1. ## Distance Problem

Two different routes between two cities differ by 21 mi. Two people made the trip between the cities in exactly the same time. One traveled the shorter route at 50 mi/hr. and the other traveled the longer route at 55mi/hr. Find the length of each route.

2. Hello reiward
Originally Posted by reiward
Two different routes between two cities differ by 21 mi. Two people made the trip between the cities in exactly the same time. One traveled the shorter route at 50 mi/hr. and the other traveled the longer route at 55mi/hr. Find the length of each route.
Use ratios. In one hour travelling at $50$ and $55$ mph, the distances travelled are (of course!) $50$ and $55$ miles, a ratio of $50:55 = 10:11$.

So the two routes between the cities are in the ratio $10:11$. In other words, the longer route is $\tfrac{1}{10}$ longer than the shorter route. This is $21$ miles. The shorter route is therefore $10\times21=210$ miles, and the longer is $\tfrac{11}{10}\times210 = 231$ miles.

3. Hello, reiward!

Another approach . . .

We use: . $\text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\rightarrow\quad T \:=\:\frac{D}{S}$

Two different routes between two cities differ by 21 miles.
Two people made the trip between the cities in exactly the same time.
One traveled the shorter route at 50 mi/hr. and the other traveled the longer route at 55mi/hr.
Find the length of each route.

Let $D$ = length of the longer route.
Then $D-21$ = length of the shorter route.

One person drove the shorter route at 50 mph.
. . This took: . $\frac{D-21}{50}$ hours.

The other drove the longer route at 55 mph.
. . This took: . $\frac{D}{55}$ hours.

The two times are equal: . $\frac{D-21}{50} \:=\:\frac{D}{55}$

Cross-multiply: . $55(D-21) \:=\:50D \quad\Rightarrow\quad 55D - 1155 \:=\:50D \quad\Rightarrow\quad 5D \:=\:1155$

Therefore: . $\begin{array}{cccc}D \;=\; 231\text{ miles} &&\text{longer route} \\ D-21 \;=\; 210\text{ miles} && \text{shorter route} \end{array}$