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1 vote
A diesel train traveled to the repair yards and

back. It took two hours longer to go there than
it did to come back. The average speed on the
trip there was 70 km/h. The average speed on
the way back was 80 km/h. How many hours
did the trip there take?

A) 15 hours
C) 25 hours
B) 16 hours
D) 10 hours


User Gmauch
by
5.6k points

1 Answer

2 votes

Answer:

B. 16 hrs

Explanation:

Distance = rate × time

The best way to do this is to make a table with the info. We are concerned with the trip There and the Return trip. Set it up accordingly:

d = r × t

There

Return

The train made a trip from A to B and then back to A again, so the distances are both the same. We don't know what the distance is, but it doesn't matter. Just go with it for now. It'll be important later.

d = r × t

There d

Return d

We are also told the rates. There is 70 km/hr and return is 80 km/hr

d = r × t

There d = 70

Return d = 80

All that's left is the time column now. We don't know how long it took to get there or back, but if it took 2 hours longer to get There than on the Return, the Return trip took t and the There trip took t + 2:

d = r × t

There d = 70 × t+2

Return d = 80 × t

The distances, remember, are the same for both trips, so that means that by the transitive property of equality, their equations can be set equal to each other:

70(t + 2) = 80t

70t + 140 = 80t

140 = 10t

14 = t

That t represents the Return trip's time. Add 2 hours to it since the There trip's time is t+2. So 14 + 2 = 16.

B. 16 hours

User Mark Brittingham
by
5.6k points