How much water should be added to 10 liters of a 20%-solution of alcoh

In original Content of 10 litre, there is 2 litre alcohol and 8 litre water.

Question ask to reduce concentration of alcohol to 5% (reduce concentration from 20% by 75%).

So lets assume X litre water is added.

So alcohol remains 2, water is 8+X and total content is 10+X.

Now 2*100/(10+X) = 5 => X = 30

Answer C.

very quickly: 2/(10+x)=1/20, x=30

Hi Bunuel,

Could you please clarify 20%-solution of alcohol to reduce the concentration of alcohol in the solution by 75%?. I am getting confusion in this term.

Regards

The question asks about the amount of water we should add to 10 liters of a 20%-solution of alcohol for the solution to become 5%-solution of alcohol (20% reduced by 75% is 5%).

The quantity of alcohol in the actual mixture is 2 litres (20% of 10 litres). Now it is said that the concentration is reduced by 75% or 3/4th of 20% which is 5%. Now in the new mixture this 2 litres alcohol is 5% of the total mixture therefore the total mixture would be 2 *20=40 litres. So 30 litres is added. (Note 5% is 1/20)

10 litres is containing 20% alcoholol...means 2 litres of alcohol and 8 litres of water.
Now the alchol percentage is to reduced by 75%..that means 75% of 20%...that means in the new solution, the content of alcohol should be 5%. (Reduction of 15%).
So that means, the 2 litres of alcohol will make 5% of new solution..
5%-->2 litres
100%-->40 litres..
That means an additional 30 litres (10 litres was initial solution)...and this additional 30 litres is only water.
To cross check--> in new solution..40 litres , 2 litres is alcohol and 38 litres is water.
2 litres = 5% and 38 litres=95%.

Regards
Naresh

Let x ltr water to be added
2ltr alcohol to be represented as (20(1-3/4(new soln.=10+x)))

2=5%*(10+x)-------->x=30
Ans C

One iteration of a weighted average equation here is
(Concentration of A)(Vol A) + (Concen of B)(Vol B) = (Desired Concen)(Total vol = A+B)

This question is a bit unusual.

First, often, the desired concentration is given. Second, it might be hard to discern which value to reduce by 75 percent. In the equation above for this question,

A = 10 liters of soltion with 20% (=.20) alcohol
B = x liters of water with 0% (=0.0) alcohol

If water is zero percent alcohol, its percentage can't be reduced any further. The question therefore asks that 20 percent concentration be reduced by 75 percent:

(.20)(.25) = .05. The latter is the desired concentration of resultant solution.

In the equation, you don't have to insert second term on LHS. Doing so just makes clear that you are adding x, but its concentration of alcohol has no weight. So

(.20)10 + (0)x = (.05)(10 + x)

2 + 0x = .5 + .05x

1.5 = .05x

\(\frac{1.5}{.05}\)= x

x = 30

Answer C

Concentration of alcohol = 20% = C1
A reduction by 75% means the concentration of alcohol should become 5% = Cavg
Concentration of alcohol in water = 0% = C2

10/w2 = (0 - 5) / (5 - 20) = 1/3

w2 = 30 lts

Answer (C)

Check out Weighted Avg and Mixtures Basics here:
https://anaprep.com/arithmetic-weighted-averages/
https://anaprep.com/arithmetic-mixtures/

and these videos:
https://www.youtube.com/watch?v=_GOAU7moZ2Q
https://www.youtube.com/watch?v=VdBl9Hw0HBg

Trying to be consistent and use a table to solve all mixture problems. Having some trouble....Bunuel could you critique me on where i am going wrong:

Amount(L) Soln
20% (alcohol) 10 20(10)
100%(water) x 100x
5% (alcohol) 10 + x 5(10 + x)

20(10) + 100x = 5(10 +x)
200 - 50 +100x = 5x
150 = -95x
Answer = some negative....

As mentioned above i see logically it is 150/5 = x........x= 30

How much water should be added to 10 liters of a 20%-solution of alcohol to reduce the concentration of alcohol in the solution by 75%

Hi FitnessforFriends
the above highlighted part should be
20(10) = 5(10 +x)

as the amount of alcohol remains same ...u have to calculate % of alcohol in new solution

thus 20(10) = 5(10 +x) ==> 200= 50+5x ==> 150/5=x===>30

hope it helps

The solution attached as asked in the chat.

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