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rschunti
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Post subject: At a certain store, each notepad costs x dollars and each Posted: Sun Mar 02, 2008 10:48 pm |
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At a certain store, each notepad costs x dollars and each marker costs y dollars. If $10 is enough to buy 5 notepads and 3 markers, is $10 enough to buy 4 notepads and 4 markers instead?
Each notepad costs less than $1
$10 is enough to buy 11 notepads
How to solve above problem quickly?
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RonPurewal
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Post subject: Posted: Mon Mar 03, 2008 6:11 am |
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Posts: 7146
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this is a weird problem.
try to attack the problem from a conceptual standpoint first. there are 2 issues:
- the absolute prices of the items (if they're dirt cheap, then $10 will be good enough regardless)
- whether a marker is more expensive than a notepad (notice that you're 'trading' one notepad for one marker, and wondering whether that transaction will cause you to go over the $10)
notice that choice (2), which means that each notepad costs $10/11 = $0.91 or less, IMPLIES choice (1). that makes the following choices impossible, without even looking at the problem:
* A is impossible (because if (1) is sufficient then (2) must also be sufficient)
* C is impossible (because (1) and (2) together is the same thing as just (2))
let's just consider EXTREME cases:
(a) make the markers really expensive
- let's make notepads cost a penny each (which definitely satisfies both criteria), and make markers $3 each. then $10 is not enough for four of each.
(b) make everything really cheap
- if everything costs a penny, then $10 will buy you whatever you want.
answer = e
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luc2r4
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Post subject: Re: At a certain store, each notepad costs x dollars and each Posted: Tue Mar 09, 2010 7:43 pm |
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Posts: 14
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May you explain a little bit more on why the ST2 is not sufficient??
Thanks
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akhp77
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Post subject: Re: At a certain store, each notepad costs x dollars and each Posted: Sat Mar 27, 2010 1:26 pm |
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Given 5x + 3y < 10 Is 4x + 4y < 10 ??
Statement 1: x < 1 assume x = 0.9 4.5 + 3y < 10 y < 1.83 x + y < 0.9 + 1.83 = 2.73 4(x + y) < 10.92 4(x + y) may take 9, 10, or 10.9. So, $10 may or may not be sufficient. Insufficient
Statement 2: x < 10/11 x < 0.91 assume x = 0.9 as in statement 1 Insufficient
Statement 1 and 2 x < 1 and x < 0.91 implies x < 0.91 as in statement 2 Insufficient
Ans E
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StaceyKoprince
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Post subject: Re: At a certain store, each notepad costs x dollars and each Posted: Tue Apr 27, 2010 12:48 pm |
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Posts: 6077 Location: San Francisco
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luc2r4, can you explain what you don't understand about Ron's explanation? That will help us to make sure we address whatever the problem is.
Note: you might be making a bad assumption. When the statement says "$10 is enough to buy 11 notepads" that only tells us that we can buy at least 11 notepads for our $10. It doesn't also mean that we can buy at most 11 notepads for $10. I might also be able to buy 20 notepads for $10.
_________________ Stacey Koprince Instructor Director of Online Community ManhattanGMAT
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alok2
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Post subject: Re: At a certain store, each notepad costs x dollars and each Posted: Tue Nov 02, 2010 10:29 pm |
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Can this be solved by combining the "if" inequality with the "?" inequality?
5x + 3y <= 10 (if) 4x + 4y <= 10 (?) subtract second from first to yield new "?" inequality: x <= 0 ?
(stmt 1) x < 1 , not sufficient because we want x <= 0
(stmt 2) 11x <= 10 x <= 10/11, not sufficient because we want x <= 0
(stmt 1 + 2) x < 1 and x < 10/11 means x < 10/11, not sufficient because we want x <=0
Therefore, answer is E.
Please provide feedback on the logic of my method. In data sufficiency, is it possible to combine an "if" equation, inequality, or value with a "?" equation, inequality, or value?
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gregoryssmith
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Post subject: Re: At a certain store, each notepad costs x dollars and each Posted: Wed Nov 03, 2010 2:32 pm |
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Posts: 4
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alok2,
I don't think that you can quite make that jump to have x <= 0 because then your 'y' term goes away instead of becoming a '- y'.
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jnelson0612
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Post subject: Re: At a certain store, each notepad costs x dollars and each Posted: Sun Nov 07, 2010 5:22 pm |
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gregoryssmith wrote: alok2,
I don't think that you can quite make that jump to have x <= 0 because then your 'y' term goes away instead of becoming a '- y'. alok2, gregoryssmith is right--you have completely eliminated y in your result, but if you subtract one from the other you will still have -y. Ron's method is the most efficient way to attack the problem; please let us know if you have more specific questions. Thank you,
_________________ Jamie Nelson ManhattanGMAT Instructor
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