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A chemist mixes one liter of pure water with x liters of a 60% salt so

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A chemist mixes one liter of pure water with x liters of a 60% salt so  [#permalink]

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New post 25 May 2016, 03:55
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D
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Difficulty:

  45% (medium)

Question Stats:

68% (01:48) correct 32% (02:02) wrong based on 232 sessions

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A chemist mixes one liter of pure water with x liters of a 60% salt solution, and the resulting mixture is a 15% salt solution. What is the value of x?

A) 1/4
B) 1/3
C) 1/2
D) 1
E) 3

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Re: A chemist mixes one liter of pure water with x liters of a 60% salt so  [#permalink]

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New post 25 May 2016, 04:26
1
.15 = .6x /(x+1)
=>(3/20)*(x+1) = (6/10) x
=> (3/20)x + 3/20 = 12/20 x
=> (9/20)x = 3/20
=>9x = 3
=> x = 1/3

Answer B
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Re: A chemist mixes one liter of pure water with x liters of a 60% salt so  [#permalink]

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New post 25 May 2016, 04:45
1
1
Concentration of salt in pure solution = 0
Concentration of salt in salt solution = 60%
Concentration of salt in the mixed solution = 15%

The pure solution and the salt solution is mixed in the ratio of --> (60 - 15)/(15 - 0) = 3/1

1/x = 3/1

x = 1/3

Answer: B
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Re: A chemist mixes one liter of pure water with x liters of a 60% salt so  [#permalink]

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New post 25 May 2016, 06:30
Bunuel wrote:
A chemist mixes one liter of pure water with x liters of a 60% salt solution, and the resulting mixture is a 15% salt solution. What is the value of x?

A) 1/4
B) 1/3
C) 1/2
D) 1
E) 3


The best way to solve MIX questions is weighted mix method, as also shown above by Vyshak...
there is a decrease of 60-15 = 45%, while there is an increase of 15-0=15% in the second...
this would depend on their respective weights/quantity, so \(\frac{x}{1} = \frac{15}{45} = \frac{1}{3}\)
B
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Re: A chemist mixes one liter of pure water with x liters of a 60% salt so  [#permalink]

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New post 25 May 2016, 07:43
Bunuel wrote:
A chemist mixes one liter of pure water with x liters of a 60% salt solution, and the resulting mixture is a 15% salt solution. What is the value of x?

A) 1/4
B) 1/3
C) 1/2
D) 1
E) 3


Solved it through allegation method:-

1 liter of 100% pure water is mixed with 40% water solution to give 85% water solution:-

1/x= (85-40)/(100-85)= 45/15

x=1/3

B is the answer
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Re: A chemist mixes one liter of pure water with x liters of a 60% salt so  [#permalink]

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New post 27 Oct 2016, 17:26
1
1+0.40x=0.85(1+x)
1+0.40x=0.85+0.85x
0.15=0.45x
x=1/3
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Re: A chemist mixes one liter of pure water with x liters of a 60% salt so  [#permalink]

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New post 27 Oct 2016, 19:05
3
1
Bunuel wrote:
A chemist mixes one liter of pure water with x liters of a 60% salt solution, and the resulting mixture is a 15% salt solution. What is the value of x?

A) 1/4
B) 1/3
C) 1/2
D) 1
E) 3


There are two ways of making equation is this type of questions

1) Water in First solution + Water in Second solution = Water in Final Mixture
i.e. (100/100)*1 + (40/100)*x = (85/100)*(1+x)


OR

2) Salt in First solution + Salt in Second solution = Salt in Final Mixture
i.e. (0/100)*1 + (60/100)*x = (15/100)*(1+x)


Solve anyone of them and find the answer i.e. x = 1/3

Answer: Option B
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Re: A chemist mixes one liter of pure water with x liters of a 60% salt so  [#permalink]

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