Author Topic: PIPES AND CISTERN PROB  (Read 5589 times)

kumar halijol

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« on: February 28, 2014, 01:32:20 AM »
Q1. Pipe A takes 16 min to fill a tank. Pipes B and C, whose cross-sectional circumferences are in the ratio 2:3, fill
another tank twice as big as the first. If A has a cross-sectional circumference that is one-third of C, how long will
it take for B and C to fill the second tank? (Assume the rate at which water flows through a unit cross -sectional
area is same for all the three pipes.

plz provide soln this problem


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« Reply #1 on: March 02, 2014, 12:20:23 AM »
Hey Kumar :) This is a really good problem :) 8) Im not really sure if my solution is right :) But this is what I think :)

Alright so Pipe A takes 16 min to fill a tank. Now Pipe B and C have cross sectional circumferences in ratios 2:3 .. Circumferences means 2*pi*r  .
However when talking about pipes, we are interested in flow per unit area. Meaning the AREA is important for us. So since we know that
Area = pi*r*r , we will have to square the ratios to get the actual area ratios. Hence in terms of Area, B and C have cross sectional areas in the ratio 4 :9 .

But they've given that A has a one third circumference of C. So the ratio can be 1:3 ...Now combining all these statements we can say that the

Ratios of Circumference of A: B :C = 1 : 2 : 3

AND Ratios of Area of A : B : C = 1 :4 :9  :o

So that means Pipe B fills tank 1,  4 times faster than Pipe A &
Pipe C fills tank 1, 9 times faster than Pipe A.

Now coming to the 2nd tank... we know that Since its twice as big as the first, A will take 32 min to fill this tank right ?

So every min, Pipe A would have to do  (1/32) of the work :)

But Pipe B and Pipe C will together do the work 13 ( 4 + 9) times faster than Pipe A.

So every minute, Pipe B and C do 13 * ( 1/32) = 13/32 of the work.

So basically if we Invert this by "RULE of FLIP", then we know that Pipe B & C will take (32/13) mins or Just under 2.5 mins to fill the tank together :)

Once again, Im not really sure of this solution :) But Im pretty sure this is the way to approach this problem :)

Hopefully we can discuss this a little further and find whether the solution is right or no :) ;D
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