What volume of antifreeze must be added to 4 liters of water to produce a mixture that is 60% anti freeze?

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You have to think:

What we want?

50% antifreeze solution.

In other words

“to obtain final solution that contains an equal volume of antifreeze and water”

We can translate this to an equation:  

 Vf,anti =Vf,water

We have two different initial solutions that will be mixed:

 8 litters with 70 % of antifeeze

and

 X liters with 40 % of antifreeze.    (X is the variable we want to calculate)

translating to an equations we have:

for the 8 liter:

Vi,ant i= 8  x  0.7 = 5.6     liters of antifreeze and

Vi,water = 8  x  0.3 =2.4    liters of Water   

for the 40%

Vadd, anti  = X  x  0.4        liters of antifreeze 

Vadd, water =  X  x  0.6         liters of water

The final total volume of water and antifreeze that we have afer mixter is 

  Vf,ant i=  Vadd,anti +  Vi,anti

 Vf,water = Vadd,water + Vi,water

sbstutuing the  variables

Vf,anti = 8 x 0.7 + X x 0.4

Vf,water= 8 x 0.3 + X  x0.6

So now we determinated the Variables os the first equation:

 subtituing the variable in the first equation we have:

8 x 0.7 + X x 0.4 = 8 x 0.3 + X x 0.6

Isolate the X variable.

Xx0.4 - Xx0.6=8*0.3-8x0.7

X(0.4-0.6)= 2.4-5.6

Multiplying the equation by -1

0.2X=3.2

X=3.2/0.2

X=16 liters

Are you asking for 4 Liters of pure (100%) antifreeze? Here is a lesson which fits your (this) form:

Two-Part Mixture: One material Quantity Unknown