Question

How many gallons each of 20​% alcohol and 10​% alcohol should be mixed to obtain 10gal of 11​% ​alcohol?

Answer 1

Answer:

20​% alcohol = 1 gal

10​% alcohol = 9 gal

Step-by-step explanation:

Let

x = 20​% alcohol

y = 10​% alcohol

x + y = 10 (1)

0.20x + 0.10y = 10*0.11

0.20x + 0.10y = 1.1 (2)

From (1)

x = 10 - y

Substitute x = 10 - y into (2)

0.20x + 0.10y = 1.1

0.20(10 - y) + 0.10y = 1.1

2 - 0.20y + 0.10y = 1.1

- 0.20y + 0.10y = 1.1 - 2

-0.10y = -0.9

y = -0.9/-0.10

y = 9 gal

substitute y = 9 into (1)

x + y = 10

x + 9 = 10

x = 10 - 9

x = 1 gal

20​% alcohol = 1 gal

10​% alcohol = 9 gal