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

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Director
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a 30% solution of alcohol is mixed with a 50% solution of [#permalink] New post 20 Sep 2007, 04:27
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.

2.0 liters
2.5 liters
2.7 liters
3.0 liters
3.2 liters

__________________________________________________________




I'm doing this and coming up empty.
30(L)/100 + 50(10-L)/100 = (10)45/100
Director
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Re: what the %&*@ am i doing wrong? [Spoiler below red l [#permalink] New post 20 Sep 2007, 04:41
ggarr wrote:
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.

2.0 liters
2.5 liters
2.7 liters
3.0 liters
3.2 liters

__________________________________________________________




I'm doing this and coming up empty.
30(L)/100 + 50(10-L)/100 = (10)45/100


It is actually quite simple. You want to find the ratio/weighted for each:

the answer should be B

dunno if I can explain this well.. there were these 'killer' notes that killersquirrel provided some time ago ;)

30% + 50 % = 45% of a 10 liter solution
ratio of 30% : 50% ---> 1 : 3
so 30% = 1/4 or 25% of 10L solution = 2.5L
Director
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 [#permalink] New post 20 Sep 2007, 04:50
beckee,

yeah. thanks. I have the notes. I'm just trying to solve the old fashioned way. i'd like to understand the traditional methods before moving on to the shortcuts

can anyone tell me what I'm doing wrong?
Quote:
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.

2.0 liters
2.5 liters
2.7 liters
3.0 liters
3.2 liters

__________________________________________________________




I'm doing this and coming up empty.
30(L)/100 + 50(10-L)/100 = (10)45/100
Director
Director
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Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 42 [0], given: 0

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 [#permalink] New post 20 Sep 2007, 04:56
ggarr wrote:
beckee,

yeah. thanks. I have the notes. I'm just trying to solve the old fashioned way. i'd like to understand the traditional methods before moving on to the shortcuts

can anyone tell me what I'm doing wrong?
Quote:
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.

2.0 liters
2.5 liters
2.7 liters
3.0 liters
3.2 liters

__________________________________________________________




I'm doing this and coming up empty.
30(L)/100 + 50(10-L)/100 = (10)45/100


I think you were getting there..but you have to set up simultaneous eqns. I would do it this way:

x + y = 10, or y = 10 - x

30x + 50 y = 45 (x + y)
30x + 50 (10-x) = 45 (10)
30x + 500 - 50x = 450
-20x = -50
20x = 50 -> 50/20
x = 2.5

see... still works out the old fashioned algebra way!
Intern
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Re: what the %&*@ am i doing wrong? [Spoiler below red l [#permalink] New post 20 Sep 2007, 11:11
beckee529 wrote:
ggarr wrote:
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.

2.0 liters
2.5 liters
2.7 liters
3.0 liters
3.2 liters

__________________________________________________________




I'm doing this and coming up empty.
30(L)/100 + 50(10-L)/100 = (10)45/100


It is actually quite simple. You want to find the ratio/weighted for each:

the answer should be B

dunno if I can explain this well.. there were these 'killer' notes that killersquirrel provided some time ago ;)

30% + 50 % = 45% of a 10 liter solution
ratio of 30% : 50% ---> 1 : 3 so 30% = 1/4 or 25% of 10L solution = 2.5L


This intrigues me - I don't understand the ratio of 1:3. Can anyone provide a link to the killer notes? I'd like to learn this.
Manager
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 [#permalink] New post 20 Sep 2007, 11:20
someone had posted this file and i found it extremely helpful...took me all of 5 seconds to solve the problem :-)
Attachments

File comment: mixture problems made easy
MixtureProblemsMadEasy.pdf [49.94 KiB]
Downloaded 53 times

To download please login or register as a user

CEO
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 [#permalink] New post 17 Oct 2007, 11:35
beckee529 wrote:
ggarr wrote:
beckee,

yeah. thanks. I have the notes. I'm just trying to solve the old fashioned way. i'd like to understand the traditional methods before moving on to the shortcuts

can anyone tell me what I'm doing wrong?
Quote:
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.

2.0 liters
2.5 liters
2.7 liters
3.0 liters
3.2 liters

__________________________________________________________




I'm doing this and coming up empty.
30(L)/100 + 50(10-L)/100 = (10)45/100


I think you were getting there..but you have to set up simultaneous eqns. I would do it this way:

x + y = 10, or y = 10 - x

30x + 50 y = 45 (x + y)
30x + 50 (10-x) = 45 (10)
30x + 500 - 50x = 450
-20x = -50
20x = 50 -> 50/20
x = 2.5

see... still works out the old fashioned algebra way!


i like this way better. more intuitive
Director
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Schools: University of Chicago, Wharton School
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Re: what the %&*@ am i doing wrong? [Spoiler below red l [#permalink] New post 17 Oct 2007, 16:04
ggarr wrote:
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.

2.0 liters
2.5 liters
2.7 liters
3.0 liters
3.2 liters


B. o.3x + 0.5 (1-x) = .45
x = 25%

so 30% alchohal has 25% of 10 lt = 2.5 lt.
VP
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Location: London
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 [#permalink] New post 17 Oct 2007, 18:19
gowani wrote:
someone had posted this file and i found it extremely helpful...took me all of 5 seconds to solve the problem :-)



Appreciate gowani,
it will take me now 3 seconds to solve this kinda PS! :)
  [#permalink] 17 Oct 2007, 18:19
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