SOLUTION: a 3% salt solution is mixed with a 9% salt solution. how many grams of the 9% salt solution were used to make 150 g of a 4% salt solution? ___________ ___________ ___________ ___

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Question 653234: a 3% salt solution is mixed with a 9% salt solution. how many grams of the 9% salt solution were used to make 150 g of a 4% salt solution?
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I don't know where to start in order to solve this. There will be 2 questions like this on my test on Thursday and I have a lot of trouble figuring out how to solve these. Do you have any tips or tricks for how to get equations out of word problems? It is very confusing for me, and my teacher isn't very good at explaining anything. :/ Thank you very much!
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
a 3% salt solution is mixed with a 9% salt solution.
how many grams of the 9% salt solution were used to make 150 g of a 4% salt solution?
:
We convert the per cent to the decimal equiv.
In this case we write a amt of salt equation, the amt of salt of the two
solutions, equal amt in the resulting solution
:
Let x = grams of 9% solution required
the result will be 150 grams, therefore
(150-x) = grams of 3% solution
:
.09x + .03(150-x) = .04(150)
.09x + 4.5 - .03x = 6
.09x - .03x = 6 - 4.5
.06x = 1.5
x = 1.5/.06
x = 25 grams of 9% solution required, which is what they asked
then we would like to know
150 - 25 = 125 grams of 3% solution
:
:
We can confirm this solution by finding the actual amt of salt in each solution,
see if it adds up to the amt of salt in the resulting solution
.09(25) + .03(125) = .04(150)
2.25 + 3.75 = 6; proves we have good answer
:
:
This is a typical mixture equation, can be applied to most mixture problems.
Always a good idea, when you get the answer, to check it as we did here.
:
Did this help you understand? C

Answer by MathTherapy(10552) (Show Source):
You can put this solution on YOUR website!
a 3% salt solution is mixed with a 9% salt solution. how many grams of the 9% salt solution were used to make 150 g of a 4% salt solution?

Let amount of the 9% salt to be mixed be T
Since 150 g is to be made, then amount of 3% salt to be mixed = 150 � T

Equation:
Amount of 9% salt, plus amount of 3% salt equals 150 g of a 4% mixture, OR
.09(T) + .03(150 � T) = .04(150)
.09T + 4.5 - .03T = 6
.09T - .03T = 6 � 4.5
.06T = 1.5

T, or amount of 9% salt to be mixed = , or g

Not needed:
Since total mixture is 150 g, then amount of 3% salt to be mixed = 150 � 25, or g

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SOLUTION: a 3% salt solution is mixed with a 9% salt solution. how many grams of the 9% salt solution were used to make 150 g of a 4% salt solution? ___________ ___________ ___________ ____

SOLUTION: a 3% salt solution is mixed with a 9% salt solution. how many grams of the 9% salt solution were used to make 150 g of a 4% salt solution? _____ _ ___