A souvenir stand began the day Wednesday with a ratio of 13 hats for

A souvenir stand began the day Wednesday with a ratio of 13 hats for every 7 shirts. If the stand received no new hats or shirts during the day Wednesday, and the only items that left the stand were those that were sold, did the stand end the day Wednesday with more shirts than hats?

Given that Hats:Shirts = 13x:7x.

(1) During the day Wednesday, the stand sold a total of 10 hats. If there were 13 hats and 7 shirts and no shirt was sold, then by the end of the day there would be 3 hats and 7 shirts left (S > H) but if for the same case all shirts were sold, then by the end of the day there would be 3 hats and 0 shirts left (S < H). Not sufficient.

(2) The stand began the day Wednesday with an even number of hats. This implies that x is even, so Hats:Shirts = 13x:7x = 26:14 = 52:28 = 78:42... We don't know how many of the items were sold. Not sufficient.

(1)+(2) From (2) we know that the least difference between the number of hats and shirts could be is 12 (if there were 26 hats and 14 shirts). (1) says that only 10 hats were sold, so even if 0 shirts were sold, the number of hats must be at least 2 more than the number of shirts. Thus the answer to the question is NO Sufficient.

Answer: C.

Hope it's clear.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

A souvenir stand began the day Wednesday with a ratio of 13 hats for every 7 shirts. If the stand received no new hats or shirts during the day Wednesday, and the only items that left the stand were those that were sold, did the stand end the day Wednesday with more shirts than hats?

(1) During the day Wednesday, the stand sold a total of 10 hats.

(2) The stand began the day Wednesday with an even number of hats.

This is a '2by2' question, most common type of question in GMAT math.

There are 3 variables (k,b,c) but only 2 equations are given from the 2 conditions, so there is high chance (E) will be our answer.
Looking at the conditions together,
13k:7k=13:7=26:14=39:21=52:28.. From this the situations in which the numbers become even are 26:14=52:28... In all of these cases, even if 10 hats are sold, the answer becomes 'no' so this is sufficient, and the answer becomes (C).

For cases where we need 3 more equation, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

I say it is C, regardless of how many shirts were sold.

1. not sufficient: hats=13, shirts=7. after 3 hats, shirts=7. or hats 26, shirts 14. 10 sold, so hats 16. H>S. not sufficient.

2. not sufficient.

1+2
hats minimum 26.
so if we start with at least 26, then shirts must be 14 minimum. 26-10=16. >than shirts.
if hats =52, then shirts =28. again, if -10 hats, we have 42 hats and 28 shirts.

H>S.

so 1+2 sufficient - C.

VERITAS PREP OFFICIAL SOLUTION:

It is fairly easy to see in this problem that each statement alone is not sufficient. With statement (1) the stand could have started with 13 hats and 7 shirts, sold 10 hats, and there would be more shirts than hats at the end of the day giving you an answer of yes. But the stand could have also sold 5 shirts in the same scenario giving you a no answer. For statement (2) you have no idea how many of each type were sold so that is clearly not sufficient. When you take the two statements together, it might still seem that you do not have enough information about the total number of items at the beginning or the end of the day, but keep in mind several facts:

-only 10 total hats were sold (per statement 1)

-no new shirts were added, so the total of shirts could only go down from the beginning value (per the given information)

-the smallest total relationship (per statement 2) is 26 hats and 14 shirts

Put all of that together and you should see that there is no way for the stand to end the day with more shirts than hats, as even with the smallest possible difference between hats and shirts, 10 sold hats don't even out the totals (that would be 16 hats and 14 shirts). Therefore, the statements together are sufficient to guarantee the answer "no" and the correct answer is (C).
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This is a nice problem... once you realize the minimum will be 26 hats you are done with this problem... Its important to be waiting this kind of contrarian problems on the GMAT...

1st statement- 13x-10>7x?
6x>10? ..... is x>1?
2nd statement- x=2 min.
combined suff.

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