Question 1151537: A 4% salt solution is mixed with a 16% salt solution to obtain 600ml of 10% salt solution. How much of the 4% salt solution was used
Found 4 solutions by Theo, Alan3354, josgarithmetic, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = number of ml of the 4% salt solution.
y = number of ml of the 16% salt solution.
you get two equations that need to be solved simultaneously.
they are:
x + y = 600
.04 * x + .16 * y = .10 * 600
the first equation gets yu the total milliliters used.
the second equation gets you the total amount of salt in the mixture.
simplify the equations to get:
x + y = 600
.04 * x + .16 * y = 60
multiply both sides of the first equation by .04 and leave the second equation as is to get:
.04 * x + .04 * y = 24
.04 * x + .16 * y = 60
subtract the first equation from the second to get:
.12 * y = 36
solve for y to get:
y = 36 /.12 = 300
since x = y = 600, then x must be equal to 300.
you have x = 300 and y = 300
you also have .04 * 300 + .16 * 300 = 12 + 48 = 60
both equations are true when x = 300 and y = 300
first equation is x + y = 600 which becomes 300 + 300 = 600 which is true.
second equation becomes .04 * 300 + .16 * 300 = .10 * 600 which becomes 12 + 48 = 60 which is also true.
your solution is that 300 milliliters of the 4% solution was used.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! A 4% salt solution is mixed with a 16% salt solution to obtain 600ml of 10% salt solution. How much of the 4% salt solution was used
============
10 is the average of 4 & 16
---> equal amounts
Answer by josgarithmetic(39617)  (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
A traditional method for solving this kind of problem using formal algebra....
The thing we are looking for is the amount of 4% salt solution to use, so we can use that as our variable.
let x = amount of 4% salt solution
then 600-x = amount of 16% salt solution (because the total amount is 600ml)
The x amount of 4% salt solution, combined with the (600-x) amount of 16% salt solution, yields 600ml of 10% salt solution:
Solve using basic algebra; I leave that to you.
Here is a much easier and faster way to solve this kind of mixture problem involving two ingredients.
Think of starting with the 4% solution and adding some of the 16% solution, stopping when the mixture reaches 10%.
Now model that by starting at 4% on a number line and moving towards 16%, stopping at 10%.
The distance from 4% to 16% is 12; the distance from 4% to 10% is 6. In moving from 4% towards 16% and stopping at 10%, the fraction of the total distance you have moved is 6/12 = 1/2.
That fraction 1/2 is then the fraction of the mixture that needs to be the 16% salt solution that you are adding.
So half of the mixture should be the 16% salt solution and the other half should be the 4% salt solution.
So the 600ml mixture should be made using 300ml of each ingredient.
Note an experienced problem solver could answer this problem almost immediately. Since the 10% is halfway between the 4% and the 16%, it is common sense that the mixture should use equal parts of the two ingredients.
All the words of explanation make this look like a long a tedious method for solving the problem; but it is in fact very fast and simple. Without all the words, here is the complete solution:
16-4 = 12; 10-4 = 6; 6/12 = 1/2
1/2 of the 600ml is the 16%; the other half is the 4%
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