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

Its given, C + 12 = 2 R and we need to find Is C < R ?

(1) Rafael has more than 5 tapes.

Here, R >5.
Minimum possible value of R is 6 as the no of tapes should be an integer.
Case 1:
R =6
C + 12 = 2 *6
C =0
Is C < R ? => is 0 < 6 ? YES

Case 2:
R =20 as R could be any integer greater than 5
C + 12 = 2 *20
C = 40 -12 = 28
Is C < R ? => is 28 < 20 ? NO

Since we don't have a definite YES or NO for the question stem, Statement 1 alone is insufficient.


(2) Carmen has fewer than 12 tapes.
It's given C < 12.
Maximum possible value of C is 11.
Case 1: C = 11
C + 12 = 2 R
11 + 12 = 2R
R = 23/2 = 11.5 which is not possible as no of tapes should be an integer.

Case 2 :C = 10
C + 12 = 2 R
10 + 12 = 2R
R = 22/2 = 11

is C < R ? YES

Case 3:
if C = 0
C + 12 = 2 R
0 + 12 = 2R
R = 6
is C < R ? YES

Also we can conclude that C should an even integer less than 12 in-order to get a integer value for R .
6 <= R <=11
As you can see, for every possible values of C, R will be greater.
is C < R ? YES
Statement 2 alone is sufficient .

Option B is the answer.

Thanks,
Clifin J Francis,
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Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
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Re: If Carmen had 12 more tapes, she would have twice as many [#permalink]
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.

My option was D. (It is shown as wrong in answer btw).
From option (1), rafael can have tapes as 6, 7, 8 and so on.
if rafael have 6 tapes, then carmen would have 0 tapes which is also a probability (Why we are assuming that everyone should have atleast a tape). And this fits in the equation too as if rafael has 6 tapes then carmen would have 0 tapes, and adding 12 to 0 would give 12 which is exactly twice of tapes that rafael has. Other no of tapes like 7, 8 and so on with rafael will give tapes no as negative with carmen which is not a possibility here. Thus in my opinion option (1) is sufficient too.

From option (2), i.e carmen has less than 12 tapes, it would be 10, 8, 6, 4, and 2 as odd no of tapes would give rafael tapes which are in fractions which is not a possibility.

From the no of tapes with carmen, i.e 10,8,6,4,and 2, with the equation c+12=2r (c=no of tapes with carmen, r=no of tapes with rafael), no of tapes with carmen will come out to be fewer than rafael. Hence option (2) is also sufficient.
Hence option D.

Any take on this.
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If Carmen had 12 more tapes, she would have twice as many [#permalink]
[quote="Bunuel"]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.

Hi Bunuel, my take as below:

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.

My option was D. (It is shown as wrong in answer btw).
From option (1), rafael can have tapes as 6, 7, 8 and so on.
if rafael have 6 tapes, then carmen would have 0 tapes which is also a probability (Why we are assuming that everyone should have atleast a tape). And this fits in the equation too as if rafael has 6 tapes then carmen would have 0 tapes, and adding 12 to 0 would give 12 which is exactly twice of tapes that rafael has. Other no of tapes like 7, 8 and so on with rafael will give tapes no as negative with carmen which is not a possibility here. Thus in my opinion option (1) is sufficient too.

From option (2), i.e carmen has less than 12 tapes, it would be 10, 8, 6, 4, and 2 as odd no of tapes would give rafael tapes which are in fractions which is not a possibility.

From the no of tapes with carmen, i.e 10,8,6,4,and 2, with the equation c+12=2r (c=no of tapes with carmen, r=no of tapes with rafael), no of tapes with carmen will come out to be fewer than rafael. Hence option (2) is also sufficient.
Hence option D.

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