document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197896>Question 1197896</A>: <I><SPAN STYLE=\"word-break: break-all;\"> It takes a smaller hose twice as long to fill a small swimming pool as it does a larger hose. It takes both hoses working together 20 minutes to fill the swimming pool. How long will it take the smaller hose to fill the pool by itself?<BR>\n" );
document.write( "Do not do any rounding.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #831341 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=831341\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Method 1\r<BR>\n" );
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document.write( "x = time it takes the larger hose to fill the pool if working alone<BR>\n" );
document.write( "2x = time it takes the smaller hose to fill the pool if working alone\r<BR>\n" );
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document.write( "The jump from x to 2x is twice as much to reflect it takes twice as long.\r<BR>\n" );
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document.write( "1/x and 1/(2x) are the unit rates for each hose\r<BR>\n" );
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document.write( "They add to...<BR>\n" );
document.write( "1/x + 1/(2x)<BR>\n" );
document.write( "2/(2x) + 1/(2x)<BR>\n" );
document.write( "(2+1)/(2x)<BR>\n" );
document.write( "3/(2x)<BR>\n" );
document.write( "This is the combined unit rate for both hoses.\r<BR>\n" );
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document.write( "(unit rate)*(time) = amount done<BR>\n" );
document.write( "( 3/(2x) )*( 20 min ) = 1 job<BR>\n" );
document.write( "60/(2x) = 1<BR>\n" );
document.write( "30/x = 1<BR>\n" );
document.write( "x = 30\r<BR>\n" );
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document.write( "If x = 30, then 2x = 2*30 = <font color=red>60 minutes</font>\r<BR>\n" );
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document.write( "It takes the larger hose 30 min to do the job alone.<BR>\n" );
document.write( "It takes the smaller hose <font color=red>60 min</font> to do the job alone.<BR>\n" );
document.write( "1/30 + 1/60 = 2/60 + 1/60 = 3/60 = 1/20 is the combined unit rate.<BR>\n" );
document.write( "(unit rate)*(time) = amount done<BR>\n" );
document.write( "(1/20 job per min)*(20 min) = 1 job<BR>\n" );
document.write( "This confirms our answer is correct.\r<BR>\n" );
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document.write( "Method 2\r<BR>\n" );
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document.write( "x = time it takes for the larger hose working alone<BR>\n" );
document.write( "2x = time it takes for the smaller hose working alone\r<BR>\n" );
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document.write( "The two hoses, when working together, need T = 20 minutes<BR>\n" );
document.write( "1/x + 1/(2x) = 1/T<BR>\n" );
document.write( "1/x + 1/(2x) = 1/20<BR>\n" );
document.write( "20x * [ 1/x + 1/(2x) ] = 20x*(1/20)<BR>\n" );
document.write( "20x*(1/x) + 20x*( 1/(2x) ) = 20x*(1/20)<BR>\n" );
document.write( "20 + 10 = x<BR>\n" );
document.write( "30 = x<BR>\n" );
document.write( "x = 30<BR>\n" );
document.write( "2x = 2*30 = <font color=red>60 minutes</font>\r<BR>\n" );
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document.write( "This method is very similar to the first section, just a slight rephrasing.\r<BR>\n" );
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document.write( "Method 3\r<BR>\n" );
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document.write( "Let m and n represent the time values for each hose when they work individually.<BR>\n" );
document.write( "Their respective unit rates are 1/m and 1/n<BR>\n" );
document.write( "Add them up to find the combined unit rate<BR>\n" );
document.write( "1/m + 1/n<BR>\n" );
document.write( "n/(mn) + m/(mn)<BR>\n" );
document.write( "(n+m)/(mn)<BR>\n" );
document.write( "(m+n)/(mn)\r<BR>\n" );
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document.write( "Apply the reciprocal because<BR>\n" );
document.write( "time = (amount done)/(rate)<BR>\n" );
document.write( "time = (1 job)/(rate)<BR>\n" );
document.write( "Effectively informally saying<BR>\n" );
document.write( "time = 1/rate\r<BR>\n" );
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document.write( "The reciprocal of (m+n)/(mn) is (m*n)/(m+n)\r<BR>\n" );
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document.write( "Therefore this formula<BR>\n" );
document.write( "x = (m*n)/(m+n)<BR>\n" );
document.write( "gives the time it takes both hoses to work together given the input time values m and n for each hose working alone.\r<BR>\n" );
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document.write( "Refer to method 3 of the link below to see an example<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.1197891.html\">https://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Polynomials-and-rational-expressions.faq.question.1197891.html</a>\r<BR>\n" );
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document.write( "Let <BR>\n" );
document.write( "m = amount of time for the larger hose<BR>\n" );
document.write( "n = 2m = amount of time for the smaller hose (that takes twice as long)\r<BR>\n" );
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document.write( "So we can further say<BR>\n" );
document.write( "x = (m*n)/(m+n)<BR>\n" );
document.write( "x = (m*2m)/(m+2m)<BR>\n" );
document.write( "x = (2m^2)/(3m)\r<BR>\n" );
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document.write( "From here we replace x with 20 since we know the two hoses work together needing only 20 minutes. Then solve for m.<BR>\n" );
document.write( "x = (2m^2)/(3m)<BR>\n" );
document.write( "20 = (2m^2)/(3m)<BR>\n" );
document.write( "20*3m = 2m^2<BR>\n" );
document.write( "60m = 2m^2<BR>\n" );
document.write( "60m - 2m^2 = 0<BR>\n" );
document.write( "2m(30 - m) = 0<BR>\n" );
document.write( "2m = 0 or 30-m = 0<BR>\n" );
document.write( "m = 0 or m = 30\r<BR>\n" );
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document.write( "Ignore m = 0 as it's a nonsensical answer.<BR>\n" );
document.write( "m = 30 fits with the conclusion reached in the earlier sections where the larger hose needs 30 minutes if working alone. <BR>\n" );
document.write( "The smaller hose then needs 2m = 2*30 = <font color=red>60 minutes</font> when working alone.\r<BR>\n" );
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document.write( "Answer: <font color=red>60 minutes</font> (aka 1 hour)<BR>\n" );
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