document.write( "Question 772960: Mila drives at 50 kph on her way to her office. Ten Minutes after she left, her husband who was still at home, discovers that she had left important papers. he tries to catch her before she gets to her office. If he drives at 70 kph, how long will he drive before he can catch up with her? \n" ); document.write( "

Algebra.Com's Answer #471217 by mananth(16946) $\"\"$ $\"About$

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let the catchup distance be x\r
\n" ); document.write( "\n" ); document.write( "time taken by MIla is x/50\r
\n" ); document.write( "\n" ); document.write( "time taken by Husband = x/70\r
\n" ); document.write( "\n" ); document.write( "diffrenece in their starting time = 10 mins = 10/60 = 1/6 hours\r
\n" ); document.write( "\n" ); document.write( "time taken by Mila - time taken by husband = 1/6 hours\r
\n" ); document.write( "\n" ); document.write( "x/50 -x/70 =1/6\r
\n" ); document.write( "\n" ); document.write( "70x-50x= 70*50/6\r
\n" ); document.write( "\n" ); document.write( "20x= 70*50/6\r
\n" ); document.write( "\n" ); document.write( "x= (70*50)/(6*20)\r
\n" ); document.write( "\n" ); document.write( "x= 29.17\r
\n" ); document.write( "\n" ); document.write( "He will catch up after he drives 29.17 km.\r
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