Question 1025706: A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 45% salt and Solution B is 70% salt. She wants to obtain 70 ounces of a mixture that is 55% salt. How many ounces of each solution should she use?
solution A = x ounces
Solution B = x ounces
thank you to whoever answers :)
Found 5 solutions by josgarithmetic, lwsshak3, MathTherapy, ikleyn, greenestamps:
Answer by josgarithmetic(39818)   (Show Source):
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 45% salt and Solution B is 70% salt. She wants to obtain 70 ounces of a mixture that is 55% salt. How many ounces of each solution should she use?
solution A = x ounces
Solution B = x ounces
let x=Amt of solution A used
70-x=Amt of solution B used
45%x+70%(70-x)=55%*70
.45x+49-.7x=38.5
.25x=10.5
x=12
70-x=58
Amt of solution A used=12 ounces
Amt of solution B used=58 ounces
Answer by MathTherapy(10837)  (Show Source):
You can put this solution on YOUR website!
A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 45% salt and Solution B is 70% salt. She wants to obtain 70 ounces of a mixture that is 55% salt. How many ounces of each solution should she use?
solution A = x ounces
Solution B = x ounces
thank you to whoever answers :)
Answer by ikleyn(53846)  (Show Source):
You can put this solution on YOUR website! .
A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt.
She knows that Solution A is 45% salt and Solution B is 70% salt. She wants to obtain 70 ounces of a mixture
that is 55% salt. How many ounces of each solution should she use?
~~~~~~~~~~~~~~~~~~~~~~
Calculations in the post by @lwsshak3 are incorrect, giving wrong answer.
I came to bring a correct solution.
solution A = x ounces
Solution B = (70-x) ounces
let x = Amt of solution A used
70-x = Amt of solution B used
0.45x + 0.7*(70-x) = 0.55*70
0.45x + 49 - 0.7x = 38.5
0.25x = 10.5
x = 10.5/0.25 = 42.
70-x = 28
Amt of solution A used = 42 ounces
Amt of solution B used = 28 ounces
CHECK for the final concentration  = 0.55, or 55%. ! precisely correct !
Solved correctly.
Answer by greenestamps(13351)  (Show Source):
You can put this solution on YOUR website!
Here is a solution using a non-traditional method that can be used to solve any 2-part mixture problem. This method can be especially fast and easy if the numbers in the problem are "nice" (which in this problem they are....)
(1) Use a number line if it helps to observe/calculate that 55% is 10/25 = 2/5 of the way from 45% to 70%.
(2) That means 2/5 of the mixture must be the 70% salt solution.
2/5 of 70 ounces is 28 ounces.
ANSWER: 28 ounces of the 70% solution and 70-28 = 42 ounces of the 45% solution.
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