When one litre of water is added to a mixture of acid and water,

Question

When one litre of water is added to a mixture of acid and water, the new mixture contains 20% acid. When one litre of acid is added to the new mixture, then the resulting mixture contains \(33 \frac{1}{3}\) acid. The percentage of acid in the original mixture was

A. 20%
B. 22%
C. 23%
D. 24%
E. 25%

A. 20%
B. 22%
C. 23%
D. 24%
E. 25%

IanStewart · Answer

Suppose the fraction of the original mixture that is acid is p, and we start with T litres in total. Then we have pT litres of acid, and the rest, T - pT litres, must be water.

When we add one litre of water, we still have pT litres of acid, and now we have T - pT + 1 litres of water. We know this is now 20% acid (and so is 80% water), so acid and water are now in a 1 to 4 ratio, and T - pT + 1 = 4pT.

When we next add one litre of acid, we still have T - pT + 1 litres of water, but now we have pT + 1 litres of acid. If we now have a solution which is 1/3 acid, then acid and water are in a 1 to 2 ratio, and T - pT + 1 = 2(pT + 1).

Since we now know T - pT + 1 is equal both to 4pT and to 2(pT + 1), those must equal each other, so

4pT = 2pT + 2
pT = 1

and we started out with 1 litre of acid. We know after adding 1 litre of water, the solution is 20% acid, and since we have 1 litre of acid, we must then have 4 litres of water. So in the initial situation, before adding the 1 litre of water, we had 3 litres of water and 1 litre of acid, and a 25% acid solution.

The OA currently listed in the first post is "C", which is not correct.

WarriorWithin · Answer

Answer is corrected to E. Here is the explanation -

If there be 1 litre of acid in 4 litres of mixture, then in

case I,

Percentage of acid =  \frac{1}{4+1}×100  = 20%

Case ll,

Percentage of acid =  \frac{2}{6} ×100=  \frac{100}{3 } = 33\frac{1}{3}%

Therefore, Percentage of acid in original 1 mixture =  \frac{1}{4}×100  = 25%

IanStewart · Answer

That is a strange solution -- naturally if we just guess the right answer in advance, that we started with 4 litres in total, then it becomes easy to check that we started with 4 litres in total. But how did we decide we started with 4 litres? That seems to me the whole point of the problem. We're not free to pick an arbitrary starting number here, and if we pick any other starting total besides 4 litres, the percentages won't work out correctly.

What is the source?

WarriorWithin · Answer

Hi  IanStewart

I think assuming values will take you to answer more in less but we can also reach to answer by solving through equation method - Here is how I came to the right answer using equations -

Let a be the acid in liters and w be the water in liters,

solving we get - 
3a + 3 = a+w+2 
2a +2 = w + 1 = 4a ------------( substituting from 1) 
a = 1 , so w = 3

Hence, original concentration of Acid =  \frac{1}{4}*100  = 25%

GMATGuruNY · Answer

At the end of the first stage, the amount of acid is equal to 1/5 of the total mixture:
 A = \frac{1}{5}T 
 5A = T

In the second stage, the amount of acid and the total each increase by 1 liter, with the result that the increased amount of acid is equal to 1/3 of the new total:
 A+1 = \frac{1}{3}(T+1) 
 3A+3 = T+1 
 3A+2 = T

Since 5A=T and 3A+2=T, we get:
5A = 3A+2
2A = 2
A = 1
T = 5A = 5

Since the total at the end of the first stage is 5 liters, the original amount of mixture -- before the addition of one liter of water -- is 4 liters.
Thus:
 \frac{A}{original-mixture} = \frac{1}{4} =  25%

E

pearljiandani · Answer

Hi IanStewart

Can you please explain the highlighted bit above? I understood till the part where we get pT = 1. I'm confused with the rest.

Thanks

IanStewart · Answer

pT in my solution was the amount of acid we started with. So when I find pT = 1, that means we started with 1 litre of acid. Say we also had w litres of water to start with.

We know when we add 1 litre of water, which will give us 1 litre of acid and w+1 litres of water, we get a solution that is 20% acid, or 1/5 acid. If the solution is 1/5 acid, the ratio of acid to water is 1 to 4. We have 1 litre of acid, so we must have 4 litres of water, and w + 1 = 4, so w = 3, and we started with 3 litres of water.

chetan2u · Answer

Let the initial weights be a and w.
When one litre of water is added to a mixture of acid and water, the new mixture contains 20% acid 
=> a:(w+1)=20:80=1:4
4a=w+1.....(I)

When one litre of acid is added to the new mixture, then the resulting mixture contains  33 \frac{1}{3}  % acid.
Be careful of the word  NEW  mixture.
=> (a+1):(w+1)=1:2
2a+2=w+1.....(II)

From I and II, we get 4a=2a+2 => a=1, and, therefore, w=3.
% of a in a+w =  100*\frac{1}{1+3}=25 %

E

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

When one litre of water is added to a mixture of acid and water,

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

When one litre of water is added to a mixture of acid and water,

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards