document.write( "Question 638573: One pipe can fill a cistern in 12 hours and a smaller pipe can cill the cistern in 15 hours. If both pipes are used, how many hours are needed? \n" ); document.write( "

Algebra.Com's Answer #402300 by DrBeeee(684) $\"\"$ $\"About$

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In general, the number of gallons delivered is the rate of delivery in gallons per hour, times the number of hours.
\n" ); document.write( "Let G = total number of gallons of water that the cistern holds.
\n" ); document.write( "The rate that the large pipe can deliver G gallons is
\n" ); document.write( "(1) rl = G/12
\n" ); document.write( "The rate that the smaller pipe delivers water is
\n" ); document.write( "(2) rs = G/15
\n" ); document.write( "When we fill the cistern with both pipes, the rate is the sum of the two,
\n" ); document.write( "(3) R = (rl + rs|
\n" ); document.write( "Using the general flow rate fomula gives, for both pipes filling the cistern
\n" ); document.write( "(4) G = (rl + rs)*T, where T is the new time to fill the cistern when both pipes are being used.
\n" ); document.write( "Solve (4) for T, yields
\n" ); document.write( "(5) T = G/(rl + rs)
\n" ); document.write( "Now substitute (1) and (2) into (5) yields
\n" ); document.write( "(6) T = G/[G/12 + G/15]
\n" ); document.write( "Now factor G from the denominator of (6)
\n" ); document.write( "(7) T = G/[G(1/12 + 1/15)]
\n" ); document.write( "Cancel G from (7) yields the final equation for T
\n" ); document.write( "(8) T = 1/(1/12 + 1/15)
\n" ); document.write( "Evaluate (8) gives us the value of T
\n" ); document.write( "T = 20/3 hrs
\n" ); document.write( "Answer: It will take 6 hours and 40 minutes to fill the cistern when we use both pipes. Note: we did not need to know the capacity, G, of the cistern \n" ); document.write( "

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