How many liters of pure water should be mixed with a 14liter solution of 80% acid to produce a mixture that is 60% water?

Question

I need to have an integer as an answer or a decimal 

Answer 1

80% acid is 20% water

60% acid is 40% water

w represents the amount of pure water

 

0.2(14) + w = 0.4(14 + w)

 

Can you solve for w?

Answer 2

14 liters of an 80% acid solution 

let x=# liters of water to add to produce a solution that is 60% water

how much water do you have now ?

20% * 14=0.2*14=2.8 L of water is what you have now

(2.8+x)/(14+x)=60% (amount of water/total amount of solution must equal 60%)

(2.8+x)/(14+x)=60/100

simplify the fraction 60/100 to 3/5

(2.8+x)/(14+x)=3/5

cross multiply

5(2.8+x)=3(14+x)

14+5x=42+3x

5x-3x=42-14

2x=28

x=28/2

x=14 L of pure water must be added

check:

14+2.8=16.8 and 14+14=28

16.8/28=168/280=42/70=6/10=60/100=60%