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Could you explain how you derived the second equation?
One in which pipe Y is closed
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Can you explain the question?
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Pipe X pours a mixture of acid and water, and pipe Y pours pure water into a bucket. After 1 hour, the bucket got filled and the concentration of acid in the bucket was noted to be 8%. If pipe Y was closed after 30 minutes and pipe X continued to pour the mixture, concentration of acid in the bucket after 1 hour would have been 10%.

What is the ratio of acid to the water in the mixture coming out of pipe X?

Let the ratio of acid to total in the mixture coming out of pipe X be k and rate of Pipe X and Pipe Y filling bucket be x & y liters/hour

After 1 hour, the bucket got filled and the concentration of acid in the bucket was noted to be 8%.
After 1 hour: -
Pipe X: acid = kx liters
Pipe X: water = (1-k)x liters
Pipe Y: water = y liters
Acid = kx liters
Water = (1-k)x + y liters
Total = x + y liters

Acid % = 8% = kx/(x+y) = .08; kx = .08(x+y); (k-.08)x = .08y; x = .08y/(k-.08)

If pipe Y was closed after 30 minutes and pipe X continued to pour the mixture, concentration of acid in the bucket after 1 hour would have been 10%.
Pipe X: acid = kx
Pipe X: water = (1-k)x
Pipe Y: water = y/2
Total = x+ y/2
Acid % = 10% = .1 = kx/(x+y/2) ; kx = .1(x+y/2); (k-.1)x = .1y/2 = .05y; x = .05y/(k-.1)

x = .08y/(k-.08) = .05y/(k-.1)
.08/(k-.08) = .05/(k-.1)
8/(k-.08) = 5/(k-.1)
5k - .4 = 8k - .8
3k = .4
k = .4/3 = 4/30 = 2/15

Acid: Acid + Water = 2:15
Acid: Water = 2: 13

IMO E
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Here's an intuitive approach:

Let quantity of bucket = 100L
Quantity of acid = 8L
Quantity of water by Pipe 1 = x
Quantity of water by Pipe 2 = y

With this info, we can create following 2 equations:

\(x + y = 92\). ----- (1)

\(\frac{8}{(8 + x + y/2)} = \frac{1}{10} = \frac{8}{80}\).

\(x + \frac{y}{2} = 72\) ----- (2)

Now just solve these like any other linear equation and you will get your ratio as 2 : 13.
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