Lesson Word problems on mixtures for antifreeze solutions

Word problems on mixtures for antifreeze solutions

Problem 1

Let "x" be the volume of pure antifreeze to be added, in gallons.
Then the total volume of the solution will be 12+x gallons.

The amount of the pure antifreeze in the new solution will be 0.1*12 + x gallons.

The percentage concentration equation is

 = 0.4.

Simplify it multiplying both sides by (12+x):

1.2 + x = 0.4*12 + 0.4x,

1.2 + x = 4.8 + 0.4x,

x - 0.4x = 4.8 - 1.2

0.6x = 3.6,

x =  = 6.

Check:   =  =  = 0.4.   OK, check is done.

Answer. 6 gallons of pure antifreeze is needed.

Problem 2

Let "v" be the volume  of water to be added, in liters.

The amount of the antifreeze in 26 liters of the 15% solution is 0.15*26 = 3.9 liters.
The amount of the antifreeze will not change after adding the water, but the volume of the solution will change and will get (26+v) liters.

The percentage equation for the antifreeze content after adding v liters of water is

 = .

To solve it, multiply both sides by (v+26) and simplify:

3.9 = 0.12*(v + 26),


3.9 = 0.12v + 3.12,

0.12v = 0.78,

v =  = 6.5.

So, 6.5 liters of water must be added.

Check:   =  =  = 0.12.   OK, check is done.

Answer.  6.5 liters of water must be added.

Problem 3

Let "v" be the amount (the volume) of the 50% antifreeze solution needed, in gallons.
Then the amount (the volume) of the 40% antifreeze solution needed is (50-v) gallons.

v gallons of the 50% antifreeze solution will contribute  0.5*v  gallons of the pure antifreeze to the resulting solution.
(50-v) gallons of the 40% antifreeze solution will contribute  0.4*(50-v)  gallons of the pure antifreeze to the resulting solution.
In all, the resulting solution of 50 gallons will contain  0.5*v + 0.4*(50-v)  gallons of the pure antifreeze.

The percentage equation is

 = 0.46.    (1)

This is your equation to determine the unknown value of v.

To solve (1), first multiply both sides by 50 to rid of the denominator. You will get 

(0.5*v + 0.4*(50-v) = 0.46*50  --->  0.5v + 20 - 0.4v = 23  --->   0.1v = 23 - 20  --->  0.1.v = 3  --->  v =  = 30.

So, 30 gallons of the 50% antifreeze solution must be mixed with 50-30 = 20 gallons of the 40%  antifreeze solution.

Check:   =  =  =  =  = 0.46.   OK,  check is done.

Answer.  30 gallons of the 50% antifreeze solution must be mixed with 20 gallons of the 40%  antifreeze solution.

Problem 4

Let "x" be (an unknown) volume of the 30% antifreeze solution which need to be drained out and replaced, 
in accordance with the condition.

After draining, you will have (10-x) quarts of the antifreeze solution in the radiator 
(assuming that initially the radiator was full and contained 10 quarts of the solution).

The amount of the pure antifreeze in the (10-x) quarts of the 30% solution is 0.3*(10-x).

After replacing of "x" quarts by the pure antifreeze by "x" quarts of the pure antifreeze 
the amount of the pure antifreeze in the radiator is  0.3*(10-x) + x quarts.

To get the antifreeze concentration of 65%, the volume "x" should satisfy the equation

 = .   (1)

First step to solve this equation is to multiply both sides by 10 to rid of the denominator. You will get

0.3(10-x) + x = 6.5,   or

3 - 0.3x + x = 6.5  --->  0.7x = 6.5 - 3  --->  0.7x = 3.5  --->  x = 5.

Answer.  The amount of the 65% antifreeze solution that needs to be drained and replaced by the pure antifreeze is 5 quarts.