# David drives to school in rush hour traffic and averages 32 mph. He returns home in mid-afternoon when there is less traffic and averages 48 mph.

slaggingV 2021-09-04 Answered
David drives to school in rush hour traffic and averages 32 mph. He returns home in mid-afternoon when there is less traffic and averages 48 mph. What is the distance between his home and school if the total traveling time is 1 hr 15 min?
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## Expert Answer

crocolylec
Answered 2021-09-05 Author has 100 answers

The table above summarizes the information of the given problem

Using $D=rt$ or the formula for the relationship of distance, rate,and time, then $d=32{t}_{1}$
$\frac{d}{32}={t}_{1}$ and $d=48{t}_{2}$
$\frac{d}{48}={t}_{2}$ Since the total time is given as 1 hour and 15 minutes $\left(\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}1\frac{15}{60}=1\frac{1}{4}=\frac{5}{4}hours\right)$ , then ${t}_{1}+{t}_{2}=\frac{5}{4}$
$\frac{d}{32}+\frac{d}{48}=\frac{5}{4}$
$96\left(\frac{d}{32}+\frac{d}{48}\right)=\left(\frac{5}{4}\right)96$ (multiply by the LCD)
$3\left(d\right)+2\left(d\right)=5\left(24\right)$
$3d+2d=120$
$5d-120$
$\frac{5d}{5}=\frac{120}{5}$
$d=24$
Hence, the distance, d, between house and school is 24 miles.

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