B1. Alice has five real numbers $a<b<c<d<e$. She adds each pair of these numbers and records the ten results. The three smallest results turn out to be 32, 36, and 37, while the two largest results are 48 and 51. Determine $e$.
B1. Of the ten outcomes $32,36,37, \ldots, 48,51$, $d+e$ is the largest and $c+e$ is the second largest. Therefore, it must be that $d+e=51$ and $c+e=48$. Furthermore, $a+b$ is the smallest outcome and $a+c$ is the second smallest, so $a+b=32$ and $a+c=36$.

The second smallest outcome can be either $a+d$ or $b+c$. However, we know that

$$
a+d=(a+c)+(d+e)-(c+e)=36+51-48=39.
$$

Apparently, $a+d$ is not the second smallest outcome. Therefore, it must be that $b+c=37$.

Combining the found outcomes gives

$$
2 e=2(c+e)-(a+c)-(b+c)+(a+b)=2 \cdot 48-36-37+32=55
$$

The answer is thus $e=\frac{55}{2}$.