SOLUTION: mike and adam left a bus terminal at the same time and traveled in opposite directions. Mike's bus was in heavy traffic and had to travel 20mi/hr slower than adams's bus. After
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Question 346433: mike and adam left a bus terminal at the same time and traveled in opposite directions. Mike's bus was in heavy traffic and had to travel 20mi/hr slower than adams's bus. After 3 hours their buses were 270 miles apart. How fast was each bus going Found 2 solutions by mananth, stanbon:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! let adam's bus speed be x mph
mike's bus speed = x-20 mph
..
They are going in opposite directions
.
the effective speed = x+x-20
2x-20 mph.
Time = distance /speed
distance = 270 miles
3= 270/2x-2
3(2x-2)=270
6x-6=270
6x=276
x=276/6
x=46 mph adam's bus speed
mike's bus speed = 46-20 =26 mph
You can put this solution on YOUR website! mike and adam left a bus terminal at the same time and traveled in opposite directions. Mike's bus was in heavy traffic and had to travel 20mi/hr slower than adams's bus. After 3 hours their buses were 270 miles apart. How fast was each bus going
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Mike DATA:
rate = x-20 mph ; time = 3 hrs ; distance = 3(x-20) miles
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Adam DATA:
rate = x mph ; time = 3 hrs ; distance = 3x miles
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Equation:
distance + distance = 270 miles
3(x-20) + 3x = 270
x-20 + x = 90
2x = 110
x = 55 mph (Adam's rate)
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x-20 = 35 mph (Mike's rate)
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Cheers,
Stan H.
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