Math problem i have no clue how to solve

Question

Juan needs to mix a 20% saline solution with a 40% saline solution to create 200 millileters of a 33% solution. How many millileters of each solution must Juan use?

Answer 1

We want to make a 33% saline solution of 200 ml. This means that, we need 200*33/100 = 66 units of salt and the rest will be water.

This salt will come from a 20% solution and a 40% solution. Let's call x the amount of 20% solution in ml and y is the amount of 40% solution in ml. Let's write the amount of salt these solution will contribute:

x*20/100 + y*40/100

This should be equal to 66. So, our first equation is

x*20/100 + y*40/100 = 66 (= 200*33/100)

The total amount should be 200 ml. Then, our second equation is

x + y = 200

Two equations, two unknowns, we can solve this:

First write x in terms of y:

x = 200 - y

Now, substitute this in the first equation:

(200-y)*20/100 + y*40/100 = 66

40 - 0.2y + 0.4y = 66 (add -40 to both sides and combine y terms)

0.2y = 26 (divide both sides by 0.2 or multiple by 5)

y = 130 ml 40% solution

x = 200 - 130 = 70 ml 20% solution

Answer 2

T=ml 20% and F=ml 40%

T+F=200

.2T+.4F=.33*200

Multiply the 2nd equation by 5 to get: T + 2F=330

Subtract the 1st equation: F=130 and T =70