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.
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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

Hello nick1816,

Why option B is not sufficient here ?

Mean of Set S is \(\frac{N+2}{3}\)
As per option B, N<3 i.e. 2.9,2,-10 etc.

If we put these values in mean we will get that Mean is less than median.

What I am missing here ?
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First we need to know how to get the Mean = Median condition, it happens when the set is evenly spaced. Hence n = 1 is right between -2 and 4 so -2, 1, 4 would allow Mean = Median. We can also have n = -8 in order to get -8, -2, 4 which also satisfies Mean = Median.

Statement 1:
n is greater than 2 so we can confirm -2 is the smallest integer in the set. Then either n or 4 is the median, depending on which is smaller. A good reference here is the evenly spaced set -2, 4, 10. When 4 < n < 10 the mean is smaller than 4 (the median), and when n > 10 the mean is bigger than 4 (the median). Since we have both cases this is insufficient.

Statement 2:
Again let us find an evenly spaced set, -8, -2, 4 for example. When n < -8 we have mean < median. Also when -8 < n < -2 we have mean > median. Then statement 2 is still insufficient.

Combined:
Combined we have 2 < n < 3, so we must have the ordering -2, n, 4. The mean is (-2 + n + 4) / 3 and the median is n. We can find they are equal when n = 1, which makes sense as -2. 1. 4 is an evenly spaced set. Thus that is the equilibrium, and when n > 1 we will have mean > median or the other way around. Since we can confirm only one case this is sufficient.

Ans: C
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Case 1 If -2<n<3, median is n

\(\frac{n+2}{3} > n\)

n+2 > 3n

n< 1

If -2<n<1, Mean > Median (Yes)

If 1<n<3, Median > Mean (No)

Hence, statement 2 is insufficient. (no need to analyze the case when n<-2)