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

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,
GMAT SME

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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.

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.

[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.