document.write( "Question 932191: It takes an older pump twice as long to drain a certain pool as it does a newer pump. Working together, it takes the two pumps 4 hours to drain the pool. How long will it take the older pump to drain the pool working alone? Do not do any rounding. \n" ); document.write( "
Algebra.Com's Answer #566043 by TimothyLamb(4379)\"\" \"About 
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r = w/t
\n" ); document.write( "t = w/r
\n" ); document.write( "---
\n" ); document.write( "x = rate of newer pump
\n" ); document.write( "y = rate of older pump = 1/( 2(1/x) )
\n" ); document.write( "y = rate of older pump = 1/( 2/x )
\n" ); document.write( "y = rate of older pump = x/2
\n" ); document.write( "---
\n" ); document.write( "time of newer pump = 1/x
\n" ); document.write( "time of older pump = 1/y
\n" ); document.write( "---
\n" ); document.write( "rate together:
\n" ); document.write( "x + y = 1/4
\n" ); document.write( "x + x/2 = 1/4
\n" ); document.write( "(2/2)x + (1/2)x = 1/4
\n" ); document.write( "(3/2)x = 1/4
\n" ); document.write( "x = (1/4)/(3/2)
\n" ); document.write( "x = (1/4)*(2/3)
\n" ); document.write( "x = (1/2)*(1/3)
\n" ); document.write( "x = 1/6
\n" ); document.write( "y = x/2
\n" ); document.write( "y = (1/6)/2
\n" ); document.write( "y = 1/12
\n" ); document.write( "---
\n" ); document.write( "answer:
\n" ); document.write( "time of newer pump = 6 hours
\n" ); document.write( "time of older pump = 12 hours
\n" ); document.write( "---
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