Pipe X pours a mixture of acid and water, and pipe Y pours pure water

Question

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?

(A) 13: 2
(B) 2 : 15
(C) 3 : 20
(D) 1 : 5
(E) 2 : 13

(A) 13: 2
(B) 2 : 15
(C) 3 : 20 
(D) 1 : 5 
(E) 2 : 13

DashingNightmare2 · Accepted Answer

Let initially there was 100 Liters of solution
=> 8 Liter of acid, w1 liter of water from solution 1 and w2 liter of water from solution 2
=>8 + w1 + w2 =100 -------- 1

Now, Pipe Y was closed after 30 mins, and acid is now 10 %
acid is still 8 liters but it now constitutes 10 % of solution => total =80 Liters solution is there

=>8+w1+ (0.5 * w2) =80 --------- 2

solving 1 and 2
0.5 *w2 =20 => w2 = 40
 
Putting in equation 1
8+ w1+ 40 =100 => w1=52

Hence E.
Required ratio = 8 :52 = 2:13

chetan2u · Answer

The percentage of acid will be 8% at any time as the speed of the two pipes have to be constant.  
Let the quantity poured in half an hour be 100ml, and the water content be  w_m  in mixture and  w_p  in pure water. 
So  8+w_m+w_p=100

In next half an hour,  8+w_m  is poured, so we have  100+8+w_m=108+w_m  after one hour out of which 8+8 or 16 is acid. 
% of acid =  \frac{16}{108+w_m}=\frac{10}{100}…………160=108+w_m………w_m=52

Ratio of acid to  w_m=\frac{8}{52}=\frac{2}{13}

E

Ruchihirawat · Answer

chetan2u  
Could you explain how you derived the second equation? 
One in which pipe Y is closed

valeriacastro11 · Answer

Can you explain the question?

Kinshook · Answer

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?

(A) 13: 2
(B) 2 : 15
(C) 3 : 20
(D) 1 : 5
(E) 2 : 13

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

ZIX · Answer

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.

Pipe X pours a mixture of acid and water, and pipe Y pours pure water

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Pipe X pours a mixture of acid and water, and pipe Y pours pure water

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