How many liters of 10% alcohol solution must be mixed in with how many liters of alcohol solution to to make 10 liters of a 12% solution?

Answer 1

I am assuming the second solution is 100% alcohol.

Then the alcohol concentration is:

0.10 X + 1*Y = 0.12*10 = 1.2 <--- 100% = 1; 10% = 0.1; 12% = .12

X + 10Y = 12 <--- multiplies everything by 10 the clear away the decimals

BUT

X+Y = 10 <--- amount of each solution must be 10 liters

Subtracting these equations:

9Y = 2

Y = 2/9

So 10 - 2/9 = 9 and 7/9 = 88/9 liters of the 10% alcohol solution

AND 2/9 liters of the 100% alcohol solution

CHECK:

10% of X + 100% of Y =

10/100 * 88/9 + 2/9 =

880/900 + 200/900 =

1080/900 =

(12/100 )(90/9) =

(12/100)(10) =

12% of 10