Question 239947: At 8:00 am the smiths left a campground driving at 48 mi/hour. At 8:20 the Garcias left the same campground and followed the same route driving at 60 mi/hour. At what time did they overtake the Smiths?
I thought that I had to take 20/60 to get 1/3, to make minutes into hours, and add it to 60t. Then have 48t equal that. (Shown Below)
48t = 60(t+1/3)
Then, I distributed, and got:
48t = 60t + 20
I subtracted 60t from 48t, and got this:
-28t = 20
I divided 28 by 20, and got 1.4, but I'm not sure if my answer is right. Can you help me?
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! you've got the times and speeds reversed
they overtake when they have travelled the same DISTANCE (rate x time)
the shorter time goes with the faster speed
60 t = 48 (t + 1/3) ___ the Smiths had a 20 min head start
12 t = 16 ___ t = 4/3 or 1:20
8:20 + 1:20 = 9:40
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