Two pipes together drain a wastewater holding tank in 6 hours. If used alone to empty the tank, one takes 2 hours longer than the other. How long does each take to empty the tank when used alone?
+26393 Two pipes together drain a wastewater holding tank in 6 hours.
If used alone to empty the tank, one takes 2 hours longer than the other.
How long does each take to empty the tank when used alone?
\(\begin{array}{|rcll|} \hline \text{Let } q_1 &=& \text{Volumetric flow pipe } 1 \\ \text{Let } q_2 &=& \text{Volumetric flow pipe } 2 \\ \text{Let } V &=& \text{Volume of the tank} \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline & q_1 \cdot t + q_2 \cdot t &=& V \\ & (q_1 + q_2) \cdot t &=& V \quad & | \quad t = 6\ hours \\ (1) & (q_1 + q_2) \cdot 6 &=& V \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline (2) & q_1\cdot t_1 &=& V \\ & t_1 &=& \frac{V}{q_1} \\ \hline & q_2 \cdot t_2 &=& V \quad & | \quad t_2 = t_1 + 2\ hours \\ (3) & q_2 \cdot (t_1 + 2) &=& V \\ & q_2 \cdot (t_1 + 2) &=& V \quad & | \quad t_1 = \frac{V}{q_1} \\ & q_2 \cdot (\frac{V}{q_1} + 2) &=& V \\ & \dots \\ (4) & V &=& \frac{2q_1q_2}{q_1-q_2} \\ \hline \end{array} \)
(1) = (4):
\(\begin{array}{|lrcll|} \hline & V = 6\cdot (q_1+q_2) &=& \frac{2q_1q_2}{q_1-q_2} \\ & 6\cdot (q_1+q_2) &=& \frac{2q_1q_2}{q_1-q_2} \quad & | \quad : 2\\ & 3\cdot (q_1+q_2) &=& \frac{q_1q_2}{q_1-q_2} \quad & | \quad \cdot (q_1-q_2) \\ & 3\cdot (q_1+q_2)(q_1-q_2) &=& q_1q_2 \\ & 3\cdot (q_1^2-q_2^2) &=& q_1q_2 \\ & 3q_1^2-3q_2^2-q_1q_2 &=& 0 \\ & 3q_1^2-q_1q_2-3q_2^2 &=& 0 \\\\ & q_1 &=& \frac{q_2\pm \sqrt{q_2^2-4\cdot 3 \cdot (-3q_2^2) } } {2\cdot 3} \\ & q_1 &=& \frac{q_2\pm \sqrt{ q_2^2+36q_2^2 } } {6} \\ & q_1 &=& \frac{q_2\pm \sqrt{ 37q_2^2 } } {6} \\ & q_1 &=& \frac{q_2\pm q_2\sqrt{37} } {6} \\ & q_1 &=& \frac16(1\pm\sqrt{37} ) q_2\\\\ & \mathbf{q_1 = \frac16(1+\sqrt{37} ) q_2 } &\text{ or }& q_1 = \frac16(1-\sqrt{37} )q_2 \\ & & & \text{ no solution } q_1 < 0 ! \\ (5) & \mathbf{\frac{q_1}{q_2} = \frac16(1+\sqrt{37} ) } \\ \hline \end{array}\)
(2) = (3):
\(\begin{array}{|lrcll|} \hline & V = q_1\cdot t_1 &=& q_2 \cdot (t_1 + 2) \\ & q_1\cdot t_1 &=& q_2 \cdot (t_1 + 2) \\ (6) & \frac{q_1}{q_2 } &=& \frac{t_1 + 2}{t_1} \\ \hline \end{array} \)
(5) = (6):
\(\begin{array}{|rcll|} \hline \frac{q_1}{q_2} = \frac16( 1+\sqrt{37} ) &=& \frac{t_1 + 2}{t_1} \\ \frac16( 1+\sqrt{37} ) &=& \frac{t_1 + 2}{t_1} \\ (1+\sqrt{37} )\cdot t_1 &=& 6\cdot(t_1 + 2) \\ (1+\sqrt{37} )\cdot t_1 &=& 6\cdot t_1 + 12 \\ (1+\sqrt{37} -6)\cdot t_1 &=& 12 \\ (\sqrt{37} - 5)\cdot t_1 &=& 12 \\ t_1 &=& \frac{12}{\sqrt{37} - 5} \\ t_1 &=& \frac{12}{6.08276253030- 5} \\ t_1 &=& \frac{12}{1.08276253030} \\ \mathbf{ t_1 } &\mathbf{=}& \mathbf{11.0827625303\ hours }\\\\ t_2 &=& t_1 + 2\ hours \\ \mathbf{ t_2 } &\mathbf{=}& \mathbf{13.0827625303\ hours }\\ \hline \end{array}\)
+118687 Two pipes together drain a wastewater holding tank in 6 hours. If used alone to empty the tank, one takes 2 hours longer than the other. How long does each take to empty the tank when used alone?
