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pavanvarmads
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Post subject: Re: how many hardcover books did Juan buy?  Posted: Wed Jan 04, 2012 10:18 am |
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Posts: 1
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Plug the minimmum values for H and P
at 8P + 25H < 260,
Eg. ( P, H ) at 8P + 25H < 260,
( 11, 6 ) , 238 < 260, VALID
( 11, 7 ) 263 < 260, WRONG
( 12, 7 ) 271 < 260, WRONG
--- why isn't below case considered
( 12, 6 ) , 246 < 260, VALID .
Am I missing anything here?
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jnelson0612
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Post subject: Re: how many hardcover books did Juan buy? Posted: Wed Jan 11, 2012 10:32 pm |
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| ManhattanGMAT Staff |
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Posts: 1857
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pavanvarmads wrote: Plug the minimmum values for H and P
at 8P + 25H < 260,
Eg. ( P, H ) at 8P + 25H < 260,
( 11, 6 ) , 238 < 260, VALID
( 11, 7 ) 263 < 260, WRONG
( 12, 7 ) 271 < 260, WRONG
--- why isn't below case considered
( 12, 6 ) , 246 < 260, VALID .
Am I missing anything here? The case of 12 paperbacks and 6 hardbacks sounds fine to me, since that would be $96 + $150, or $246. The question, however, is how many hardbacks he buys, not how many paperbacks he buys. As you have shown, given all the constraints between the question and the two statements, he can only buy 6 hardbacks. Thus, the answer is C.
_________________
Jamie Nelson
ManhattanGMAT Instructor
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pawank_18
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Post subject: Re: how many hardcover books did Juan buy? Posted: Mon Feb 06, 2012 1:39 pm |
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luc2r4 wrote: From Stem : Price Paperback = $ 8
Price Hardcover = $ 25 , Number paperback >10 , P = { 11, 12, ....}
P and H are integers
St1>
25* Number Hardcover >=150
Number Hardcover > =6 Then , H = { 6, 7, ....}
Since question requires a unique value, INSUFFICIENT
St2> 8P + 25H < 260
Since Greatest Common Factor of { 8, 25, 260} is different from 1,
More than one solution , INSUFFICIENT
From St1 and St2 >
While P and H can only be integers
Plug the minimmum values for H and P
at 8P + 25H < 260,
Eg. ( P, H ) at 8P + 25H < 260,
( 11, 6 ) , 238 < 260, VALID
( 11, 7 ) 263 < 260, WRONG
( 12, 7 ) 271 < 260, WRONG
From this point, all other combinations will NOT comply the inequality.
Then , there is a unique value for H = 6
answer = C Eg. ( P, H ) at 8P + 25H < 260, ( 11, 6 ) , 238 < 260, VALID ( 11, 7 ) 263 < 260, WRONG ( 12, 7 ) 271 < 260, WRONG Can you tell me why did you not consider (12,6) as it complies with all the condition.. this would make E as the correct Sorry the Answer in C .. Forgot to see the last post .. Thank you ..
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RonPurewal
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Post subject: Re: how many hardcover books did Juan buy? Posted: Thu Feb 09, 2012 8:05 pm |
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| ManhattanGMAT Staff |
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Posts: 7146
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yep -- this is probably the single thing that causes people to miss DS problems more than anything else: not solving for the correct information.
when you solve data sufficiency problems, you MUST pay very careful attention to exactly what you are trying to find in the problem.
if a DS problem only asks for one specific piece of information, and you try to find ALL of the information in the problem, you will almost always get the wrong answer.
what's especially evil about this is that it has nothing to do with your knowledge and mastery of actual mathematics! in other words, you can be an amazingly good at all of the math on the gmat, and still blow DS questions left and right because you simply aren't looking for the right information.
don't let it happen.
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aps_asks
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Post subject: Re: how many hardcover books did Juan buy? Posted: Sun Apr 01, 2012 8:21 am |
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Posts: 134
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Hi Ron ,
Can U throw some light on Jeffrey's approach?...
What is he doing wrong ?
It will be great if we can solve this problem by using Inequalities ...
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RonPurewal
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Post subject: Re: how many hardcover books did Juan buy? Posted: Wed Apr 04, 2012 3:08 pm |
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| ManhattanGMAT Staff |
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Posts: 7146
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aps_asks wrote: It will be great if we can solve this problem by using Inequalities ... you can't solve the problem with algebra. in fact, that's the whole point of problems like this one. if the possibilities are restricted to WHOLE NUMBERS, then YOU CANNOT TRUST ALGEBRA, and YOU MUST TEST CASES.the reason is that algebra has no tools for distinguishing between whole numbers and non-whole numbers -- so, by definition, any problem depending on that distinction can't be solved with it. this goes for both inequalities and equations, by the way. for instance, consider the equation 5x + 7y = 48. if that's just a random algebraic equation, then it has lots and lots of solutions. however, if the solutions have to be positive integers, then there's only one solution (x = y = 4) -- a fact that's impossible to figure out with algebra.
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