looks easy, but... (1 Viewer)

zemaj

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Oct 17, 2002
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Here's a question. Totally got me stumped and everyone in my house. I'm pretty good at maths, so this is wierd.

Here we go:

Solve using a suitable quadratic:

There are two pipes filling up a tank. Together they take 80mins. One pipe is larger than the other and takes 120mins less time to fill up the tank, if it were to do this by itself. How long would it take for each pipe were to individually fill up the tank?
That's the whole thing. I've looked back at the question about 5 times just to make sure.

Anyone? It was in an extention section of easy algebra. I've never found the book to have an answer wrong.

-zemaj
 

spice girl

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Aug 10, 2002
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Suppose it took x minutes to fill up the tank with the faster pipe (pipe A), so it took x + 120 minutes to fill with the second pipe (pipe B)

So call the rate of water from pipe A = dA/dt, pipe B = dB/dt, volume of tank be V

dA/dt = V / x
dB/dt = V / (x + 120)

Together, the two pipes get the job done in 80 minutes

dA/dt + dB/dt = V / 80

So V/80 = V / x + V / (x + 120)
V/80 = [ V *(x + 120) + V *(x) ]/ x(x + 120)
V/80 = V *(2x + 120) / x(x + 120)

1/80 = (2x + 120)/ x(x + 120)
x(x + 120) = 80*(2x + 120)
x^2 - 40x - 9600 = 0

x = (40 +- sqrt(1600 + 38400)) / 2
x = (40 +- 200)/2
x = 240/2 = 120 (since x > 0)

So it takes 120 minutes with the faster one, and 240 minutes with the slower one.
 

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