A chemist has 20% and 50% solutions of acid available. How many liters of each solution should be mixed to obtain 75 liters of 28% acid solution?

we will mix 55 liters of 20% solution and 20 Liters of 50% solutions.

Step-by-step explanation:

let the 20% solution = x liters

let the 50% solution = y liters

total solution of the mixture of the two solutions in % = 28%

total solution of the mixture of the two solutions in liters = 75 liters

considering the total solutions given, we will have the following equations if we mix the solutions in equal proportion;

20x + 50y = 28 (75)

0.2x + 0.5y = 0.28(75)

0.2x + 0.5y = 21 equation 1

Also,

x + y = 75 equation 2

y = 75 - x  

(substitute this into equation 1)

0.2x + 0.5(75 - x) = 21

0.2x + 37.5 - 0.5x = 21

0.2x - 0.5x = 21 - 37.5

-0.3x = -16.5

x = -16.5 / -0.3

x = 55 liters

Recall, y = 75 - x

y = 75 -55

y = 20 Liters

Thus, we will mix 55 liters of 20% solution and 20 Liters of 50% solutions.

damm bro the struggle

step-by-step explanation:

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p