algebra word problems?
A 4% salt solution is mixed with an 8% salt solution to obtain 400 grams of a 5% solution
a.how many grams of 4% salt solution are needed?
b. how many grams of 8% salt solution are needed?
please answer asap,, THANK YOU!!!
A 4% salt solution is mixed with an 8% salt solution to obtain 400 grams of a 5% solution
a.how many grams of 4% salt solution are needed?
b. how many grams of 8% salt solution are needed?
please answer asap,, THANK YOU!!!
Comments
You need to find the ratio of 4% solution to 8% solution that gives 5% solution when it is mixed.
Assume the proportion of 4% solution is x, then the proportion of 8% solution is 1-x (and the amount of 5% solution you end up with is obviously 1).
then 4x +8(1-x) = 5 * 1
4x + 8 - 8x = 5
-4x = -3
x = ¾
Thus the final solution comprises ¾ of 4% solution and ¼ of 8% solution
You end up with 400 grams of 5% solution.
The amount of 4% solution = 400 * ¾ = 300 grams.
The amount of 8% solution = 400 * ¼ = 100 grams
Let the amount of 8% salt solution be x grams
Amount of 4% salt solution = (400 - x) grams
0.08x + 0.04 ( 400 - x) = 0.05 ( 400)
0.08x + 16 -0.04x = 20
0.04x = 4
x = 4 /0.04 = 100 grams
ANSWER a) 300 grams b) 100 grams
CHECK
0.04 X 300 = 12
0.08 X 100 = 8
12 + 8 = 20
20/400 X 100 = 5%
Grams of 4% solution = x
Grams of 8% solution = y
x + y = 400
0.04x + 0.08y = 0.05(400)
x + y = 400
0.04x + 0.08y = 20
-0.08x - 0.08y = -32
0.04x + 0.08y = 20
-0.04x = -12
x = 300
If x = 300...and x + y = 400...that means y = 100.
Grams of 4% solution = x = 300
Grams of 8% solution = y = 100
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