How many gallons each of 30​% alcohol and 10​% alcohol should be mixed to obtain 20 gal of 14​% ​alcohol?

Question

How many gallons of 3030% alcohol should be in the mixture?           gal

Answer 1

0.10(20 - x) + 0.30x = 0.14(20)

10(20 - x) + 30x = 14(20)

200 - 10x + 30x = 280

 

Can you solve for x?

Answer 2

first we can make table to present given things and find out

 

 

 

 

                                      quantity                  percent                             total

 

 

30% alcohol                       x                             30%                             x*0.3

 

10 % alcohol                      y                             10%                             y*0.1

 

total                                 20                            14%                             20*0.14

 

 

from above table we can set up 2 equation

 

x+y= 20     -----------eq 1

 

0.3x+0.1y =20(0.14)

 

0.3x+0.1y=2.8  -----------eq 2

 

 

eq 1) x+y=20

 

         x=20-y

 

substitute value of x in eq 2

 

0.3x+0.1y =2.8

 

0.3(20-y)+0.1y=2.8

 

 6-0.3y+0.1y=2.8

 

6-0.2y=2.8

 

-0.2y=-3.2

 

y= 16

 

so x=20-y

  

 

    x=20-16=4

 

 

4 gallon of 30% alcohol and 16 gallon of 10% alcohol required