Paul travel to the lake and back. the trip took 3 hours and the trip back took 4 hours he averaged 10 mph faster on the trip there than on the return trip. what was paul's average speed on the outbound trip

  40 mph

Step-by-step explanation:

We assume "outbound" refers to the trip to the lake. The ratio of speeds is inversely proportional to the ratio of times, so ...

  outbound speed : inbound speed = 4 : 3

These differ by one ratio unit, so that one ratio unit corresponds to the speed difference of 10 mph. Then the 4 ratio units of outbound speed will correspond to ...

  4×10 mph = 40 mph

Paul's average speed on the outbound trip was 40 mph.

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The distance to the lake was 120 mi.

Paul's average speed on the outbound trip is 40mph

Step-by-step explanation:

Let x represent Paul's outbound trip which is the trip to the lake.

The trip to the lake took 3 hours.

Distance travelled = speed × time

It means that

Distance covered on the trip to the lake would be

3 × x = 3x

the trip back took 4 hours. He averaged 10 mph faster on the trip there than on the return trip. It means that his speed would be

x - 10

Therefore, distance travelled on return trip would be

4(x - 10) = 4x - 40

Since the distance travelled is the same, it means that

3x = 4x - 40

4x - 3x = 40

x = 40

ok

step-by-step explanation:

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%