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There are three pipes A,B & C which can fill empty tank in 30 min, 20 min, and 10 min. All three pipes are opened and the pipes discharge chemical P, Q & R. What is the proportion of solution P, Q & R in the liquid in the tank after 3 min?

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Are you just wanting us to do your homework? Show some effort and then we might help! –  Pragabhava Oct 3 '12 at 5:57
    
Marked as homework, as was not able to mark as ratio...!! –  user1575536 Oct 3 '12 at 5:59
    
You can always use the "algebra-precalculus" tag. –  Gerry Myerson Oct 3 '12 at 6:00
    
What happened to waiting for one answer and then thinking about it before posting another question? –  Gerry Myerson Oct 3 '12 at 6:01
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Thats not the point. As it is, it seems your intention is for us to do your homework. What have you done to solve the problem? Where are you stuck? What is it that you dont understand? –  Pragabhava Oct 3 '12 at 6:02
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1 Answer 1

Let me begin by saying that it was very naughty of you for getting us to do your homework. But because you are a new user, I have decided I will post this to help you just this once:

Let x be the rate at which water exits pipe C. (volume/time)
Let y be the rate at which water exits pipe B.
let z be the rate at which water exits pipe A.

From the above information, we can deduce that x = 3z. This is because a fixed volume is filled in 1/3 time when pipe C is used than when pipe A is used. Since Rate has time on the denominator, the rate water exits pipe C must be 3 times faster than pipe A.

Similarly, y = 1.5z.

If all pipes are oppened at once, water enters the system at a rate equal to:
x + y + z = 3z + 1.5z + z = 5.5z.

Therefore, proportion of liquid from pipe A = z/5.5z = 1/5.5
proportion of liquid from pipe B = 1.5z/5.5z = 1.5/5.5
proportion of liquid from pipe C = 3z/5.5z = 3/5.5

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HI Chris...thanks man but did my work my way... And yes Gerry Myerson & Pragabhava tahnsk for forcing me to go through all my incomplete chapters. Actually man that day i was in really need of help but anyway completed my all way of work..thanks all.. –  user1575536 Oct 12 '12 at 1:23
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