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