How many ml of a 20% acid mixture and a 80% acid mixture should be mixed to get 120ml of a 35% acid mixture?

Volume of Mixture A =90 ml

Volume of Mixture B =30 ml

Step-by-step explanation:

Let say, Mixture A + Mixture B = Mixture C

Volume of Mixture A is x

Volume of Mixture B is y

So, Volume of Mixture C is x+y = 120 ml

Now, Acid contain in Mixture A is 20% =0.2x

Acid contain in Mixture B is 80% =0.8y

Also, Acid contain in Mixture C is 35% =(0.35)(x+y) = 0.35×120=42

Now, we know that,

Acid contain of Mixture A + Acid contain of Mixture B=Acid contain of Mixture C

∴ 0.2x+0.8y=42

∴ 2x+8y=420

We get two linear equations

2x+8y=420 and x+y = 120

Solving above equation...

∴ x=120-y

Replacing x value in 2x+8y=420

∴ 2(120-y)+8y=420

∴ 240-2y+8y=420

∴ 6y=180

∴ y=30

Replacing y value in any equation

∴ x=120-y=120-30=90

∴ x=90

Thus,

Volume of Mixture A is x=90 ml

Volume of Mixture B is y=30 ml

you would have to sell 24 bracelets.

step-by-step explanation:

84/3.5 = 24