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Re: stumped over DS problem, please help [#permalink]
25 Feb 2012, 08:42

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If Carmen had 12 more tapes, she would have twice as many as Rafael. Does Carmen have fewer tapes than Rafael?

Given \(c+12=2r\), question is \(c<r\)?

(1) Rafael has more than 5 tapes --> \(r>5\). If \(r=6>5\) then \(c=0\) and \(c<r\) BUT if \(r=14>5\) then \(c=16\) and \(c>r\). Two different answers. Not sufficient.

(2) Carmen has fewer than 12 tapes --> \(c<12\). Max number of tapes Carol can have is 10 (if \(c=11\) then \(r=11.5\neq{integer}\), which is not possible since \(c\) and \(r\) represent # of tapes and must be integers). So, \(c_{max}=10\) and \(r=11\) (from \(c+12=2r\)), hence \(c<r\). Sufficient.

Since even for \(c_{max}\) we got that \(c<r\), then for all other possible values of \(c\), \(c<r\) will also hold true.

Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
10 Jan 2014, 09:39

nina11 wrote:

If Carmen had 12 more tapes, she would have twice as many tapes as Rafael. Does Carmen have fewer tapes than Rafael?

(1) Rafael has more than 5 tapes. (2) Carmen has fewer than 12 tapes.

The time pressure is killing me..

We're given: C + 12 = 2R

1), we don't know anything about how many tapes carmen has, she could have more or less than Rafael, so insufficient.

2) Here's why so many of us get this wrong: We forget that the number of tapes Carmen has cannot be odd, it HAS to be a multiple of 2 otherwise we get fractions, and tapes have to be integer values.

So, she has, at most 10 tapes and then Rafael has 24/2 = 11 tapes.. So he always has more tapes than her, for all even values from 2-10... So, B is sufficient..

Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
28 May 2014, 22:46

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I think algebraic approach works the best here. If Carmen had 12 more tapes, she would have twice as many as Rafael. Does Carmen have fewer tapes than Rafael?

Given c+12=2r, question is c<r?

Lets simplify the question. First lets put c in terms of r in the inequality: c<r --> 2r-12<r --> r<12? Second, lets put r in terms of c in the inequality: c<r --> c<(c/2+6) --> c<12? Thus we will have sufficient info if we know either r<12 or c<12.

St1) r>5: r could be greater, equal, or less than 12. Not Suff St2) c<12: Suff

Answer: B _________________

Please consider giving 'kudos' if you like my post and want to thank

Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
03 Jul 2015, 18:54

Hello!

What I am unable to understand in the algebraic approach is if both C and R are less than 12 then how can we derive that C is less than 12 or not. Please advice.

Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
03 Jul 2015, 19:02

sandeepkummara wrote:

Hello!

What I am unable to understand in the algebraic approach is if both C and R are less than 12 then how can we derive that C is less than 12 or not. Please advice.

Hi' your Q is not clear.. you are assuming that both C and R are less than 12.. then what do you want to derive ? you may have to rephrase the Q..

gmatclubot

Re: If Carmen had 12 more tapes, she would have twice as many
[#permalink]
03 Jul 2015, 19:02

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