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how many liters of 40% antifreeze solution must be mixed with 8 liters 70% antifreeze solution to get a 50% antifreeze?

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3 Answers

Let x equal the number of liters of 40% solution, because that is the unknown value we want to find.

We can write the equation as follows (I'm writing the percentages as decimal values - e.g. 40% = 0.4)

(x)(0.4) + (8)(0.7) = (x+8)(0.5)

The total number of liters of solution (x+8) is the sum of the two individual amounts (x liters of 40% and 8 liters of 70%)

Multiply the values together and distribute the 0.5 into the parentheses on the right:

0.4x + 5.6 = 0.5x + 4.0

Subtract 0.4x from both sides to get all the terms with an x on the right side:

0.4x - 0.4x + 5.6 = 0.5x - 0.4x + 4.0

5.6 = 0.1x + 4.0

Subtract 4.0 from both sides to get all the constant terms on the left side:

5.6 - 4.0 = 0.1x + 4.0 - 4.0

1.6 = 0.1x

Divide both sides by 0.1 to get x by itself:

1.6/0.1 = (0.1x)/0.1

16 = x

So you need 16 liters of the 40% solution.

Comments

Use the method of Alligation:

    40%            70%     Write down the concentration of each "component" of the mixture

          \            /

            50%                 Write down the concentration of the resuling solution

         /               \

    20 %            10%     Go across,find the difference of the values of components/result

(70-50 = 20)     (50 - 40 =10)

Then the 40% AF is 2 times (=20/10) the 70 % AF so the answer is 16 Litres.

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

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