Lists S and T consist of the same number of positive integers. Is the median of the
integers in S greater than the average (arithmetic mean) of the integers in T?
(1) The integers in S are consecutive even integers, and the integers in T are
consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.
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.
Pls explain your approach.
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DS- Median and Avg
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Ian Stewart
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It's clear Statement 1 is insufficient on its own.
Statement 2 is insufficient, since we are comparing the median with the average. The median of set S = {1, 2, 1,000,000} is smaller than the mean of set T = {5, 6, 7}, for example, even though the sum of the elements in S is greater than the sum of the elements in T.
So, we need to know if the statements are sufficient together. Statement 1 illustrates one of the most important facts of statistics: if S consists of consecutive even integers, the set is 'evenly spaced'- we can write the list so it increases by 2 each time. In math language, we say S is an 'arithmetic progression'. T is also evenly spaced. In any evenly spaced set, the median and the mean are equal.
So, if we know Statement 1 is true, the median of S = the mean of S, and we can rephrase the question: Is the mean of S > the mean of T? And because they have the same number of elements, that's the same as asking: Is the sum of S > the sum of T? Which is what Statement 2 tells us. C.
Statement 2 is insufficient, since we are comparing the median with the average. The median of set S = {1, 2, 1,000,000} is smaller than the mean of set T = {5, 6, 7}, for example, even though the sum of the elements in S is greater than the sum of the elements in T.
So, we need to know if the statements are sufficient together. Statement 1 illustrates one of the most important facts of statistics: if S consists of consecutive even integers, the set is 'evenly spaced'- we can write the list so it increases by 2 each time. In math language, we say S is an 'arithmetic progression'. T is also evenly spaced. In any evenly spaced set, the median and the mean are equal.
So, if we know Statement 1 is true, the median of S = the mean of S, and we can rephrase the question: Is the mean of S > the mean of T? And because they have the same number of elements, that's the same as asking: Is the sum of S > the sum of T? Which is what Statement 2 tells us. C.
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ildude02
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Ian, for statement 1, I took some numbers to see if it's sufficient and I always got the same answer where, median of S is NOT > avg of odd numbers(T)Ian Stewart wrote:It's clear Statement 1 is insufficient on its own.
Statement 2 is insufficient, since we are comparing the median with the average. The median of set S = {1, 2, 1,000,000} is smaller than the mean of set T = {5, 6, 7}, for example, even though the sum of the elements in S is greater than the sum of the elements in T.
So, we need to know if the statements are sufficient together. Statement 1 illustrates one of the most important facts of statistics: if S consists of consecutive even integers, the set is 'evenly spaced'- we can write the list so it increases by 2 each time. In math language, we say S is an 'arithmetic progression'. T is also evenly spaced. In any evenly spaced set, the median and the mean are equal.
So, if we know Statement 1 is true, the median of S = the mean of S, and we can rephrase the question: Is the mean of S > the mean of T? And because they have the same number of elements, that's the same as asking: Is the sum of S > the sum of T? Which is what Statement 2 tells us. C.
eg, considering set S= 2, 4, 6 and T= 3, 5, 7
median of S = 4 and avg of T = 5
different set, s=2, 4, 6, 8 and T= 3, 5, 7, 9(since both of them have equal number of elements). Again we see that median is less then avg. So I thought 1 was sufficient. Where did I miss ?
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Choosing numbers (or in this case, sets) can be dangerous if you don't pick enough examples. From Statement 1 alone, all we know is that S consists of consecutive even integers, and T consists of consecutive odd integers. We might have:ildude02 wrote: Ian, for statement 1, I took some numbers to see if it's sufficient and I always got the same answer where, median of S is NOT > avg of odd numbers(T)
eg, considering set S= 2, 4, 6 and T= 3, 5, 7
median of S = 4 and avg of T = 5
different set, s=2, 4, 6, 8 and T= 3, 5, 7, 9(since both of them have equal number of elements). Again we see that median is less then avg. So I thought 1 was sufficient. Where did I miss ?
S = {1,000,000; 1,000,002; 1,000,004}
T = {1, 3, 5}
or
S = {2, 4, 6}
T = {1,000,001; 1,000,003; 1,000,005}
The median of S is greater than the mean of T in the first case, and smaller in the second case.
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ildude02
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Ian, thanks for your response.
I realized my mistake after I posted the message. I made the dum mistake of not considering different possibilities. This is one thing which maks DS tricky for me when compared to PS questions, since I feel strong about PS but not so much abt DS. I do tend to take more time then I should normally take on DS questions. I don't know what would be the best way to avoid this, but I'm trying to practice more on the DS questions. Any thoughts from you are appreciated.
I realized my mistake after I posted the message. I made the dum mistake of not considering different possibilities. This is one thing which maks DS tricky for me when compared to PS questions, since I feel strong about PS but not so much abt DS. I do tend to take more time then I should normally take on DS questions. I don't know what would be the best way to avoid this, but I'm trying to practice more on the DS questions. Any thoughts from you are appreciated.
