Question

A plane traveled from California and back. It took one hour on the way out than it did on the way back. The plane's average speed out wad 300 mph. The speed on the way back was 350mph. How many hours did the trip out take?

Answer 1

6 hours

Explanation:

You can set up a chart:

R T D

O 300 1+x ~ O=out B=back R=Rate/Speed T=time D=distance

B 350 x ~

The distance is the same. On the way out the speed was 300 mph, so you can see where that is on the chart. And the speed back was 350 mph. It took an extra hour on the way out than on the way back, so you can call the way back's time x and then say the way out's time was x+1.

It's important to know that R times T = D

Therefore we can say 300(x+1)=D and 350(x)=D

Since the distance is the same, 300(x+1)=350(x)

Solve:

300(x+1)=350x

300x+300=350x

50x=300

x=6

Therefore, the trip out took 6 hours.

I hope that helps! :)

Answer 2

Answer:the trip out took 7 hours

Explanation:

Distance travelled = speed × time

Let t represent the time it took the plane to travel from California.

It took one hour on the way out than it did on the way back. This means that the time it took the plane to return would be t - 1

The plane's average speed out ward 300 mph. Distance travelled when going out would be

300 × t = 300 = 300t

The speed on the way back was 350mph. Distance travelled when returning would be

350(t - 1) = 350t - 350

Since the distance is the same, it means that

300t = 350t - 350

350t - 300t = 350

50t = 350

t = 350/50 = 7

Time taken to go out is 7 hours