B1. Arthur is driving to David's house intending to arrive at a certain time. If he drives at 60 $\mathrm{km} / \mathrm{h}$, he will arrive 5 minutes late. If he drives at $90 \mathrm{~km} / \mathrm{h}$, he will arrive 5 minutes early. If he drives at $n \mathrm{~km} / \mathrm{h}$, he will arrive exactly on time. What is the value of $n$ ?
The answer is 72 .

Solution 1: Let $d$ be the distance from Arthur to David's house in $\mathrm{km}$ and $t$ the time, in hours, for Arthur to drive to David's place driving at $n \mathrm{~km} / \mathrm{h}$. If he drives at $60 \mathrm{~km} / \mathrm{h}$, Arthur will drive for $t$ hours +5 minutes $=t+5 / 60$ hours. If he drives at $90 \mathrm{~km} / \mathrm{h}$, Arthur will drive for $t$ hours - 5 minutes $=t-5 / 60$ hours. Therefore, using the distance $=$ speed $\times$ time formula, we have
$$
d=n t=60(t+5 / 60)=90(t-5 / 60) \text {. }
$$

This simplifies to
$$
d=n t=60 t+5=90 t-\frac{15}{2},
$$

We first determine $t$. Using the right-most equation of (1), we have $30 t=5+\frac{15}{2}=\frac{25}{2}$. Therefore, $t=25 / 60$. Hence, $d=60 t+5=60(25 / 60)+5=30$. Consequently, $n=d / t=$ $30 /(25 / 60)=30 \times 60 / 25=72 \mathrm{~km} / \mathrm{h}$.

Solution 2: Let $d$ be the distance from Arthur to David's house. Note that the time it takes for Arthur to drive to David's place at $n \mathrm{~km} / \mathrm{h}$ is the average of the times it take for Arthur to drive to David's place at 60 and $90 \mathrm{~km} / \mathrm{h}$, respectively. Hence,
$$
\frac{d}{n}=\frac{\frac{d}{60}+\frac{d}{90}}{2} .
$$

Dividing both sides by $d$ and cross multiplying yields
$$
\frac{2}{n}=\frac{1}{60}+\frac{1}{90}=\frac{5}{180}
$$

Hence, $5 n=360$. Therefore, $n=72$.