document.write( "Question 930780: A big pump can pump water out of a pool three times as fast as a small pump. If both pumps are going, they can empty out a pool in 4 hours. How long would it take the faster pump working alone? \n" ); document.write( "
Algebra.Com's Answer #565301 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Let x=time it takes the small pump to empty the pool
\n" ); document.write( "Then small pump empties the pool at the rate of 1/x of the pool per hour
\n" ); document.write( "x/3=time it takes large pump to empty pool
\n" ); document.write( "Large pump empties the pool at the rate of (1/(x/3))=3/x of the pool per hour
\n" ); document.write( "Both pumps working together empties the pool at the rate of 1/4 of the pool per hour
\n" ); document.write( "Soooo 1/x + 3/x=1/4 multiply each term by 4x
\n" ); document.write( "4+12=x
\n" ); document.write( "x=16 hours ----time it takes small pump working alone to empty the pool
\n" ); document.write( "x/3=16/3=5 1/3 hours--time it takes large pump working alone to empty pool\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "In 4 hours, small pump empties (1/16)*(4)=1/4 of the pool
\n" ); document.write( "In 4 hours, large pump empties (3/16)*(4)=3/4 of the pool
\n" ); document.write( "3/4 + 1/4=1 (one pool, that is)
\n" ); document.write( "In 5 1/3 = 16/3 hours, large pump empties (3/16)*(16/3)=1 (1 pool that is)\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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