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How many liters of a solution that is 15% salt must be added

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How many liters of a solution that is 15% salt must be added [#permalink] New post 09 Oct 2006, 16:55
How many liters of a solution that is 15% salt must be added to 5 liters of a solution that is 8% salt so that the resulting solution is 10% salt?

Solution:
Let n represent the number of liters of the 15% solution. The amount of salt in the 15% solution [0.15n] plus the amount of salt in the 8% solution [(0.08)(5)] must be equal to the amount of salt in the 10% mixture [0.10 (n + 5)].

.15n + .08(5) = .10(n+5)

Answer: n = 2

Can someone help me understand the .10(n+5) part? Any help is appreciated.

Thanks!
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 [#permalink] New post 09 Oct 2006, 17:18
New mixture total volume is N + 5 and it has 10% salt and by conservation salt amount must equal
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 [#permalink] New post 09 Oct 2006, 17:29
(n + 5) is the total resulting solution and it has 10% salt, so the amount of salt in the resulting solution is: .10(n+5)
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 [#permalink] New post 09 Oct 2006, 23:05
Folks, this would be simple if u apply alligation technique.........

First variety contains 15% salt
Second variety contains 8 % salt.
The resultant contains 10% salt.

Do it this way

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 [#permalink] New post 10 Oct 2006, 03:10
Neat trick Cicerone! :lol:
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 [#permalink] New post 10 Oct 2006, 06:52
0.15*x + 0.08*5 = 0.1*(x+5)
0.15*x + 0.4 = 0.1*x + 0.5
0.05*x = 0.1
x = 2
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 [#permalink] New post 10 Oct 2006, 07:42
Cicerone,

I like your allocation technique...could that be applied to other mixture problems such as one another member posted...which is as follows:

John has 20 ounces of a 20% of salt solution. How much salt should he add to make it a 25% solution?
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 [#permalink] New post 10 Oct 2006, 08:51
axl169 wrote:
Cicerone,

I like your allocation technique...could that be applied to other mixture problems such as one another member posted...which is as follows:

John has 20 ounces of a 20% of salt solution. How much salt should he add to make it a 25% solution?


yes it can be applied........
Look at this..............

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 [#permalink] New post 10 Oct 2006, 09:10
Thanks cicerone ..Really cool
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 [#permalink] New post 10 Oct 2006, 09:25
very cool tip cicerone, nice one.
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 [#permalink] New post 10 Oct 2006, 20:32
based on Cicerone's technique..wouldn't the following solution be 2?.. I got this question from an earlier post, and apparently the OA is D? Can someone explain?

How many gallons of water must be mixed with 1 gallon of a 15% salt soltion to obtain a 10% salt solution?
A)0.5
B)0.67
C)1.0
D)1.5
E) 2.0
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 [#permalink] New post 10 Oct 2006, 20:56
amorica wrote:
based on Cicerone's technique..wouldn't the following solution be 2?.. I got this question from an earlier post, and apparently the OA is D? Can someone explain?

How many gallons of water must be mixed with 1 gallon of a 15% salt soltion to obtain a 10% salt solution?
A)0.5
B)0.67
C)1.0
D)1.5
E) 2.0


hey, the answer will be o.5. I guess u made a silly mistake.......
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Answer [#permalink] New post 10 Oct 2006, 21:11
It can be applied as explained in the doc
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Answer   [#permalink] 10 Oct 2006, 21:11
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