0 (zero). Only possible with the ( 3r+2-s) = 0 why not Sufficient?
IOM C
Cheers
y is s and x is r, now rewrite everything --> s=3r+2 OR 3r+2-s=0In 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.
/I know you smile toost(1) BECOMES ridiculously simple ( 3r+2-s)(4r+9-s) = 0 is Not Sufficient, because we have also (4r+9-s) can be 0
st(2) heh, (4r-6-s)(3r+2-s) =0; you know everything here --> (3r+2-s) =0 ? Not Sufficient
may be Combined ? st(1&2): both ( 3r+2-s)(4r+9-s) = 0 AND (4r-6-s)(3r+2-s) =0 can be zero if one of common factors is 0 (zero). Only possible with the ( 3r+2-s) = 0 why not Sufficient?
IOM C
Cheers
earnest10 wrote:
