A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189, 320, 287, 264, x, y$. Find the greatest possible value of: $x + y$.
By assuming the elements to be $a, b, c, d$ and by making $6$ different equations, I solved for $x+y$. The result came to be $793$. I am not sure if this is correct or not. Can anyone help me out?
On
You are off to a good start. But you want a stronger assumption. Otherwise, you will have a lot of cases to consider. For example, if $a+b=189$ should $a+c = 264$ or $c+d = 264$?
Try to assume that $a \le b \le c \le d$ instead.
This will still run into some choices. But those choices won't matter if you focus on solving for maximum value of $x + y$ so don't actually need to calculate $a,b,c,d$ individually.
Additional hint:
Can you see why to get the maximum value of $x + y$ then $x + y = b + c + 2d$?
Taking the hint from @Dubs, you want d to be the largest, b and c the next larger than a. For this to happen, you get ($d+a = 320, b+c = 287, b+a = 264 and c+a = 189$).
From the last two weg get $b-c = 75$ and from this and the second one we get $2b = 362$, giving you $b = 181, a =83, c = 106$ and $d= 237$ and hence $x+y = 761$