Word Problem

Question

How many gallons of a90% antifreeze solution must be mixed with 6060 gallons of 20% antifreeze to get a mixture that is 80% antifreeze? Use the six-step method.

Answer 1

If you meant 60 gallons rather than 6060 then:

 

Letting x = # gallons of 90% solution

 

.9x + .2(60) = .8(x + 60)

.9x + 12 = .8x + 48

.9x - .8x = 48 - 12

.1x = 36

x = 360

 

If you actually meant 6060 gallons of 20% solution then replace

60 with 6060 on left side and replace (x+60) with (x+6060) on

right side and follow same steps

.9x + .2(6060) = .8(x+6060)

Answer 2

Answer deleted due to error of misreading the problem statement.

Answer 3

To get you started, let x = the number of gallons of 90% antifreeze

 

0.90x + .20(6060) = .80(x + 6060)

 

Note that: 90% = 0.90 ,  20% = 0.20 ,  80% = 0.80

 

The left hand-side of the equation represents the concentration and individual volumes of each antifreeze solution

 

The right-hand side of the equation is the resulting sum and concentration of the mixture

 

To solve the equation for x, you can multiply both sides of the equation by 100 to clear the decimals. 

 

You should be able to handle the rest, Derrick.