Mike drives to work and back home using the same route.
On his way, there is traffic and he averages a speed of 40 miles per hour (mph). On his way back, when there is no traffic, he averages 60 mph. If the total driving time going to and from work is 1 hour and 10 minutes,
how far, in miles, is his workplace from his home?
The general rate equation is given as the following:
$$ d=rt $$
Where
\(\\ d = \text{ distance} \\ \)
\( r = \text{ rate or speed} \\ \)
\( t = \text{ time} \)
Since there is a different rate going to and from work, we need to take two cases:
To work
$$d=rt$$
$$ d = 40t_1 $$
$$ t_1=\frac{d}{40} $$
From work
$$d=rt$$
$$ d = 60t_2 $$
$$ t_2=\frac{d}{60} $$
We know that the time will add up to 1 hour and 10 minutes, which is the same as \( 1\frac{1}{6} \) hours.
$$ t_1+t_2=1\frac{1}{6} $$
$$ \frac{d}{40}+\frac{d}{60} = 1\frac{1}{6}$$
$$ \frac{3d}{120}+\frac{2d}{120}=\frac{7}{6}$$
$$ \frac{3d+2d}{120} = \frac{7}{6} $$
$$ \frac{5d}{120}=\frac{7\cdot20}{6\cdot20} $$
$$ 5d=140 $$
$$ d=\boxed{28}$$
We can test the answer choices. For example, here is how we would test the answer, \( \boxed{28} \).
To work
$$ d=rt $$
$$ 28=40t $$
$$ \frac{28}{40} = t $$
$$ \frac{7}{10} =t $$
From work
$$ d=rt $$
$$ 28=60t $$
$$ \frac{28}{60} = t $$
$$ \frac{7}{15} =t $$
$$ \text{ Total time}=\frac{7}{10}+\frac{7}{15} $$
$$ \frac{7}{6} = \frac{7}{10} + \frac{7}{15} $$
$$ \frac{7}{6} = \frac{21}{30} + \frac{14}{30} $$
$$ \frac{7}{6} = \frac{35}{30} $$
$$ \frac{7}{6} =\frac{7}{6} \checkmark $$