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

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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.
[Reveal] Spoiler: OA

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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.

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
Bumping for review and further discussion.
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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 2-10... 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 --> 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

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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=2R-12

so...
Is 2R-12 < 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'
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

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Re: If Carmen had 12 more tapes, she would have twice as many   [#permalink] 23 Feb 2017, 05:53
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