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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]

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10 Jan 2014, 10:39

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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]

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28 May 2014, 23:46

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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
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Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]

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03 Jul 2015, 19:54

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.

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..
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If Carmen had 12 more tapes, she would have twice as many [#permalink]

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23 Sep 2016, 12:29

This is quite tricky

From the question we have that C+12 = 2R which can be rewritten as R = C/2 + 6. Before even start thinking about this you should already be aware that C has to be even and multiple of 2 since number of tapes has to be integer.

(i) R>5: Resolve the equation to [b]C = 2 ( R - 6) [/b]

Here start plugging numbers so you have

R = 6 gives C =0 R = 7 gives C = 2 etc etc R = 12 gives C =12 (btw this is not a option as we allready know that one of the two has more) R = 13 gives C =14

So as you can see R > C for 6<= R <12 AND R < C for R >=13 hence insufficient. It would be sufficient if and only if we could see that R>C for R>5.

(ii) C>12: Use the original equation [b]R = C/2 + 6 (for the more "mathy" people this can be seen as y = 0.5x+6) [/b]. Again start plugging numbers BUT the twist is that you care only for the even values defined by the space 0 < R <12

Start plugging numbers again

C = 10 gives R = 11 C = 8 gives R =10 C = 6 gives R = 9 C = 4 gives R = 8 C = 2 gives R = 6

Notice that for all values of C the variable R IS ALWAYS GREATER THAN C therefore sufficient

On a side note:

Bunuel wrote:

(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.

imho the take away msg from this question is to be able to understand the trend of equation for all values of x given by the question. Essentially boils down to the point to be able to undestand that in (i) the equation changes while in (ii) constantly y < x.

I am pretty sure people with solid math background just by looking the two equations could figure out the right answer rather than plug in number etc.