I always have an awful problem getting my head around questions like this so I made the question easier to see how it would work.
I said:
Let the tank be 36L and let the times to drain be 4 hours and 6 hours respectively
So tank 1 will flow at 36/4 = 9L/hour
and tank 2 will flow at 36/6 = 6L/hour
if they are both draining then the water will flow at (9+6) L/hour = 15L/hour
so
\(\frac{15L}{1hour}=\frac{15L*\frac{36}{15}}{1hour*\frac{36}{15}}=\frac{36L}{2.4hours}\)
It will take 2.4 hours to drain the tank if both pipes are emptying it.
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Ok now I can do the same thing for your more difficult problem.
Let the tank be V litres
Pipe1 will take t hours to drain it so that is \(\frac{V}{t}\;\;litres/hour\)
Pipe2 will take (t+2) hours to drain it so that is \(\frac{V}{t+2}\;\;litres/hour\)
So the 2 pipes together will drain
\(\frac{V}{t}+\frac{V}{t+2}\;\;litres/hour\\ =V[\frac{1}{t}+\frac{1}{t+2}]\;\;litres/hour\\ =V[\frac{(t+2)+t}{t(t+2)}]\;\;litres/hour\\ =\frac{V\left[\frac{2t+2}{t^2+2t}\right]\;\;litres}{1hour}\\ =\frac{V\left[\frac{2t+2}{t^2+2t}\right]\left[\frac{t^2+2t}{2t+2}\right]\;\;litres}{\left[\frac{t^2+2t}{2t+2}\right]hour}\\ =\frac{V\;\;litres}{\left[\frac{t^2+2t}{2t+2}\right]hour}\\\)
So it will take
\(\frac{t^2+2t}{2t+2}\;\;\text{hours to drain the tank}\\~\\ now\\ \frac{t^2+2t}{2t+2}=6\\ t^2+2t=6(2t+2)\\ t^2+2t=12t+12\\ t^2-10t-12=0\\ t=\frac{10\pm\sqrt{100+48}}{2}\\ t=\frac{10\pm\sqrt{148}}{2}\\ t=\frac{10\pm2\sqrt{37}}{2}\\ t=5\pm\sqrt{37}\\ \text{t cannot be negative so }\\t=5+\sqrt{37}\approx 11.08276253\approx 11hours\;\;4\;minutes \;and\;58\;seconds\\\)
Which is near enough to 11 hours and 5 minutes
So the pipes will take 11 hours and 5 minutes and 13 hours and 5 min respectively to empty the tank on their own.
Since they take 6 hours together that sounds reasonable
+26393 Two pipes together drain a wastewater holding tank in 6 hours.
If used alone to empty the tank, one takes 2 hours longer than the other.
How long does each take to empty the tank when used alone?
\(\begin{array}{|rcll|} \hline \text{Let } q_1 &=& \text{Volumetric flow pipe } 1 \\ \text{Let } q_2 &=& \text{Volumetric flow pipe } 2 \\ \text{Let } V &=& \text{Volume of the tank} \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline & q_1 \cdot t + q_2 \cdot t &=& V \\ & (q_1 + q_2) \cdot t &=& V \quad & | \quad t = 6\ hours \\ (1) & (q_1 + q_2) \cdot 6 &=& V \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline (2) & q_1\cdot t_1 &=& V \\ & t_1 &=& \frac{V}{q_1} \\ \hline & q_2 \cdot t_2 &=& V \quad & | \quad t_2 = t_1 + 2\ hours \\ (3) & q_2 \cdot (t_1 + 2) &=& V \\ & q_2 \cdot (t_1 + 2) &=& V \quad & | \quad t_1 = \frac{V}{q_1} \\ & q_2 \cdot (\frac{V}{q_1} + 2) &=& V \\ & \dots \\ (4) & V &=& \frac{2q_1q_2}{q_1-q_2} \\ \hline \end{array} \)
(1) = (4):
