Question

Jacob went on a bike ride. After 10 miles he got a flat tire and had to jog back home. He jogs 5 mph slower than he bikes, so the jog took 1 hour longer than the bike ride. At what rate did he travel each way?

Answer 1

Answer: He traveled 10 km/hr through bike and 5km/hr by jogging.

Explanation:

Let the speed of bike be 'x'.

Let the speed of his jogging be 'x-5'.

Distance covered = 10 miles

So the jog took 1 hour longer than the bike ride.

According to question, we get that

`(10)/(x-5)-(10)/(x)=1\\\\10(x-x+5)/(x(x-5))=1\\\\(50)/(x^2-5x)=1\\\\50=x^2-5x\\\\x^2-5x-50=0\\\\x^2-10x+5x-50=0\\\\x(x-10)+5(x-10)=0\\\\(x+5)(x-10)=0\\\\x=10\ km/hr`

Hence, he traveled 10 km/hr through bike and 5km/hr by jogging.