Question 246079: at 9:00am, a car and a bus start from manila and head for the north. by 12:00noon the car, already 85km ahead of the bus, suffers a total breakdown. At what time will the bus pass by the car if the bus just maintain its speed which four-fifths that of the car?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! We are told the bus travels at 4/5 speed of the car. So the car travels at 5/4 speed of bus.
b = 4/5*c
c = 5/4*b
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We are told the car drives for 3 hr at the speed c = 5/4*b.
d = rt
d = 5/4b*3 = 15/4b
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We are told that at that time the car had traveled 85 km more than the bus.
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d = 3b + 85
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If a=b and b=c, then a=c.
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15/4*b = 3b+ 85
or
15/4*b = 12/4*b + 85
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Subtract 12/4*b from both sides
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3/4b = 85
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b = (4*85)/3 = 340/3 = 113 1/3 = 113.3333333...
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So at the time the car broke down, the bus had traveled
d = 3*113.333 = 340 km
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340 km is 85 less than the car travelled, so the car had travelled 340+85 = 425.
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The bus will cover the 85 km in t time
d = rt
t = d/r = 85/113.333
t = 0.75 hr = 45 min
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So the bus will pass the car at 12:45 pm.
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We can check this solution by checking to see if the bus can travel 425 km in 3.75 hr.
3.75*113.333333333333 = 425.
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Done.
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