document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #179821 by oberobic(2304)\"\" \"About 
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We are told the bus travels at 4/5 speed of the car. So the car travels at 5/4 speed of bus.
\n" ); document.write( "b = 4/5*c
\n" ); document.write( "c = 5/4*b
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\n" ); document.write( "We are told the car drives for 3 hr at the speed c = 5/4*b.
\n" ); document.write( "d = rt
\n" ); document.write( "d = 5/4b*3 = 15/4b
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\n" ); document.write( "We are told that at that time the car had traveled 85 km more than the bus.
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\n" ); document.write( "d = 3b + 85
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\n" ); document.write( "If a=b and b=c, then a=c.
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\n" ); document.write( "15/4*b = 3b+ 85
\n" ); document.write( "or
\n" ); document.write( "15/4*b = 12/4*b + 85
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\n" ); document.write( "Subtract 12/4*b from both sides
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\n" ); document.write( "3/4b = 85
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\n" ); document.write( "b = (4*85)/3 = 340/3 = 113 1/3 = 113.3333333...
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\n" ); document.write( "So at the time the car broke down, the bus had traveled
\n" ); document.write( "d = 3*113.333 = 340 km
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\n" ); document.write( "340 km is 85 less than the car travelled, so the car had travelled 340+85 = 425.
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\n" ); document.write( "The bus will cover the 85 km in t time
\n" ); document.write( "d = rt
\n" ); document.write( "t = d/r = 85/113.333
\n" ); document.write( "t = 0.75 hr = 45 min
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\n" ); document.write( "So the bus will pass the car at 12:45 pm.
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\n" ); document.write( "We can check this solution by checking to see if the bus can travel 425 km in 3.75 hr.
\n" ); document.write( "3.75*113.333333333333 = 425.
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\n" ); document.write( "Done.
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