document.write( "Question 1145725: Tina and Joan averaged 6km/h walking from their house to school, and they arrived 15 min late. If they had averaged 9km/h, they would have arrived on time. How far is it from their house to the school? \n" ); document.write( "

Algebra.Com's Answer #766952 by ikleyn(52810) $\"\"$ $\"About$

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document.write( "Let \"d\" be the distance to the school.\r\n" );
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document.write( "Then the time in the first scenario is    hours.\r\n" );
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document.write( "     The time in the second scenario is    hours.\r\n" );
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document.write( "The difference is 15 minutes, or  of an hour :\r\n" );
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document.write( "It gives you this \"time\" equation\r\n" );
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document.write( "     -  = .\r\n" );
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document.write( "Multiply both sides by 36 to rid off the denominators.\r\n" );
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document.write( "You will get\r\n" );
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document.write( "    6d - 4d = 9\r\n" );
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document.write( "    2d      = 9\r\n" );
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document.write( "     d      = 9/2 = 4.5 miles.\r\n" );
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document.write( "ANSWER.  The distance to the school is  4.5 miles.\r\n" );
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document.write( "CHECK.  a)   =  of an hour;\r\n" );
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document.write( "        b)   =   of an hour.\r\n" );
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document.write( "        The difference is   -  =  of an hour.  ! Correct !\r\n" );
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\n" ); document.write( "\n" ); document.write( "Look into the lesson\r
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\n" ); document.write( "\n" ); document.write( "And find there the exact analogue (TWIN problem) to yours.\r
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