document.write( "Question 645148: It takes 3 pumps 8 hours to pump water out of a flooded basement. Find the time it takes each of them to pump water out of the basement by itself if it is known that the first pump is
\n" ); document.write( "twice as fast as the second pimp and three times as fast as the third pump.
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Algebra.Com's Answer #405410 by ankor@dixie-net.com(22740)\"\" \"About 
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It takes 3 pumps 8 hours to pump water out of a flooded basement.
\n" ); document.write( " Find the time it takes each of them to pump water out of the basement by
\n" ); document.write( " itself if it is known that the first pump is twice as fast as the second
\n" ); document.write( " pimp and three times as fast as the third pump.
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\n" ); document.write( "I am not sure how effective a pimp would be in this task.
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\n" ); document.write( "Three pumps with times of: a, b, c
\n" ); document.write( "Let the completed job = 1 (an empty basement)
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\n" ); document.write( "Write a shared work equation
\n" ); document.write( "\"8%2Fa\" + \"8%2Fb\" + \"8%2Fc\" = 1
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\n" ); document.write( "\"the first pump is twice as fast as the second\"
\n" ); document.write( "b = 2a (b take 2 times as long as a)
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\n" ); document.write( "three times as fast as the third pump.
\n" ); document.write( "c = 3a (c takes 3 times as long as a)
\n" ); document.write( ":
\n" ); document.write( "Replace b and c in the shared work equation
\n" ); document.write( "\"8%2Fa\" + \"8%2F%282a%29\" + \"8%2F%283a%29\" = 1
\n" ); document.write( "Multiply by 6a to clear the denominators, results:
\n" ); document.write( "6(8) + 3(8) + 2(8) = 6a
\n" ); document.write( "48 + 24 + 16 = 6a
\n" ); document.write( "88 = 6a
\n" ); document.write( "a = 88/6
\n" ); document.write( "a = 14\"2%2F3\" hrs working alone
\n" ); document.write( "then
\n" ); document.write( "b = 2(14\"2%2F3\") = 29\"1%2F3\" hrs working alone
\n" ); document.write( "and
\n" ); document.write( "c = 3(14\"2%2F3\") = 44 hrs working alone
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