MIXTURE PROBLEMS

  For A Detailed Explantion of a Mixture Problem Click HERE

 

Problem:

A chemistry experiment calls for a 30% sulfuric acid solution. If the lab supply room has only 50% and 20% sulfuric acid solutions on hand, how much of each should be mixed to obtain 12 liters of a 30% solution?

 

Solution:

Let x be the number of liters of the 50% solution to be used.

Let y be the number of liters of the 20% solution to be used.

Note that x + y = 12 because we want a total of 12 liters. We can use this equation anytime we want to for now for simplicity lets keep using y.

12 liters which is 30% sulfuric acid contains (.3)(12) liters of sulfuric acid.

x liters which is 50% sulfuric acid contains (.5)x liters of sulfuric acid.

y liters which is 20% sulfuric acid contains (.2)y liters of sulfuric acid.

Note we will mix x liters with y liters to obtain 12 liters.

This completes the analysis of the problem and we are now in a position to model the mixture problem with an equation.

(.3)(12) = (.5)x + (.2)y .

Now replace y with 12 - x to obtain

(.3)(12) = (.5)x + (.2)(12 x)
as the mathematical model for this mixture problem.

Now we need only solve this equation for x and then calculate y

Multiply both sides by 10 and do some of the multiplications and additions to obtain

36 = 5x + 24 2x 12 = 3x x = 4 and then it follows that y = 8.

 

Problem:

How many gallons of a 3% salt solution must be mixed with 50 gallons of a 7% solution to obtain a 5% solution?

 

Solution:

Let x be the number of gallons of the 3% solution to be added to the 50 gallons of 7% solution.

Note that the result will be (x + 50) gallons.

The desired amount of salt in this (x + 50) gallons is 5% of (x + 50) or (.05)(x + 50)

The amount of salt contributed by the two components of the mixture will be:

3% of x gallons or (.03)x and 7% of 50 gallons or (.07)(50)

The amount of salt contributed by the two components must equal the desired amount of salt in the final mixture.

This completes the analysis of the problem and we are now in a position to model the mixture problem with an equation:

(.05)(x + 50) = (.03)x + (.07)(50)
as the mathematical model for this mixture problem.

Now we simply solve this equation. Begin by multiplying both sides of the equation by 100 to get rid of the decimals

(.05)(x + 50) = (.03)x + (.07)(50) 5(x + 50) = 3x +(7)(50) 5x + 250 = 3x + 350 2x = 100 x = 50


Problem:

To make low fat cottage cheese, milk containing 4% butterfat is mixed with 10 gallons of milk containing 1% butterfat to obtain a mixture containing 2% butterfat. How many gallons of the richer milk is used.

 

Solution:

Let x be the number of gallons of 4% milk to be used. The final mixture will then be (10 + x) gallons and we want it to contain 2% butterfat.

The amount of butterfat in the final solution must be 2% of (10 + x) gallons or (.02)(10 + x)

The amount of butterfat contributed by the 10 gallons of 1% milk is (.01)(10)

The amount of butterfat contributed by the x gallons of 4% milk is (.04)(x)

This completes the analysis of the problem and we are now in a position to model the mixture problem with an equation

(.02)(10 + x) = (.01)(10) + (.04)(x)
as the mathematical model for this mixture problem.

Now we simply solve this equation. Begin by multiplying both sides of the equation by 100 to get rid of the decimals

(.02)(10 + x) = (.01)(10) + (.04)(x) 2(10 + x) = 1(10) + 4x 20 + 2x = 10 + 4x 10 = 2x x = 5

 


Problem:

A 100% concentrate is to be mixed with a mixture having a concentration of 40% to obtain 55 gallons of a mixture with a concentration of 75%. How much of the 100% concentrate will be needed?

 

Solution:

Let x be the amount of the 100% concentrate to be used.

Then 55 x is the amount of the 40% solution to be used.

The amount of concentrate contributed by the 100% solution is 100% of x or simply x.

The amount of concentrate contributed by the 40% solution is 40% of (55 x) or simply (.4)(55 x).

The total amount of concentrate is the sum of these two contributions, so the total is x + (.4)(55 x).

However, the total amount of concentrate in the final solution is required to be 75% of 55 gallons or (.75)(55).

This completes the analysis of the problem and we are now in a position to model the mixture problem with an equation

x + (.4)(55 x) = (.75)(55)
as the mathematical model for this mixture problem.

Now we solve this equation. Begin by multiplying both sides of the equation by 100 to get rid of the decimals.

x + (.4)(55 x) = (.75)(55) 100x + (40)(55 x) = (75)(55) 100x + 2200 40x = 4125 60x = 1925 x = 32.08 gallons


Problem:

A grocer mixes peanuts that cost $2.49 per pound and walnuts that cost $3.89 per pound to make 100 pounds of a mixture that costs $3.19 per pound. How much of each kind of nut is put into the mixture?

 

Solution:

Let x be the amount of peanuts to be put into the mixture.

Then 100 x is the amount of walnuts put into the mixture.

The cost of the peanuts in the mixture is 2.49x

The cost of the walnuts in the mixture is 3.89(100 x)

The total cost contributed by the peanuts and walnuts is 2.49x + 3.89(100 x).

However, the total cost of the final mixture is required to be (3.19)(100) or 319.

This completes the analysis of the problem and we are now in a position to model the mixture problem with an equation

2.49x + 3.89(100 x) = 319
as the mathematical model for this mixture problem.

Now we solve this equation. Begin by multiplying both sides of the equation by 100 to get rid of the decimals.

2.49x +3.89(100 x) = 319 249x + 389(100 x) = 31900 249x + 38900 389x =31900

-140x = - 7000 x = 50.

So 50 pounds of peanuts and 50 pounds of walnuts must be used to obtain 100 pounds costing $3.19 per pound.


Problem:

A forester mixes gasoline and oil to make 2 gallons of mixture for his two-cycle chain-saw engine. This mixture is 32 parts gasoline and 1 part two-cycle oil. How much gasoline must be added to bring the mixture to 40 parts gasoline and 1 part oil?

 

Solution:

Let x be the amount of gasoline to be added.

The total amount of mixture will then be x + 2 gallons.

The concept of 32 parts gasoline and 1 part oil in the original mixture, quite naturally suggests that we should consider each gallon as being divided into 33 equal parts. The total amount of gasoline in the two gallons of the original mixture is then 32 + 32 = 64 parts.

When we add x gallons of gasoline to the mixture we will have gallons of gasoline in the final mixture.

This completes the analysis of the problem and we are now in a position to model the mixture problem with an equation

 

as the mathematical model for this mixture problem.

Now we solve this equation.

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