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If Carmen had 12 more tapes, she would have twice as many [#permalink]
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25 Feb 2012, 09:33
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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.
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25 Feb 2012, 09:42
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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. Answer: B. Hope it's clear.
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Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
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26 Feb 2012, 10:53
yes, definitely....
thanks Bunuel :D :D



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Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
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20 Aug 2013, 02:33



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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 210... So, B is sufficient..



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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 > 2r12<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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If Carmen had 12 more tapes, she would have twice as many [#permalink]
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17 Dec 2014, 19:38
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.
Is C<R?
Given C+12=2R
ISOLATE R C + 12 = 2R C=2R12 so... Is 2R12 < R Is R<12 ?
ISOLATE C C + 12 = 2R R = (C+12) / 2 so... Is C < (C+12)/2 ? Is C<12 ?
Statement 1 > Insufficient Statement 2 > Sufficient



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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.



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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, 20:02
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..
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If Carmen had 12 more tapes, she would have twice as many [#permalink]
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03 Nov 2015, 18:35
A few thoughts: Start by translating the question into the formula c = 2r  12
1) Define the value of r at which c = 0, i.e. the least possible value of r: 0 = 2r  12 ==> r = 6 2) Define the value or at which both c and r are equal: c = r = 2r  12 ==> r = 12
3) Define the values of r at which c > r: c = r + (1 to infinity) = 2r 12 ==> 12 + (1 to infinity) = r ==> c > r for values of r >= 13
4) Define the values of r at which 0 <= c < r: c = r  (1 to 6) =2r 12 ==> 12  (1 to 6) = r ==> 0 <= c < r for values of r = 6 to 11 inclusive
From the above: statement 1 is insufficient and statement 2 is sufficient.



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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.



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Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
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23 Feb 2017, 05:53
Prompt analysis Let the number of tapes with carmen and rafael be x and y respectively.
x+12 = 2y
Superset Value of x and y will be a whole number.
Translation St 1: y =x +5. Solving 2 equations we get x = 2, y =7. ANSWER St 2: x<12. Cannot determine the exact value of x. INSUFFICIENT
Option A




Re: If Carmen had 12 more tapes, she would have twice as many
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