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 the

Question

allochka39001 [22]

2 years ago

14

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 the

re than on the return trip. What was Paul's average speed on the outbound trip

Mathematics

2 answers:

ollegr [7] · Answer 1 · 2022-05-30T12:08:42+03:00

Answer:

40 mph

Step-by-step explanation:

We assume "outbound" refers to the trip <em>to the lake</em>. 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.

___

The distance to the lake was 120 mi.

BARSIC [14] · Answer 2 · 2022-05-30T14:24:01+03:00

Answer: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