SOLUTION: A mixture containing 8% salt is to be mixed with 40 mL of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the first solution must be used?

SOLUTION: A mixture containing 8% salt is to be mixed with 40 mL of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the first solution must be used?

Algebra.Com

Question 215279: A mixture containing 8% salt is to be mixed with 40 mL of a mixture that is 20% salt, in order to obtain a solution that is 12% salt. How much of the first solution must be used?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.08x+.20*40=.12(40+x)
.08x+8=4.8-.12x
.08x-.12x=4.8-8
-.04x=-3.2
x=-3.2/.04
x=80 ml of 8% salt is used.
Proof:
.08*80+.20*40=.12(40+80)
6.4+8=.12*120
14.4=14.4

RELATED QUESTIONS
a mixture containing 8% salt is to be mixed with 40mL of a mixture that is 20% salt, in... (answered by RAY100,Theo)

A mixture containing 8% salt is added to be mixed with 4 ounces of a mixture that is 20%... (answered by josgarithmetic)

chemistry. A mixture containing 8% salt is to be mixed with 4 ounces of mixture that is... (answered by josgarithmetic)

A mixture containing 10% salt is to be mixed with 4 ounces of a mixture that is 20% salt, (answered by josgarithmetic)

How many quarts of a 20% salt solution must be mixed with 15 quarts of a 35% salt... (answered by CharlesG2)

A mixture containing 11 % salt is to be mixed with 4 ounces of a mixture that is 20 %... (answered by ewatrrr,josmiceli)

1 ounce of the mixture containing 6% salt is to be mixed with 2 ounces of a mixture which (answered by Fombitz)

a solution containing 8% salt is to be mixed with a solution containing 16% salt to... (answered by ikleyn)

Solve This Problem Interactively Customize This Problem A mixture containing 6%... (answered by Fombitz,ankor@dixie-net.com)