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Re: How much water should be added to 10 liters of a 20%-solution of alcoh [#permalink]

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22 Jul 2015, 08:34

marcodonzelli wrote:

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% ? 25 liters 27 liters 30 liters 32 liters 35 liters

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.

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% ? 25 liters 27 liters 30 liters 32 liters 35 liters

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%).
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Re: How much water should be added to 10 liters of a 20%-solution of alcoh [#permalink]

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22 Jul 2015, 12:08

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)

Re: How much water should be added to 10 liters of a 20%-solution of alcoh [#permalink]

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26 Jul 2015, 21:46

sudipnayak93 wrote:

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%.

Re: How much water should be added to 10 liters of a 20%-solution of alcoh [#permalink]

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26 Jul 2017, 21:37

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Re: How much water should be added to 10 liters of a 20%-solution of alcoh [#permalink]

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26 Jul 2017, 22:05

marcodonzelli wrote:

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% ?

A. 25 liters B. 27 liters C. 30 liters D. 32 liters E. 35 liters

Originally, the solution had 2L of alcohol and 8L of water. To reduce the concentration of alcohol by 75% means to bring down the alcohol level to 2L * 25% = 0.5L, so that the solution would contain 9.5L water and 0.5L alcohol.

\(\frac{8+x}{10+x} = \frac{95}{100}\)

\(800 + 100x = 950 + 95x\)

\(5x = 150\)

\(x = 30\). Ans - C.
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How much water should be added to 10 liters of a 20%-solution of alcoh [#permalink]

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30 Jul 2017, 11:55

2

This post received KUDOS

marcodonzelli wrote:

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% ?

A. 25 liters B. 27 liters C. 30 liters D. 32 liters E. 35 liters

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
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