Question

If x ml pure alcohol is added to 400 ml of a 15% solution and after adding its strength becomes 32%, then the value of x is

Answer 1

first, convert percents to decimals. Then, use the fact that (concentration1)*(volume1)+(concentration2)*(volume2) = (concentrationF)(volume1+volume2) where volume is measured in milliliters and concentration is the decimal equivalent of percent. I got x=100 mL, which logically makes sense because it has a high concentration but a low amount, so its not going to increase the final concentration by an extreme amount
This formula is based on the chemistry formula M1V1=M2V2, relating molarity and volume in acid-base titrations

Answer 2

Amount Equation:
x + 400 = y

Alcohol Equation:
x + .15(400) = .32y

Isolate x in both:
x = y - 400
x = .32y - .15(400)

Set x = x
y - 400 = .32y - 60

Solve for y:

y - .32y = 400 - 60
.68y = 340
y = 340/.68
y = 500

solving for x using one of the previous equations:
x = y - 400
x = 500 - 400
x = 100

Answer 3

By the way, this was solved using systems of equations