How many gallons of a 90% antifreeze solution must be mixed with 100 gallons of 10% antifreeze to get a mixture that is 80% antifreeze

Question

How many gallons of a 90% antifreeze solution must be mixed with 100 gallons of 10% antifreeze to get a mixture that is 80% antifreeze

Answer 1

Let x represent the 90% antifreeze solution to be added

therefore, the amount of antifreeze in this solution will be (90%)(x)

The amount of antifreeze in the 100 gallons solutions is (10%)(100)

The amount of antifreeze in the total solutions is (80%)(x+100)

 

Now write the equation from the information provided,

 

(90%)x   + (10%)(100)= (x+100)(80%)

        0.9x + (0.1)(100)=(x+100)(0.8)

        0.9x + 10 = 0.8x+ 80

          0.1x =70

              x = 700

700 gallons of 90% antifreeze solution should be added.

 

Check,

(90%)(700)+ (10%)(100)= (700+100)(80%)

630 + 10 = 800(0.80)

        640 = 640

 

Answer 2

Hello Crystal,

 

The possible solution follows. 

First, convert percentage to decimal (i.e. 90% as 0.9 and 10% as 0.1) and then write the linear equation for unknown x (gallons of 90% antifreeze):

0.9x + 0.1*100 = 0.8*(100+x)

0.9x + 10 = 80 + 0.8x

0.1x + 10 = 80

 

This simple linear equation can be solved manually or using online Linear Equation Solver, which can be helpful in other cases: http://examn8.com/LinearEquations.aspx

 

Doing it manually, simplify and solve:
0.1x = 70
x = 700

 

The answer is 700 gallons of 90% antifreeze.

 

To check the number, let's calculate the antifreeze percentage in resulted mixture:

 

P = 100*(700*0.9 + 100*0.1)/(700+100) = 80% ( as required)

 

Hope this will help. Best regards,