Nate is driving to see his grandmother. If he drives at a constant speed of $40 \mathrm{~km} / \mathrm{h}$, he will arrive 1 hour late. If he drives at a constant speed of $60 \mathrm{~km} / \mathrm{h}$, he will arrive 1 hour early. At what constant speed should he drive to arrive just in time?
(A) $56 \mathrm{~km} / \mathrm{h}$
(B) $80 \mathrm{~km} / \mathrm{h}$
(C) $54 \mathrm{~km} / \mathrm{h}$
(D) $48 \mathrm{~km} / \mathrm{h}$
(E) $58 \mathrm{~km} / \mathrm{h}$
Solution 1

Suppose that when Nate arrives on time, his drive takes $t$ hours.

When Nate arrives 1 hour early, he arrives in $t-1$ hours.

When Nate arrives 1 hour late, he arrives in $t+1$ hours.

Since the distance that he drives is the same in either case and distance equals speed multiplied by time, then $(60 \mathrm{~km} / \mathrm{h}) \times((t-1) \mathrm{h})=(40 \mathrm{~km} / \mathrm{h}) \times((t+1) \mathrm{h})$.

Expanding, we obtain $60 t-60=40 t+40$ and so $20 t=100$ or $t=5$.

The total distance that Nate drives is thus $(60 \mathrm{~km} / \mathrm{h}) \times(4 \mathrm{~h})=240 \mathrm{~km}$.

When Nate drives this distance in 5 hours at a constant speed, he should drive at $\frac{240 \mathrm{~km}}{5 \mathrm{~h}}$ which equals $48 \mathrm{~km} / \mathrm{h}$.

## Solution 2

Suppose that the distance that Nate drives is $d \mathrm{~km}$.

Since driving at $40 \mathrm{~km} / \mathrm{h}$ causes Nate to arrive 1 hour late and driving at $60 \mathrm{~km} / \mathrm{h}$ causes Nate to arrive 1 hour early, then the difference between the lengths of time at these two speeds is 2 hours.

Since time equals distance divided by speed, then

$$
\frac{d \mathrm{~km}}{40 \mathrm{~km} / \mathrm{h}}-\frac{d \mathrm{~km}}{60 \mathrm{~km} / \mathrm{h}}=2 \mathrm{~h}
$$

Multiplying both sides of the equation by $120 \mathrm{~km} / \mathrm{h}$, we obtain

$$
3 d \mathrm{~km}-2 d \mathrm{~km}=240 \mathrm{~km}
$$

which gives us $d=240$.

Thus, the distance that Nate drives is $240 \mathrm{~km}$.

At $40 \mathrm{~km} / \mathrm{h}$, the trip takes 6 hours and Nate arrives 1 hour late.

To arrive just in time, it should take 5 hours.

To drive $240 \mathrm{~km}$ in 5 hours, Nate should drive at a constant speed of $\frac{240 \mathrm{~km}}{5 \mathrm{~h}}=48 \mathrm{~km} / \mathrm{h}$.

ANSWER: (D)