A list of six positive integers $p, q, r, s, t, u$ satisfies $p<q<r<s<t<u$. There are exactly 15 pairs of numbers that can be formed by choosing two different numbers from this list. The sums of these 15 pairs of numbers are:

$$
25,30,38,41,49,52,54,63,68,76,79,90,95,103,117
$$

Which sum equals $r+s$ ?
(A) 52
(B) 54
(C) 63
(D) 68
(E) 76
Consider all possible sums of pairs of numbers that include $p: p+q, p+r, p+s, p+t$ and $p+u$.

We see that $p$ is included in 5 different sums (once with each of the other 5 numbers in the list).

Similarly, each of the numbers will be included in 5 sums.

If the sums of all 15 pairs are added together, each of $p, q, r, s, t$, and $u$ will have been included 5 times.

Therefore, $5 p+5 q+5 r+5 s+5 t+5 u=25+30+38+41+\ldots+103+117=980$ or $5(p+q+r+s+t+u)=980$.

Dividing this equation by 5 , we obtain $p+q+r+s+t+u=\frac{980}{5}=196$.

Since $p$ and $q$ are the smallest two integers, their sum must be the smallest of all of the pairs; thus, $p+q=25$.

Similarly, $t$ and $u$ are the largest two integers and their sum must be the largest of all of the pairs; thus, $t+u=117$.

Now, $(p+q)+r+s+(t+u)=196$ becomes $25+r+s+117=196$ or $r+s=196-142=54$.