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M17-11

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M17-11  [#permalink]

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New post 16 Sep 2014, 01:01
2
2
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A
B
C
D
E

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  35% (medium)

Question Stats:

74% (02:01) correct 26% (01:57) wrong based on 76 sessions

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Re M17-11  [#permalink]

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New post 16 Sep 2014, 01:01
Official Solution:

If there are 10 liters of a 20%-solution of alcohol, how much water should be added 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


Amount of alcohol in the original solution is \(0.2*10=2\) liters;

A 20% concentration of alcohol reduced by 75% means that the new solution should be a \(20-0.75*20=5 \%\) alcohol solution;

Since the amount of alcohol in the new solution is not changed, then 2 liters of alcohol must be 5% of the new solution, hence \(0.05*x=2\), which gives \(x=40\) liters.

So, we should add \(40-10=30\) liters of water.


Answer: C
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Re: M17-11  [#permalink]

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New post 21 Nov 2016, 09:15
Bunuel:

Let w be the no. of litres of water

0.2(10) + w = 0.05 (10+w)

What is wrong with this equation?
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Re: M17-11  [#permalink]

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New post 21 Nov 2016, 09:23
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Re: M17-11  [#permalink]

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New post 25 Apr 2017, 07:13
The solution contains 20% alcohol. Therefore, the amount of alcohol is 2l. We need to find out as to how much water must be added to reduce the concentration of alcohol by 75% or make the concentration 25% of the original one.

2/(10+x)=(1/4)*(1/5)
Solving for x , you will get the value as 30.
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Re: M17-11  [#permalink]

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New post 06 May 2017, 14:45
% after reduction= 20% * (100-75)%=5%
water added=x
before adding water, A=10 lit* 20%=2, w=10-2=8 lit
atq
2/(8+x) = 5/95
x=30 lit
ans:c
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New post 20 Aug 2017, 13:14
I think this is poorly worded. It is not very clear whether it is 10liters of alcohol or 10 liters in the entire solution
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New post 21 Aug 2017, 02:48
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New post 24 Sep 2018, 08:48
Bunuel, I tried to convert this into a equation : 20% [10 Lit] = 25% [ 10 +W] and it is giving incorrect answer. After looking at clarification I can understand the solution but appreciate if you can advise what is wrong in my equation.
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Re: M17-11  [#permalink]

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New post 16 Oct 2018, 21:21
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NAvinash wrote:
Bunuel, I tried to convert this into a equation : 20% [10 Lit] = 25% [ 10 +W] and it is giving incorrect answer. After looking at clarification I can understand the solution but appreciate if you can advise what is wrong in my equation.


Hello NAvinash

IMO, you are slightly incorrect there. See the highlighted portion in RED. You correctly understood that you have to reduce the percentage of alcohol to 25% (or by 75%), but that reduction has to be in the given percentage of alcohol, that is, 25% of 20% of alcohol in the solution = 5% of alcohol in the solution.

So, your equation will be;
20% [10 L] = 5% [10+W]
Which will get you W = 30L.

I hope it helped.
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Re: M17-11  [#permalink]

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New post 28 Oct 2018, 09:28
aalekhoza wrote:
NAvinash wrote:
Bunuel, I tried to convert this into a equation : 20% [10 Lit] = 25% [ 10 +W] and it is giving incorrect answer. After looking at clarification I can understand the solution but appreciate if you can advise what is wrong in my equation.


Hello NAvinash

IMO, you are slightly incorrect there. See the highlighted portion in RED. You correctly understood that you have to reduce the percentage of alcohol to 25% (or by 75%), but that reduction has to be in the given percentage of alcohol, that is, 25% of 20% of alcohol in the solution = 5% of alcohol in the solution.

So, your equation will be;
20% [10 L] = 5% [10+W]
Which will get you W = 30L.

I hope it helped.


aalekhoza Yes thanks. Catch is reduce by 75% of existing 20% which comes around 5%.
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Re M17-11  [#permalink]

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New post 27 Feb 2019, 11:52
I think this is a high-quality question and I agree with explanation.
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Re: M17-11  [#permalink]

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New post 22 Mar 2019, 17:49
To solve I set up the following:

\(\frac{2}{10+W} = \frac{1}{20}\)

We can see that we start with a 10L 20%-solution. Set up fraction with alcohol over total. Now we want to know how much water we need to add. Water has no alcohol, so anything we add to this fraction will only increase the denominator. So the denominator becomes 10+w. We set this to be 1/20 because 2/10 * 1/4 = 1/20.

Solving the equation, we get w=30.
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M17-11  [#permalink]

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New post 15 Aug 2019, 03:21
Bunuel wrote:
If there are 10 liters of a 20%-solution of alcohol, how much water should be added 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



Hi Bunuel

As per my understanding, 10 liters of 20% solution of alcohol means we have 2 liters of alcohol and 8 liters of water currently. So, let's assume X liters of water are to be added to reduce the concentration by 75% i.e to 5%. So 2/(8+X) = 0.05. This gives us 32 liters of water. What is wrong with this approach?
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M17-11   [#permalink] 15 Aug 2019, 03:21
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