document.write( "Question 530607: A boy walks at 11a.m. at 20mph and another boy starts at 1pm with 35mph on the same route. At what time , the second boy will catch the first boy? \n" ); document.write( "

Algebra.Com's Answer #350169 by mananth(16949) $\"\"$ $\"About$

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the distance traveled by both is same when one catches up with the other.
\n" ); document.write( "Let them meet at distance x
\n" ); document.write( "Boy speed= 20 mph , starts at 11:00 am
\n" ); document.write( "Another boy speed= 35 mph starts at 01:00 pm
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\n" ); document.write( "Time difference between the starting of the two= 2.00 hours
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\n" ); document.write( "Time taken by Boy = X / 20
\n" ); document.write( "Time taken by Another boy = X / 35
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\n" ); document.write( "X / 20 - X / 35 = 2
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\n" ); document.write( "LCD - 140
\n" ); document.write( "Multiply equation by 140
\n" ); document.write( "140/20 x- 140/35 x= 2 * 140
\n" ); document.write( "7 x - 4 x = 280
\n" ); document.write( "3 x = 280
\n" ); document.write( "/ 3
\n" ); document.write( "x= 93.33 miles
\n" ); document.write( "Boy speed= 20 mph
\n" ); document.write( "Distance = 93.33 miles
\n" ); document.write( "Time taken = 93.33 / 20
\n" ); document.write( "Time taken = 4.67 =
\n" ); document.write( "0.67 hours = 40.2 minutes
\n" ); document.write( "They will be together at 03:37 pm
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