Question

A chemist has a 40% alcohol solution and a 25% alcohol solution. How much of each should he mix together to get 50 liters of a 35% alcohol solution?

Answer 1

Answer:

We need 50 liters of 35% alcohol.  

We set up two equations:

A) x + y = 50

B) .40x + .25y = .35 * 50

Multiplying A) by -.40

A) -.40x -.40y = -20  then adding equation B)

B) .40x  + .25y = 17.50

-.15y = -2.50

y = 16.67

x = 33.33

So, we need 33.33 liters of 40% alcohol and

16.67 liters of 25% alcohol

Double-Check

33.33 * .40 + 16.67 * .25 = 13.33 + 4.17 = 17.50 liters of alcohol in 50 total liters

17.50 / 50 = 35% alcohol

Source: https://www.1728.org/mixture.htm

Step-by-step explanation:

Answer 2

Answer:

33 1/3 L of the 40% solution, 16 2/3 L of the 25% solution

Step-by-step explanation:

Set up two equations...

Let x represent the number of Liters of the 40% solution

Let y represent the number of Liters of the 25% solution

We need 50 liters total, so

x + y = 50

and we need the 50 L to be 35% solution, so

0.4x = 0.25y =  0.35(50)  

Solve the first equation for one variable...

x = 50 - y             (subtract y from both sides in equation 1)

Now substitute that value into the second equation...

0.4(50 - y) + 0.25y = 17.5      (x becomes 50 - y, 0.35(50) = 17.5)

Now solve for y...

20 - 0.4y + 0.25y = 17.5

  -0.15y = -2.5

      y = 16.66666667

        y = 16 2/3 L

So we need to plug that into the first equation to find 'x'

x + 16 2/3 = 50

x = 50 - 16 2/3

x = 33 1/3