Peter and her daughter Liza leave home at the same time in separate cars. Peter drives to his office, a distance of 28 km, and Liza drives to school, a distance of 36 km. They arrive at their destinations at the same time. What are their average rates if Peter’s average rate is 10km/hr less than his daughter’s?

Solution:
We need to determine the average rates of Liza and Peter; therefore, we make the following representations:

    $\mathbf{r:}$  the number of kilometers per hour in Liza’s average rate
    $\mathbf{r - 10:}$  the number of kilometers per hour in Peter’s average rate

We shall obtain an equation from the fact that the driving times of Liza and Peter are the same. From the distance formula:

\[d = r \cdot t \]

where:
    d = distance traveled
    r = rate of travel
    t = time spent to travel the distance in a given rate

Rearranging the variables, or solving the formula for t, we get,

\[ t = \frac{d}{r} \]

Draw a Table

The last column in the table indicates that the number of hours under TIME column can be represented by either $\frac{36}{r}$ or $\frac{28}{r – 10}$. Therefore, we have the following equation:

                    \[ \frac{36}{r} = \frac{28}{(r – 10)} \]

We solve the equation by first multiplying on both sides by the LCD:

\[ [r(r – 10)](\frac{36}{r}) = \frac{28}{(r – 10)}[r(r – 10)] \]
\[ (r – 10)(36) = 28r \]
\[ 36r – 360 = 28r \]
\[ 36r – 28r = 360 \]
\[ 8r = 360 \]
\[ r = \frac{360}{8} \]
\[ r = 45 \]

\[ AND \]

\[ r – 10 = 45 – 10 \]
\[ = 35 \]


Therefore, Liza’s average rate is 45 km/hr and Peter’s average rate is 35km/hr.

Check
The time for Liza to travel 36 km is 48 minutes.
$t = \frac{d}{r}$
$= \frac{36}{45}$
$=\frac{4}{5}$
$= \frac{4}{5} \cdot (60 minutes)$
$ t = 48 minutes $

The time for Peter to travel 28 km is 48 minutes.
$t = \frac{d}{r}$
$= \frac{28}{35}$
$=\frac{4}{5}$
$= \frac{4}{5} \cdot (60 minutes)$
$ t = 48 minutes$