SOLUTION: Castel drove to the lake and back. It took three hours less time to get there than it did to get back. The average speed on the trip there was 70 km/h. The average speed on the way

Question 1142950: Castel drove to the lake and back. It took three hours less time to get there than it did to get back. The average speed on the trip there was 70 km/h. The average speed on the way back was 28km/h. How many hours did the trip there take?

***NOTE:
I need a full and indepth answer. I'm really confused on how to approach the problem. I know we calculate the rate by the distance travel I. just can't figure out were to go from that point on. Please help.
Found 2 solutions by josgarithmetic, solver91311:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
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Castel drove to the lake and back. It took three hours less time to get there than it did to get back. The average speed on the trip there was 70 km/h. The average speed on the way back was 28km/h. How many hours did the trip there take?
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SPEED TIME DISTANCE TO LAKE 70 x-3 d RETURN 28 x d

Distance each way is the same. Basic RT=D formula and the equal distance each way allow the equation .

That is for the TRIP GOING BACK HOME.

Trip time TO the lake was:

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

The key to this is that the distance is the same either way (unless space aliens came down and moved the guy's starting point at some point during the trip). The basic relationship is, as you stated, .

We know that the rate for the outbound trip is 70 km/hr, and let represent the time for the outbound trip, so we can model the trip TO the lake as:

Since it took three hours less to get there than to return, we can express the time for the return trip by , and since we know the return trip speed is 28 km/hr, the return trip can be modeled as:

Now, since , we can say:

Now all you need to do is solve for

John

My calculator said it, I believe it, that settles it

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