Really Easy Median Question

[stevefeiner5]

I for some reason can not grasp this question to save my life. Any help would be most appreciated.

Given the ascending set of positive integers {a, b, c, d, e, f}, is the median greater than the mean?

(1) a + e = (3/4)(c + d)

(2) b + f = (4/3)(c + d)

[tim]

You need to rephrase the question first:

The median is (c+d)/2

The mean is (a+b+c+d+e+f)/6

Is (c+d)/2 > (a+b+c+d+e+f)/6 ?

Multiply to clear the fractions:

Is 3c+3d > a+b+c+d+e+f, or is 2c+2d > a+b+e+f ?

(1) a + e = (3/4)(c + d)

remember that b>a and f>e, so this also gives you
b+f > 3/4(c+d)

This tells you a+b+e+f > 3/2(c+d), which doesn't help

(2) b + f = (4/3)(c + d)

remember that a<b and e<f, so this also gives you
a+e < 4/3(c+d)

This tells you a+b+e+f < 8/3(c+d), which also doesn't help

Combine them though and you get
a+b+e+f = 25/12(c+d) > 2c+2d so we have a definite NO, which means the combined information is sufficient..