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Updated on: 22 Oct 2014, 04:00
Originally posted by Thoughtosphere on 22 Oct 2014, 03:15.
Last edited by Bunuel on 22 Oct 2014, 04:00, edited 1 time in total.
Last edited by Bunuel on 22 Oct 2014, 04:00, edited 1 time in total.
RENAMED THE TOPIC.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
54% (02:08) correct
46%
(01:55)
wrong
based on 344
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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.
(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.
22 Oct 2014, 04:11
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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.
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.
General Discussion
10 Nov 2015, 10:43
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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.
GCDS Thoughtosphere souvenir stand (20151107).jpg [ 23.05 KiB | Viewed 5969 times ]
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.
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.
Attachment:
GCDS Thoughtosphere souvenir stand (20151107).jpg [ 23.05 KiB | Viewed 5969 times ]
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.
