Question

A man hiked a distance of 9 miles into the country. He rode back in a car along the same route at the rate of 27 mph. If the entire trip took 3 hours and 20 min, find his rate while walking

Answer 1

Recall your d = rt, distance = rate * time.

going into the country, he went for 9 miles, coming back it was the same 9 miles.

let's say hiking his speed is "r" mph, and he took "t" hours hiking.

now, 3 hours and 20 minutes is just 3 hours and one third, or 3 and 1/3 hour.

if he took "t" hours hiking, driving in the car, it took the slack from 3 and 1/3 and "t", or 3 1/3 - t.

$\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ walking&9&r&t\\ driving&9&27&3\frac{1}{3}-t \end{array} \\\\\\ \begin{cases} 9=rt\implies \frac{9}{r}=\boxed{t}\\\\ 9=27\left(3\frac{1}{3}-t \right)\\ ----------\\ 9=27\left(3\frac{1}{3}-\boxed{\frac{9}{r}} \right) \end{cases} \\\\\\ \cfrac{9}{27}=\cfrac{10}{3}-\cfrac{9}{r}\implies \cfrac{1}{3}=\cfrac{10r-27}{3r}\implies 3r=30r-81 \\\\\\ 81=27r\implies \cfrac{81}{27}=r\implies 3=r$

Answer 2

Let n be the man's walking rate. Then:
9/n+9/27=3&1/3
243+9n=90n
243=81n
n=3
The man's walking rate was 3 mph
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