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14 Feb 2024, 01:16
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
63% (02:03) correct
37%
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wrong
based on 688
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Is the median of the a + b quiz scores greater than the mean of the a + b quiz scores?
(1) a + b = 79
(2) a = 42 and b = 37.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
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14 Feb 2024, 02:33
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DerekLin
Firstly, it's important to note that if statement (1) is sufficient, then statement (2) is automatically sufficient. Therefore, the answer cannot be A or C. It must be either B, D, or E.
However, if we step back and apply logical reasoning, it becomes immediately clear that the answer must be E. When we combine the statements, we find that we can indeed calculate the average, which would be (82a + 78b)/(a + b). However, finding the median is impossible from the information provided. The median in a set of an odd count, like 79 numbers, is the middle term when they are arranged in order, and this value cannot be deduced from the given information. To illustrate this point, let's simplify by considering a = 4 and b = 3, keeping all other numbers unchanged. Now consider the following sets:
- A = {73, 85, 85, 85} and B = {75, 75, 84}, then A + B = {73, 75, 75, 84, 85, 85, 85}
- A = {73, 85, 85, 85} and B = {74, 75, 85}, then A + B = {73, 74, 75, 85, 85, 85, 85}
Answer: E.
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10 Mar 2024, 22:47
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It is a good tricky question. We do see that info of Statement 1 is automatically obtained in Statement II too and without the break up of a and b, we cannot judge what the mean and median are of the combined set. So we can just ignore Statement 1.
With Statement 2, we can find the average of the combined set using weighted averages. It will be something close to 80, because the means are 82 and 78 with slightly more weightage to 82. We do not need to do this calculation.
We have to consider now whether median must be 80 or less or more than 80. With total 79 integers in a + b, the median will be the 40th integer.
Say
42 numbers of Set A = .., 79, 79, 79,... 85, (Median), 85, 85, 85, ... (adjustment in the first number is made to make mean 82)
37 numbers of Set B = 75, 75, 75, ... 75 (Median), 75, 75, 75... 75, Y (Y is a big big number - whatever is required to make the mean 78)
The 40th integer would be 79 or something else less than 80 here.
OR Say
42 numbers of Set A = 0, 0, 0, ... 81, 81, 81, 85, Median, 85, 85, 85, ... 85 (adjustment in the first few numbers is made to make mean 82)
37 numbers of Set B = 75, 75, 75, ..75 (Median), 81, 81, 81, 81, 81... (Adjustment is made in the last number to make mean 78)
In Set B, 19 numbers are less than 81, and in set A first few numbers are less than 81 but median can easily be 81.
Answer (E)
Check out weighted averages concept here: https://www.youtube.com/watch?v=_GOAU7moZ2Q
Blog post on another tricky question on mean, median: https://anaprep.com/sets-statistics-mean-median-range-question/
Quote:
It is a good tricky question. We do see that info of Statement 1 is automatically obtained in Statement II too and without the break up of a and b, we cannot judge what the mean and median are of the combined set. So we can just ignore Statement 1.
With Statement 2, we can find the average of the combined set using weighted averages. It will be something close to 80, because the means are 82 and 78 with slightly more weightage to 82. We do not need to do this calculation.
We have to consider now whether median must be 80 or less or more than 80. With total 79 integers in a + b, the median will be the 40th integer.
Say
42 numbers of Set A = .., 79, 79, 79,... 85, (Median), 85, 85, 85, ... (adjustment in the first number is made to make mean 82)
37 numbers of Set B = 75, 75, 75, ... 75 (Median), 75, 75, 75... 75, Y (Y is a big big number - whatever is required to make the mean 78)
The 40th integer would be 79 or something else less than 80 here.
OR Say
42 numbers of Set A = 0, 0, 0, ... 81, 81, 81, 85, Median, 85, 85, 85, ... 85 (adjustment in the first few numbers is made to make mean 82)
37 numbers of Set B = 75, 75, 75, ..75 (Median), 81, 81, 81, 81, 81... (Adjustment is made in the last number to make mean 78)
In Set B, 19 numbers are less than 81, and in set A first few numbers are less than 81 but median can easily be 81.
Answer (E)
Check out weighted averages concept here: https://www.youtube.com/watch?v=_GOAU7moZ2Q
Blog post on another tricky question on mean, median: https://anaprep.com/sets-statistics-mean-median-range-question/
