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04 Sep 2018, 02:16
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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:
45%
(medium)
Question Stats:
64% (01:36) correct
36%
(02:01)
wrong
based on 25
sessions
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Jackie is presented with two bags of chocolate bars, R and S, and told she can choose only one of the bags. Jackie knows both bags have the same number (n) of candy bars, and thus is interested in picking the bag with the largest amount of chocolate, as measured in grams. Which bag should Jackie choose?
(1) The weights of the chocolate bars in bag R are consecutive even numbers of grams, and the weights of the chocolate bars in bag S are consecutive odd numbers of grams.
(2) The median weight of chocolate bars in bag R is greater than the median weight of chocolate bars in bag S.
(1) The weights of the chocolate bars in bag R are consecutive even numbers of grams, and the weights of the chocolate bars in bag S are consecutive odd numbers of grams.
(2) The median weight of chocolate bars in bag R is greater than the median weight of chocolate bars in bag S.
04 Sep 2018, 02:16
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Official Solution:
We’ll go for ALTERNATIVE because we can use numbers to eliminate contradictory answers.
(1) The weights in R could be, say, {2, 4, 6}, or {100, 102, 104}, and the weights in S could be, say, {11, 13, 15}, so that the total weight of set S is greater than the first R set example, but lower than the second example. Insufficient. (A) and (D) eliminated.
(2) Median is the middle value, so if the weights in R are, say, {4, 5, 6}, then the median is 5. Say the weights in S are {1, 2, 100}, or {1, 2, 3} – in both cases, the median is 2, however in the first case the total weight of S is greater, and in the other case the total weight is lower. Insufficient. (B) is eliminated.
On combining the two statements, we can use the fact that for any set of consecutive numbers the median is also the average. So if the median and average are higher for one such set, and also both sets have the same number of elements (n), the set with the higher median weight would have a higher total weight. For example, if set R is {2, 4, 6} and set S is {1, 3, 5}. (E) is eliminated.
Answer: C
We’ll go for ALTERNATIVE because we can use numbers to eliminate contradictory answers.
(1) The weights in R could be, say, {2, 4, 6}, or {100, 102, 104}, and the weights in S could be, say, {11, 13, 15}, so that the total weight of set S is greater than the first R set example, but lower than the second example. Insufficient. (A) and (D) eliminated.
(2) Median is the middle value, so if the weights in R are, say, {4, 5, 6}, then the median is 5. Say the weights in S are {1, 2, 100}, or {1, 2, 3} – in both cases, the median is 2, however in the first case the total weight of S is greater, and in the other case the total weight is lower. Insufficient. (B) is eliminated.
On combining the two statements, we can use the fact that for any set of consecutive numbers the median is also the average. So if the median and average are higher for one such set, and also both sets have the same number of elements (n), the set with the higher median weight would have a higher total weight. For example, if set R is {2, 4, 6} and set S is {1, 3, 5}. (E) is eliminated.
Answer: C
04 Oct 2018, 05:40
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So, basically the chocolate and candy bar are the same thing in this question. 
