SOLUTION: A boy walks at 11a.m. at 20mph and another boy starts at 1pm with 35mph on the same route. At what time , the second boy will catch the first boy?
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Question 530607: A boy walks at 11a.m. at 20mph and another boy starts at 1pm with 35mph on the same route. At what time , the second boy will catch the first boy? Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! the distance traveled by both is same when one catches up with the other.
Let them meet at distance x
Boy speed= 20 mph , starts at 11:00 am
Another boy speed= 35 mph starts at 01:00 pm
Time difference between the starting of the two= 2.00 hours
Time taken by Boy = X / 20
Time taken by Another boy = X / 35
X / 20 - X / 35 = 2
LCD - 140
Multiply equation by 140
140/20 x- 140/35 x= 2 * 140
7 x - 4 x = 280
3 x = 280
/ 3
x= 93.33 miles
Boy speed= 20 mph
Distance = 93.33 miles
Time taken = 93.33 / 20
Time taken = 4.67 =
0.67 hours = 40.2 minutes
They will be together at 03:37 pm
