Mixture Word Problem

Question

A chemical company makes two brands of antifreeze. The first brand is 65% pure antifreeze and the second brand is 85%
pure antifreeze. In order to obtain 160 gallons of a mixture that contains 70% pure antifreeze, how many gallons of each brand of antifreeze must be used?

 

Answer 1

X is the number of gallons of 65-proof anti-freeze

Y is the number of gallons of 98-proof anti-freeze

 

X + Y = 160 --> Y - 160 - X

 

The amount of anti-freeze is 0.65*X + 0.85*Y = 160*0.70 = 112

     0.65X + 0.85 * (160 - X) = 112

 

65X + 85 (160 -X) = 11200 <-- multiplies both sides by 100 to clear the decimals

 

65X + 13600 - 85X = 11200

 

-20X + 13600 = 11200

 

-20X = 11200 - 13600

 

-20X = -2400

 

X = 120

 

120 gallons of the 65-proof antifreeze

 40 gallons of the 85-proof antifreeze

 

 

 

 

 

 

Answer 2

Since there are two unknowns, we need two equations.  Let's call the first brand A and the second brand B.  The amount of pure antifreeze in each mixture is just the percentage multiplied by the total amount.

 

A + B = 160

.65A + .85B = .70(160)

 

A = 160 - B

.65(160 - B) + .85B = 112

104 - .65B + .85B = 112

.20B = 8

B = 40

A = 160 - 40 = 120

 

So you will need to mix 120 gallons of the 65% mixture with 40 gallons of the 85% mixture.  this makes sense because 70 is closer to 65 than 85.