SOLUTION: You have two solutions; solution 1 contains 10% chemical X, and solution 2 contains 50% chemical X. (a)You mix 5 liters of solutions 1 with 3 liters of solution 2. What percent

Question 1064127: You have two solutions; solution 1 contains 10% chemical X, and solution 2 contains 50% chemical X.
(a)You mix 5 liters of solutions 1 with 3 liters of solution 2. What percent chemical X is the resulting solution?
(b)You want to make 12 liters of solution that contains 20% chemical X. how many liters of solutions 1 and 2 should you mix together?
(c)To make a solution that is 40% chemical X, what ratio of solution 1 to solution 2 should you mix? By ratio, we mean something like x parts solution 1 to y parts solution 2.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x equal the amount of liters in solution 1.
let y equal the amount of liters in solution 2.

then:

.1 * x = amount of chemical X in solution 1.
.5 * y = amount of chemical X in solution 2.

(a) You mix 5 liters of solutions 1 with 3 liters of solution 2. What percent chemical X is the resulting solution?

when x = 5 and y = 3, you get x + y = 8, and you get .1 * x + .5 * y = .1 * 5 + .5 * 3 = .5 + 1.5 = 2.

you get a total of 8 liters, 2 of which are chemical X.

the percent of chemical X in the resulting solution is 2/8 = 25%.

(b)You want to make 12 liters of solution that contains 20% chemical X. how many liters of solutions 1 and 2 should you mix together?

x + y = 12
.1 * x + .5 * y = .2 * (x + y)

since x + y = 12, these formulas become:

x + y = 12
.1 * x + .5 * y = .2 * 12 = 2.4

since x + y = 12, this means that x must be equal to 12 - y.

replace x in the equation of .1 * x + .5 * y = 2.4 with 12 - y to get:

.1 * (12 - y) + .5 * y = 2.4

simplify to get 1.2 - .1 * y + .5 * y = 2.4

combine like terms to get 1.2 + .4 * y = 2.4

subtract 1.2 from both sides of the equation to get .4 * y = 1.2

divide both sides of the equation by .4 to get y = 1.2 / .4 = 3.

since x + y = 12, this means that x must be equal to 9.

you get x = 9 and y = 3

the amount of chemical X is therefore .1 * 9 + .5 * 3 = .9 + 1.5 = 2.4.

2.4 / 12 = 20%.

you need 9 liters of solution 1 and 3 liters of solution 2 to get a solution that is 20% composed of chemical X.

what you had here are two equations that needed to be solved simultaneously.

those equation were:

x + y = 12
.1 * x + .5 * y = 2.4

if you had decided to solve these by elimination, you would have multiplied the second equation by 10 and left the second equation as is to get:

x + y = 12
x + 5y = 24

you would then have subtracted the first equation from the second to get:

4y = 12

you would then have solved for y to get y = = 3

that would then have led to x = 9.

same answer.

first solution was by substitution.
second solution was by elimination.

(c)To make a solution that is 40% chemical X, what ratio of solution 1 to solution 2 should you mix? By ratio, we mean something like x parts solution 1 to y parts solution 2.

you want .1 * x + .5 * y = .4 * (x + y)

simplify this to get:

.1 * x + .5 * y = .4 * x + .4 * y

subtract .1 * x from both sides of the equation and subtract .4 * y from both sides of the equation to get:

.5 * y - .4 * y = .4 * x - .1 * x

combine like terms to get:

.1 * y = .3 * x

divide both sides of the equation by y and divide both sides of the equation by .3 and you get:

.1 / .3 = x / y

simplify to get 1/3 = x/y

that's your ratio.

x/y = 1/3

you will need one x for every three y.

this means 1 liter of 10% solution for every 3 liters of 50% solution.

for example:

assume x = 1 and y = 3.
your resulting solution will be 4 liters total and the amount of chemical X will be .1 * 1 + .5 * 3 = 1.6.
1.6 / 4 = 40%.

assume x = 2 and y = 6.
your resulting solution will be 8 liters total and the amount of chemical X will be .1 * 2 + .5 * 6 = 3.2.
3.2 / 8 = 40%

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