document.write( "Question 775609: If a boy walks from his house to school at 4 km/h, he reaches the school 10 minutes earlier than the scheduled time. However if he walks the rate of 3 km/h he reaches 10 minutes late. The distance of the school from his house is \n" ); document.write( "
Algebra.Com's Answer #473013 by josgarithmetic(39617)![]() ![]() ![]() You can put this solution on YOUR website! The required time to make his trip is x, in hours. When he is early using 4 km/hour, time is x-1/6. When he is late at 3 km/hour time is x+1/6. Whether early or late, the trip is the same distance. \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Traveling to school, rate*time=distance \n" ); document.write( "10 minutes = 1/6 hour \n" ); document.write( "Early: \n" ); document.write( "Late: \n" ); document.write( "Equal Distance: \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "1 hour 10 minutes if he were to be \"on-time\".\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "HOW FAR FROM HOME TO SCHOOL? \n" ); document.write( "Pick either early or late expression and compute the value. \n" ); document.write( " |