\(\begin{array}{|lrcll|} \hline & V = 6\cdot (q_1+q_2) &=& \frac{2q_1q_2}{q_1-q_2} \\ & 6\cdot (q_1+q_2) &=& \frac{2q_1q_2}{q_1-q_2} \quad & | \quad : 2\\ & 3\cdot (q_1+q_2) &=& \frac{q_1q_2}{q_1-q_2} \quad & | \quad \cdot (q_1-q_2) \\ & 3\cdot (q_1+q_2)(q_1-q_2) &=& q_1q_2 \\ & 3\cdot (q_1^2-q_2^2) &=& q_1q_2 \\ & 3q_1^2-3q_2^2-q_1q_2 &=& 0 \\ & 3q_1^2-q_1q_2-3q_2^2 &=& 0 \\\\ & q_1 &=& \frac{q_2\pm \sqrt{q_2^2-4\cdot 3 \cdot (-3q_2^2) } } {2\cdot 3} \\ & q_1 &=& \frac{q_2\pm \sqrt{ q_2^2+36q_2^2 } } {6} \\ & q_1 &=& \frac{q_2\pm \sqrt{ 37q_2^2 } } {6} \\ & q_1 &=& \frac{q_2\pm q_2\sqrt{37} } {6} \\ & q_1 &=& \frac16(1\pm\sqrt{37} ) q_2\\\\ & \mathbf{q_1 = \frac16(1+\sqrt{37} ) q_2 } &\text{ or }& q_1 = \frac16(1-\sqrt{37} )q_2 \\ & & & \text{ no solution } q_1 < 0 ! \\ (5) & \mathbf{\frac{q_1}{q_2} = \frac16(1+\sqrt{37} ) } \\ \hline \end{array}\)
(2) = (3):
\(\begin{array}{|lrcll|} \hline & V = q_1\cdot t_1 &=& q_2 \cdot (t_1 + 2) \\ & q_1\cdot t_1 &=& q_2 \cdot (t_1 + 2) \\ (6) & \frac{q_1}{q_2 } &=& \frac{t_1 + 2}{t_1} \\ \hline \end{array} \)
(5) = (6):
\(\begin{array}{|rcll|} \hline \frac{q_1}{q_2} = \frac16( 1+\sqrt{37} ) &=& \frac{t_1 + 2}{t_1} \\ \frac16( 1+\sqrt{37} ) &=& \frac{t_1 + 2}{t_1} \\ (1+\sqrt{37} )\cdot t_1 &=& 6\cdot(t_1 + 2) \\ (1+\sqrt{37} )\cdot t_1 &=& 6\cdot t_1 + 12 \\ (1+\sqrt{37} -6)\cdot t_1 &=& 12 \\ (\sqrt{37} - 5)\cdot t_1 &=& 12 \\ t_1 &=& \frac{12}{\sqrt{37} - 5} \\ t_1 &=& \frac{12}{6.08276253030- 5} \\ t_1 &=& \frac{12}{1.08276253030} \\ \mathbf{ t_1 } &\mathbf{=}& \mathbf{11.0827625303\ hours }\\\\ t_2 &=& t_1 + 2\ hours \\ \mathbf{ t_2 } &\mathbf{=}& \mathbf{13.0827625303\ hours }\\ \hline \end{array}\)
+15000 Two pipes together drain a wastewater holding tank in 6 hours. If used alone to empty the tank, one takes 2 hours longer than the other. How long does each take to empty the tank when used alone?
W ist die Entleerung des Tanks.
x ist die Zeit des Einen
x-2h ist die Zeit des Anderen
\(W=(\frac{W}{x}+\frac{W}{x-2h})\times 6h\)
\(\frac{1}{6h}= \frac{1}{x}+\frac{1}{x-2h}\)
\(\frac{1}{6h}=\frac{x-2h+x}{x^2-2hx}\)
\(\frac{x^2-2hx}{6h}=2x-2h\)
\(\Large To\ be \ continued\)
!
+15000 Two pipes together drain a wastewater holding tank in 6 hours. If used alone to empty the tank, one takes 2 hours longer than the other. How long does each take to empty the tank when used alone?
W is the emptying of the tank.
x is the time of the one tube.
x-2h is the time of the other tube.
\(\large W=(\frac{W}{x}+\frac{W}{x-2h})\times 6h\)
\(\large\frac{1}{6h}= \frac{1}{x}+\frac{1}{x-2h}\)
\(\large\frac{1}{6h}=\frac{x-2h+x}{x^2-2hx}\)
\(\large\frac{x^2-2hx}{6h}=2x-2h\)
\(\large x^2-2hx-12hx+12h^2=0\)
\(\large x^2-14hx+12h^2=0\)
\(\large x=7h\pm\sqrt{49h^2-12h^2}=7h\pm6.08276\)
\(x_{a1}=13.0827h=13h \ 4 min \ 58 sec\)
\((x_{a1}-2h)_b=11h \ 4min \ 58sec \)
\(x_{a2}=0.917237h=55min \ 2sec\)
\((x_{a2}-2h)_b=-(1h \ 4min\ 58sec Flow \ rate)\)
!
