document.write( "Question 1039935: A salesman started walking from office A at 8:30 AM. at the rate of 2.50 kph. He arrived at office B 12 seconds late. Had he started at A at 8:00 AM and walked at 1.50 kph , he would have arrived at B one minute before the appointed time. At what time was he supposed to be at B? \n" ); document.write( "

Algebra.Com's Answer #654710 by ikleyn(52781) $\"\"$ $\"About$

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A salesman started walking from office A at 8:30 AM. at the rate of 2.50 kph. He arrived at office B 12 seconds late.
\n" ); document.write( "Had he started at A at 8:00 AM and walked at 1.50 kph , he would have arrived at B one minute before the appointed time.
\n" ); document.write( "At what time was he supposed to be at B?
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document.write( "This problem is on the difference of times spent by the body (by the two bodies) traveled with different rates (speeds).\r\n" );
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document.write( "Before hands, input data should be converted to the minutes units:\r\n" );
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document.write( "2.5  =  ,  1.5  =  ,  12 seconds =  of a minute.\r\n" );
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document.write( "An equation is on the difference of times\r\n" );
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document.write( " -  =  =  minutes =  minutes     (1)     or\r\n" );
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document.write( " = .     (2)\r\n" );
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document.write( "where D is an unknown distance in kilometers.\r\n" );
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document.write( "Multiply both sides of (2) by 30 to rid of the denominators. You will get\r\n" );
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document.write( "20*60D - 12*60D = 144*6,   or  1200D - 720D = 864,  or  480D = 864,  or  D =  =  km.\r\n" );
document.write( "Time to walk in the 1-st scenario is   =  = 43.2 minutes =  minutes.\r\n" );
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document.write( "Time to walk in the 2-nd scenario is   =  = 72 minutes.\r\n" );
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document.write( "The difference   =  -  =  minutes, exactly as stated in the condition.\r\n" );
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\n" ); document.write( "\n" ); document.write( "The lesson to learn from this solution: this problem is on the difference of times.\r
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\n" ); document.write( "\n" ); document.write( "For similar problem see the lesson How far do you live from school? in this site.\r
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