I have a mixture problem question?

Question

A chemist has a 35% saline solution and a 60% saline solution available. How many millitliters of each of these available solutions should she mix to obtain 700ml of a 40% solution?

Answer 1

x*.35+(700-x)*.60=700*.4=280
-.25x=-140
x=560@35% and 140@60%

Answer 2

Everyone's got the algebraic solution down pat.

You can follow it logically like this, if you like:

The weaker solution is 5% too weak, but one must use some because the strong solution will not be weakened in any other acceptable way (problem only allows this one method). The strong solution has an excess of 20% (since it must not, itself, weaken past 40%). Divide 20% by 5% and you find the solutions must combine in a 4:1 ratio, strong to weak. Given the resultant 700ml of solution, a 4:1 ratio is 560ml 35% solution and 140ml 60% solution.

Sometimes one likes a logical look at the real items involved rather than the abstract mathematical view. Especially when teasing apart complicated problems.

Answer 3

Let
x = amount 35% solution
y = amount 60% solution

We have

x + y = 700 ml
y = 700 - x

.35x + .60y = .40(700) = 280
.35x + (700 - x).60 = 280
.35x + 420 - .60x = 280
-.25x = -140
x = 560
y = 700 - x = 700 - 560 = 140

She should use 560 ml of 35% solution and 140 ml of 60% solution to get 700 ml of 40% solution.

Answer 4

a*0.35+b*0.6=700*0.40
a+b=700

b=700-a

a*0.35+700*0.6-a*0.6=700*0.40
collecting like therms

-0.25a=140
a=560
b=700-560=140

560 ml of 35%
140ml of 60%