Question 296126: Samuel left the house at noon heading to David's store, 60 miles away. David left his store at 2 pm, heading to Samuel's house and traveling at twice Samuel's speed. The two men met at 3 p.m. How fast was David traveling?
Found 2 solutions by kitkat231, MathTherapy:
Answer by kitkat231(1)   (Show Source):
Answer by MathTherapy(10553)  (Show Source):
You can put this solution on YOUR website! Samuel left the house at noon heading to David's store, 60 miles away. David left his store at 2 pm, heading to Samuel's house and traveling at twice Samuel's speed. The two men met at 3 p.m. How fast was David traveling?
Since the journey from David's store to Samuel's house is 60 miles, then the sum of the distances traveled by the 2 will total 60, or,
Samuel's distance covered + David's distance covered = 60 miles
Let Samuel's speed be S, then David's speed will be 2S, since David is traveling at twice Samuel's speed
Then Samuel's distance traveled = S(3), since it took him 3 hours to cover the distance to the meeting point, and
David's distance covered = 2S(1), since it took David 1 hour to cover the distance to the meeting point
Therefore, we'll have: 3S + 2S = 60
5S = 60
S = 
Now, since S, or Samuel's speed is 12 mph, then David's speed =  mph (2S, or 2*12).
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