MATH SOLVE

2 months ago

Q:
# A diesel train traveled to the repair yards andback. It took two hours longer to go there thanit did to come back. The average speed on thetrip there was 70 km/h. The average speed onthe way back was 80 km/h. How many hoursdid the trip there take?A) 15 hoursC) 25 hoursB) 16 hoursD) 10 hours

Accepted Solution

A:

Answer:B. 16 hrsStep-by-step 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 × tThere ReturnThe 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 × tThere d Return dWe are also told the rates. There is 70 km/hr and return is 80 km/hr d = r × tThere d = 70Return d = 80All 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 × tThere d = 70 × t+2Return d = 80 × tThe 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) = 80t70t + 140 = 80t140 = 10t14 = tThat 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