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10 Jun 2018, 04:27
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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:
75%
(hard)
Question Stats:
60% (02:39) correct
40%
(02:15)
wrong
based on 449
sessions
History
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Time
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Not Attempted Yet
If set S consists of the numbers n, -2, and 4, is the mean of set S greater than the median of set S ?
(1) n > 2
(2) n < 3
(1) n > 2
(2) n < 3
10 Jun 2018, 05:49
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Testing numbers smartly:
For (1), if n = 2 (it can't be technically, but we are figuring out the boundary), then mean = 4/3 and median = 2, so the median > mean. The reason why I tested 2 is because we can imagine our number to actually be 2 and an infinity number of 0s after the decimal point, then 1 at the very end. The result of plugging in such a number (which we can't) would be close enough to plugging a 2 in for this type of question specifically, so we can test 2. So anyways, let's try a much larger n. Much, much larger. Say 100. In that case, the mean is (-2 + 4 + 100)/3 = 102/3, or 34. In this case, our median is 4 (the moment n is greater than or equal to 4, the median will always be 4). Thus the mean here is greater than the median. Hence not sufficient.
(2) would be tested in a similar way. What if we use 3? Then we'd get mean = 5/3, and median = 3. So median > mean. What if n = -2? Mean = (-2 + -2 +4)/3 = 0. Median is -2, so mean > median. So not sufficient.
Together, we know that values between 2 and 3 would result in a mean < median. So, both (1) and (2) together are sufficient. So C is the answer.
I think the proper math calculations for it involves assuming n < -2, -2 < n < 4, and n > 4 or something, and so combining (1) and (2) gives you 2 < n < 3 which is between -2 and 4 so it can be figured out, but I a.) I don't remember all the proper math stuff because it's too long ago, and b.) am doing all this off the top of my head.
For (1), if n = 2 (it can't be technically, but we are figuring out the boundary), then mean = 4/3 and median = 2, so the median > mean. The reason why I tested 2 is because we can imagine our number to actually be 2 and an infinity number of 0s after the decimal point, then 1 at the very end. The result of plugging in such a number (which we can't) would be close enough to plugging a 2 in for this type of question specifically, so we can test 2. So anyways, let's try a much larger n. Much, much larger. Say 100. In that case, the mean is (-2 + 4 + 100)/3 = 102/3, or 34. In this case, our median is 4 (the moment n is greater than or equal to 4, the median will always be 4). Thus the mean here is greater than the median. Hence not sufficient.
(2) would be tested in a similar way. What if we use 3? Then we'd get mean = 5/3, and median = 3. So median > mean. What if n = -2? Mean = (-2 + -2 +4)/3 = 0. Median is -2, so mean > median. So not sufficient.
Together, we know that values between 2 and 3 would result in a mean < median. So, both (1) and (2) together are sufficient. So C is the answer.
I think the proper math calculations for it involves assuming n < -2, -2 < n < 4, and n > 4 or something, and so combining (1) and (2) gives you 2 < n < 3 which is between -2 and 4 so it can be figured out, but I a.) I don't remember all the proper math stuff because it's too long ago, and b.) am doing all this off the top of my head.
11 Jun 2018, 18:21
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1) not sufficient
If n>2 and <4
then it means the set is -2,n,4 and median is n.
so, median - 2<n<4.
mean is (2+n)/3
so, largest value of mean is slightly less than 2 since largest value of n is slightly less than 4.
so, mean < median
If n>4 say 20
then median = 4
and mean = -2+4+20/3=7.something
so, mean > median
NS
2)n<3
n can be anywhere. no sufficient
1+2
n is between 2 and 3.
so the set is 2,n,4 - 1st check of 1st condn
so, mean < median
answer C
If n>2 and <4
then it means the set is -2,n,4 and median is n.
so, median - 2<n<4.
mean is (2+n)/3
so, largest value of mean is slightly less than 2 since largest value of n is slightly less than 4.
so, mean < median
If n>4 say 20
then median = 4
and mean = -2+4+20/3=7.something
so, mean > median
NS
2)n<3
n can be anywhere. no sufficient
1+2
n is between 2 and 3.
so the set is 2,n,4 - 1st check of 1st condn
so, mean < median
answer C
