Rheecaidn
Answered

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Robert leaves home on a motorcycle and travels towards his aunt’s house at an average speed of 75 kilometers per hour (kph or km/h). While on the road, the motorcycle breaks down and he decides to walk the rest of the way to his aunt at an average speed of 5 kph. If he reaches her aunt’s house after 1 hour and 20 minutes and the distance from his home to his aunt’s house is 30 kilometers, how far has he gone before the motorcycle breaks down?

Sagot :

Mlcparra16idn
So the formula for distance is:
[tex]Distance= {Rate}*{Time}[/tex]

So we compute for the distance he traveled by walking and we get:
[tex]5* \frac{4}{3} = \frac{20}{3} [/tex]
(if you are wondering how we got 4/3 we are expressing 1 hour and 20 minutes as 4/3 an hour)

Computing for the distance he traveled before his motorcycle broke down is\
[tex]30- \frac{20}{3} = \frac{70}{3}[/tex]

Therefore he had traveled [tex] \frac{70}{3} [/tex]km before his motorcycle broke down
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