SOLUTION: At 09 00 car A starts its journey and traveling at 70 km/h at 10 30.Car B started from the same place and traveled steadily on the same road. Car B took 3 1/2 h to catch up with ca

Question 1110002: At 09 00 car A starts its journey and traveling at 70 km/h at 10 30.Car B started from the same place and traveled steadily on the same road. Car B took 3 1/2 h to catch up with car A.
At 15 24, car B reached its destination. When did car A reach the same destination?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39615) (Show Source): You can put this solution on YOUR website!
Last part, is a little unclear.

Time quantity from 9:00 to 10:30 MIGHT be 1 and 1/2 hours.

CAR SPEED TIME DISTANCE(catch-up) A 70 x+3.5 d B r 3.5 d Difference 1.5

x should be 1.5 hours.

CAR SPEED TIME DISTANCE(catch-up) A 70 5 d B r 3.5 d Difference 1.5

You can find d.

and you can then find r:

------speed for car B.
Answer by ikleyn(52769) (Show Source): You can put this solution on YOUR website!
.
(1) At 09:00 car A started its journey and traveled at 70 km/h.
(2) At 10:30 car B started from the same place and traveled steadily on the same road. Car B took 3 1/2 h to catch up with car A.
(3) At 15:24, car B reached its destination.
(4) When did car A reach the same destination?
~~~~~~~~~~~~~~~~~~~~~~

Catching up moment was at 10:30 + 3 h 30 min = 14:00. From 9:00 till 14:00 the car A moved 5 hours and covered 5*70 = 350 kilometers. From 10:30 till 14:00 car B moved 3.5 hours and covered the same distance. So, its rate was = 100 km/h. After catching up moment, car B was on the way from 14:00 till 15:24, i.e 1 hour and 24 minutes = hours = 1.4 hours. and covered 100*1.4 = 140 kilometers. The car A took = 2 hour to cover the same 140 km. So, car A arrived to the destination point at 14:00 + 2 hours = 16:00.

Answer. The car A reached the same destination point at 16:00.

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