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a 30% solution of alcohol is mixed with a 50% solution of

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a 30% solution of alcohol is mixed with a 50% solution of [#permalink] New post 17 Jul 2007, 14:22
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a 30% solution of alcohol is mixed with a 50% solution of alcohol to form a 10 liter solution of 45% alchohol. How much of the 30% solution was used.
:SPOILER BELOW LINE::

2.0 liters
2.5 liters
2.7 liters
3.0 liters
3.2 liters




_______________________________________-

so i was able to find the answer; but how do i know which solution to assign the 10-x variable to (other than trial and error)

i.e.

30%x+50%(10-x)=45%10liters leads you to 2.5 as an answer (which is correct)

but 30%(10-x) +50%(x)... leads you to 7.5. Obviously that isn't in the answer choices, so I switched them, but it wasted valuable time
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Re: PS - mixture problem - question about assigning variable [#permalink] New post 17 Jul 2007, 14:36
you were correct in both cases but you defined both as x where as you need to difine which is x and is (10-x)?

suppose: x = 30% solution
30% x + 50% (10 - x) = 45%
x = 2.5 liters.


suppose: x = 50% solution
then 30% solution = 10 - x

30% (10-x) + 50% x = 45%
x = 7.5 liters.
then 30% solution = 10 - x = 2.5 liters.


hth
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Re: PS - mixture problem - question about assigning variable [#permalink] New post 17 Jul 2007, 14:45
Himalayan wrote:
you were correct in both cases but you defined both as x where as you need to difine which is x and is (10-x)?

suppose: x = 30% solution
30% x + 50% (10 - x) = 45%
x = 2.5 liters.


suppose: x = 50% solution
then 30% solution = 10 - x

30% (10-x) + 50% x = 45%
x = 7.5 liters.
then 30% solution = 10 - x = 2.5 liters.


hth


oohhh... yeah... ::duh::

thank you himalayan. it is easy to lose track when you are rushing like crazy.... and it is hard to keep very detailed notes when you are working a problem. i thought i was losing my mind for a second... :oops:
Re: PS - mixture problem - question about assigning variable   [#permalink] 17 Jul 2007, 14:45
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