document.write( "Question 362407: At 7 am Joe starts jogging at 6mi/hr. At 7:10 am Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 am? \n" ); document.write( "
Algebra.Com's Answer #258271 by mananth(16946)\"\" \"About 
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At 7 am Joe starts jogging at 6mi/hr. At 7:10 am Ken starts off after him. How fast must Ken run in order to overtake him at 7:30 am?
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\n" ); document.write( "Joe --- 7.00 am 6mph
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\n" ); document.write( "Ken ----2.10am---x mph
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\n" ); document.write( "difference in speed = x-6 mph
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\n" ); document.write( "in 10 minutes joe has jogged 1/6*6 = 1mile.
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\n" ); document.write( "time = distance /speed
\n" ); document.write( "time = 20 minutes ( 7.10 to 7.30)=2/3 hours
\n" ); document.write( "speed = x-6. ( Ken catches up x-6 mph.
\n" ); document.write( "he has to catch up 1 mile
\n" ); document.write( "1/(x-6)=2/3
\n" ); document.write( "2(x-6)=3
\n" ); document.write( "2x-12=3
\n" ); document.write( "2x=12+3
\n" ); document.write( "2x=15
\n" ); document.write( "x= 7.5 mph.
\n" ); document.write( "ken should run at 7.5 mph
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\n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( "
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