+37155 One more chime in......
x = time of one pipe x+2 = time of other pipe
Much like two resistors in parallel: r= (r1xr2)/(r1+r2)
x * (x+2) / (x +(x+2)) = 6
(x^2 +2x)/ (2x+2) = 6
x^2 +2x = 12x + 12
x^2-10x-12 = 0 Now use quadratic formula
x = (10+- sqrt(100+48))/2 x = 5+- 6.08 x = 11.08 hrs the other pipe is x+2 = 13.08 hrs
+129933 Work = Rate per hr * Time ....so....
Rate of one pipe * hrs worked + Rate of second pipe * hrs worked = Whole job done
Call x the number of hours it takes one pipe to fill the tank...its rate per hr = 1/x
And the second pipe takes x + 2 hrs.......its rate per hour = 1 /[ x + 2]
Whole job done = 1
So we have
[1/x] * 6 + 1/ [x + 2] *6 = 1
( [6 ( x + 2)] + 6x ) / [ x( x + 2) ] = 1 multiply both sides by x (x + 2)
6x + 12 + 6x = x (x + 2)
12x + 12 = x^2 + 2x rearrange as
x^2 - 10x - 12 = 0 taking the postiive solution for x and we have
x = 5 + √37 hrs ≈ 11.083 hrs and x + 2 ≈ 13.083 hrs
+129933 Well.....that's settled !!!!!.....LOL!!!!!
We all "piped up" with the same answer.......
Guys, guys..........Remember Occam Razor's principle..........!!!
CPhill and EP got it !!!!!
1/T + 1/(T+2) =1/6, solve for T
T =5 + sqrt(37) =11.0827...... Time taken by 1 pipe to drain the tank.
11.0827 + 2 =13.0827........ Time taken the 2nd pipe to drain the tank.
+129933 Thanks, Guest.....I tried to use Occam's Razor once....man.....did he get steamed.....!!!!
+2511 Guys, guys, … ... don’t listen to this blarney bag. … … … …
Mr. Blarney Bag, your many uses of Occam's razor are always inept, and very much so in this instance. You use it as an arbiter for known correct solutions –it was never intended for this. In your plop of BS, you use the magic incantation of Occam’s razor to present a compact answer in three lines. In reality, you have used it as an excuse to trim the requirements for the proof and the related work needed for demonstration.
If Euler had used Occam’s razor he would never have derived (i) or e(iπ).
If Einstein embraced Occam’s razor philosophy, he may have said Newton’s theory of gravity is enough—no need to expand our horizons.
If Alexander Graham Bell said the telegraph is enough, then, instead of “Mr. Watson — Come here — I want to see you.” We’d have:
-- .-. .-.-.- / .-- .- - ... --- -. --..-- / ... --- -- . --- -. . / .--- ..- ... - / ..- ... . -.. / --- -.-. -.-. .- -- ... / .-. .- --.. --- .-. / --- -. / -- -.-- / -... . .-.. .-.. ...
If the progenitors of we genetically enhanced chimps embraced Occam’s razor philosophy, we would be only slightly more intelligent than the mentally encumbered, blarney-classed humans, like you --instead of the vastly superior beings that we are.
There is hope. Maybe there is a cure for your batshit-stupid strains of CDD and your arrogant self-delusions. Just use Occam’s razor to exsanguinate yourself. I’m sure Chirurgeon Tud would recommend starting with the jugular vein or the carotid artery. It is a good thing –you are removing trivial excess from the population. Don’t forget to use the magic incantation. I know this is a small hope, but hope always springs eternal, because Occam’s razor rusts away in this spring.
+118687 Hi Ginger,
I found a translator for you morse code:
http://morsecode.scphillips.com/translator.html
Your code
-- .-. .-.-.- / .-- .- - ... --- -. --..-- / ... --- -- . --- -. . / .--- ..- ... - / ..- ... . -.. / --- -.-. -.-. .- -- ... / .-. .- --.. --- .-. / --- -. / -- -.-- / -... . .-.. .-.. ...
Translates as follows
"MR. WATSON, SOMEONE JUST USED OCCAMS RAZOR ON MY BELLS"
Interesting little variation there :)) 
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Alexander Graham Bell's wife was completely deaf from age 5.
It is interesting that the man spent so much time making hearing people communicate with more ease.
Maybe he did not like his wife ://
Just something to ponder. 
+129933 All true, GA......however.....I believe that at least one of your chimps had a brief fling with Occam's Razor.....
The results got mixed reviews :
Mmmmm...he bears a faint resemblance to Sisyphus......except more handsome.....
+2511 The biographies, published anecdotes from family and friends, and personal letters indicate a great love and devotion between them. His wife lamented to him that he often worked all night and slept all morning. To me, her lament seems more like a tease than a nag.
The one thing Bell did not like was a telephone. He would disconnect it because he didn’t want to be disturbed. Joseph-Ignace Guillotin didn’t much care for his invention either. 
