Question 515751: Victor and Rich live 100 miles apart. They agree to meet at the bridge between their two houses. Victor runs at 20 mph but Rich just walks at 8 mph. If each leaves home at 11 a.m, when do they meet?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! d = distance
r = rate
t = time
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d = r*t is the basic distance equation
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You will use this to solve the problem.
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d = 100, which is known.
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The two guys travel at different rates, but the rates are known: 20 mph and 8 mph
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If they leave at the same time, then their elapsed times will be the same.
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20 mph * t hr + 8 mph * t hr = 100 miles
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t(20+8) = 100
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t = 100/28
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t= 25/7 = 3 4/7 hr
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They will meet in 3 4/7 hr.
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4/7 of an hour = 4/7 * 60 min/hr = about 34.3 minutes
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They left at 11 a.m.
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11 a.m. + 3 hr 34.3 min = about 2:34 p.m.
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We can check the distances traveled by each to be sure the answer is correct.
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25/7 hr * 20 mph = 500/7 miles
25/7 hr * 8 mph = 200/7 miles
500/7 + 200/7 = 700/7 = 100 miles
Correct.
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Done.
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