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26 May 2017, 04:42
Kudos
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At least one pencil is distributed to each of the 19 students. Did at least two students receive the same number of pencils?
(1) The number of pencils distributed to each student is less than 19.
(2) The total number of pencils distributed to students is 187.
(1) The number of pencils distributed to each student is less than 19.
(2) The total number of pencils distributed to students is 187.
26 May 2017, 04:42
Kudos
Bookmarks
Official Solution:
In the original condition, there are 19 variables (since you need to know the number of pencils distributed to each of the 19 students). In order to match the number of variables and the number of equations, there must be 19 equations. Therefore, E is most likely to be the answer. By solving con 1) & con 2),
In the case of con 1), the question claims that the number of pencils distributed to 19 students is less than 19, so if each number of pencils that was distributed to 19 students is 1,2,3,4,...... , 17,18, and 19, then the number of pencils that the 19th student got is less than 19, so it cannot be 19. Then, another student before the 19th student always gets the same number of pencils, hence yes, and sufficient.
In the case of con 2), you get \(1+2+\ldots \ldots + 18 + 19 = \frac{19(19+1)}{2} = 190\). The total number of the distributed pencils is 187 because \(187 = 190 - 3\), and the 19th (the last) student cannot get 19 pencils, so he or she has to get \(16(=19-3)\) pencils. If so, the 19th student always receives the same number of pencils as other students who received a pencil before him or her, hence yes, and sufficient. Thus, con 1)=con 2). Therefore, the answer is D.
Answer: D
In the original condition, there are 19 variables (since you need to know the number of pencils distributed to each of the 19 students). In order to match the number of variables and the number of equations, there must be 19 equations. Therefore, E is most likely to be the answer. By solving con 1) & con 2),
In the case of con 1), the question claims that the number of pencils distributed to 19 students is less than 19, so if each number of pencils that was distributed to 19 students is 1,2,3,4,...... , 17,18, and 19, then the number of pencils that the 19th student got is less than 19, so it cannot be 19. Then, another student before the 19th student always gets the same number of pencils, hence yes, and sufficient.
In the case of con 2), you get \(1+2+\ldots \ldots + 18 + 19 = \frac{19(19+1)}{2} = 190\). The total number of the distributed pencils is 187 because \(187 = 190 - 3\), and the 19th (the last) student cannot get 19 pencils, so he or she has to get \(16(=19-3)\) pencils. If so, the 19th student always receives the same number of pencils as other students who received a pencil before him or her, hence yes, and sufficient. Thus, con 1)=con 2). Therefore, the answer is D.
Answer: D
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