SOLUTION: How much 9% saline solution should Kent mix with 70 cubic centimeters (cc) of a 17% saline solution to produce a 13% saline solution?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: How much 9% saline solution should Kent mix with 70 cubic centimeters (cc) of a 17% saline solution to produce a 13% saline solution?       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1202656: How much 9% saline solution should Kent mix with 70 cubic centimeters (cc) of a 17% saline solution to produce a 13% saline solution?
Found 4 solutions by math_tutor2020, Edwin McCravy, josgarithmetic, greenestamps:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

x = amount of the 9% saline solution in cubic centimeters (cc)
0.09x = amount of pure saline from the 9% solution.

Add on 70*0.17 = 11.9 more ccs of pure saline to get 0.09x+11.9 ccs of total pure saline.

Let A = 0.09x+11.9 to represent this quantity.

B = total amount of solution
B = x + 70

A/B = concentration of saline
(0.09x+11.9)/(x + 70) = 0.13


Let's solve for x
(0.09x+11.9)/(x + 70) = 0.13
0.09x+11.9 = 0.13(x + 70)
0.09x+11.9 = 0.13x + 9.1
0.09x-0.13x = 9.1-11.9
-0.04x = -2.8
x = -2.8/(-0.04)
x = 70


Answer: 70 ccs of the 9% saline solution.

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
We mix x of the weaker solution with 70 of the stronger to get y of the medium
strength solution:

                  amt.| %  |total 
                   in |as  |amt of
                  ccs |dec.|saline
weaker          |  x  |0.09|0.09x
stronger        | 70  |0.17|0.17(70)
medium strength |  y  |0.13|0.13y

The equations comes from the first and last columns:

system%28x%2B70=y%2C0.09x%2B0.17%2870%29=0.13y%29

system%28x%2B70=y%2C0.09x%2B11.9=0.13y%29

Multiply the 2nd equation through by 100 to clear decimals:

system%28x%2B70=y%2C9x%2B1190=13y%29

Substitute x+70 for y in the 2nd equation:

9x+1190 = 13(x+70)
9x+1190 = 13x+910
    280 = 4x
     70 = x

So as it turns out, we use 70 ccs of each and get 140 ccs. of
the medium strength.  We could have told that because 13% just 
happens to be exactly half-way between 9% and 17%.  But that's 
just a coincidence.  In another problem that won't be the case. 

Edwin

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
     CONC.%         VOLUME         PURE SALT
        9             d              0.09d
        17           70              0.17*70
        13           d+70           0.09d+0.17*70

0.09d%2B%280.17%29%2870%29=0.13%28d%2B70%29
Simplify and solve for d.

0.09d%2B0.17%2A70=0.13d%2B0.13%2A70
0.17%2A70-0.13%2A70=0.13d-0.09d
%280.13-0.09%29d=70%280.17-0.13%29
d=70%28%280.17-0.13%29%2F%280.13-0.09%29%29
d=70%280.04%2F0.04%29
d=70%2A1
highlight%28d=70%29 ------------notice, the same amount of 9% as the 17%

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The formal algebraic solutions in the responses from the other tutors are fine.

But as one of the tutors notes at the end of his response, the formal algebra is not needed. Common sense tells us that, since 13% is exactly halfway between 9% and 17%, the amount of 9% saline should be equal to the amount of 17% saline.

That tutor implies that solving the problem by that method only works when the percentage of the mixture is halfway between the percentages of the two ingredients.

But that is not the case. That informal method can be used to solve any 2-part mixture problem quickly and easily. In this problem, the percentage of the mixture is 1/2 of the way from the 9% to the 17%, so half of the mixture is the 17% saline. If in a similar problem the percentage of the mixture is 2/5 of the way from the lower percentage to the higher percentage, then 2/5 of the mixture would have to be the higher percentage ingredient.

You should be able to search this website for many examples of problems like this for which I have provided responses showing solutions by this quick informal method.