Lesson More Mixture problems

More Mixture problems

This lesson is a continuation and an extension of the previous lesson Mixture problems of this module.
Here we consider the problems similar to this: "there are two solutions (two liquids of the same solute and the same solvent) with known concentrations.
How much the solutions of these two should be mixed to get the certain volume of the solution with the given concentration?"
The way to solve these problems is to reduce them to the system of two linear equations in two unknowns, and then to solve this system.

The Mixture problems that are presented in the lesson Mixture problems of this module are of another type.
That problems are solved using reduction to one linear equation in one unknown.

Problem 1. A Solution Mixture

A chemist needs to mix a 12% acid solution with a 32% acid solution to obtain an 8-liters mixture consisting of 20% acid.
How many liters of each of the acid solutions must be used?

(Concentrations here are volume concentrations, measured in [volume/volume] units).

Solution
Let us denote as liters a volume of the 12% acid solution to be used in the mixture, and
as liters a volume of the 32% acid solution to be used in the mixture.

Since liters of the 12% acid solution are mixed with liters of the 32% acid solution, the total volume of the mixture is liters.
According to the problem condition, this volume should be equal to 8 liters. This gives us the first equation
.

Further, liters of the 12% acid solution contains liters of the pure acid, and liters of the 32% acid solution contains liters of the pure acid.
Therefore, the volume of the pure acid in the liters of the mixture is equal to liters.
According to the problem condition, this volume is equal to 20% of 8 liters. This gives us the second equation
.

Thus, we have the system of two linear equations in two unknowns
.

To simplify the solution, multiply the second equation by 100. You get
.

We'll solve the last system of linear equations by the elimination method (see the lesson Solution of the linear system of two equations with two unknowns by the Elimination method) in this site.

To eliminate the variable , multiply both sides of the first equation by 12. You get
.

Now subtract the first equation from the second one. You get
,
.

Thus, the solution for is liters.

To calculate , simply substitute this value of into the very first equation. You get
.

Thus, the solution for is liters.

To check this solution, substitute the found values of and into the second equation of the system. You get
, which is exactly equal to 20% of 8 liters.

Answer. 4.8 liters of the 12% acid solution and 3.2 liters of the 32% acid solution must be used to obtain the 8-liters mixture of 20% acid.

Problem 2. Alloy

A metallurgist needs to make 2 lbs of alloy containing 50% silver.
He is going to melt and combine one metal, which is 60% silver, with another metal, which is 40% silver.
How much of each metal should he use?

(Percentages here represent [weight/weight] units or [mass/mass] units).

Solution
Let us denote as lbs a weight of the first metal (containing 60% silver) to be used in the alloy, and
as lbs a weight of the second metal (containing 40% silver).

The total weight of the new alloy is lbs.
According to the problem condition, it should be equal to 2 lbs. This gives us the first equation
.

Further, lbs of the first metal contains lbs of silver, and lbs of the second metal contains lbs of silver.
Therefore, the new alloy contains lbs of silver.
According to the problem condition, this amount of silver should be equal to 50% of 2 lbs, or 0.5*2 = 1 lb. This gives us the second equation
.

Thus, we have the system of two linear equations in two unknowns
.

To solve the system, multiply the first equation by 0.6. You get
.

Now subtract the second equation from the first one. You get
.

Subdivide both sides of this equation by 0.2. You get
.

Thus, the solution for is lbs.

To calculate , simply substitute this value of into the very first equation. You get
.

To check this solution, substitute the found values of and into the first and the second equations of the system. You get
                    for the first equation, and
  for the second equation,
which exactly satisfies the problem conditions.

Answer. The metallurgist should use 1 lb of the first metal and 1 lb of the second metal.

Problem 3. Tickets

Tickets to the local school play cost $5.00 for adults and $2.00 for children.
For the one night show 285 tickets were sold and $1065 was collected.
How many of each type of ticket were sold?

Solution
Let us denote as the numbers of tickets sold for adults, and
as the numbers of tickets sold for children.

Since totally 325 tickets were sold, the first equation is
.

The second equation is for the amount of collected money:
.

Thus, we have the system of two linear equations in two unknowns
.

To solve the system, multiply the first equation by 2. You get
.

Now subtract the first equation from the second one. You get
.

Subdivide both sides of this equation by 3. You get
.

Thus, the solution for is , which means that 165 tickets for adults were sold.

To calculate , substitute this value of into the very first equation. You get
,
.

To check this solution, substitute the found values of and into the first and the second equations of the system. You get
               for the first equation, and
  for the second equation,
which exactly satisfies the problem conditions.

Answer. 165 tickets for adults and 120 tickets for children were sold.

Problem 4. Chickens and rabbits

There are chickens and rabbits at a local farm.
There is a total of 205 heads and 676 legs.
How many chickens and how many rabbits are there at the farm?

Solution
Let us denote as the number of chickens and
as the number of rabbits at the farm.

Counting the heads, you get the first equation
.

Counting the legs, you get the second equation
.

Thus, you have the system of two linear equations in two unknowns
.

To solve the system, multiply the first equation by 2. You get
.

Now subtract the first equation from the second one. You get
.

Subdivide both sides of this equation by 2. You get
.

Thus, the solution for is , which means that there are 133 rabbits at the farm.

To calculate , substitute the found value of into the very first equation. You get
,
, which means that there are 72 chickens at the farm.

To check this solution, substitute the found values of and into the first and the second equations of the system. You get
                            for the first equation, and
  for the second equation,
which exactly satisfies the problem conditions.

Answer. 72 chickens and 133 rabbits are at the farm.

Problem 5. Investment

John invested a total of $12000 last year, part at 4% and the rest at 2%.
One year later he received $400 of interest.
How many did he invest at each rate?

Solution
Let us denote as the amount John invested at 4% and
as the amount John invested at 2%.

Then the first equation is
.

The interest John received from the first investment was .
The interest John received from the second investment was .

Thus, the second equation is
.

Thus, we have the system of two linear equations in two unknowns
.

To solve the system, multiply the first equation by 0.02. You get
.

Now subtract the first equation from the second one. You get
.

Subdivide both sides of this equation by 0.02. You get
.

So, the solution for is , which means that John invested $8000 at 4%.

To calculate , substitute the found value of into the very first equation. You get
,
, which means that John invested $4000 at 2%.

To check this solution, substitute the found values of and into the first and the second equations of the system. You get
                                            for the first equation, and
  for the second equation,
which exactly satisfies the problem conditions.

Answer. John invested $8000 at 4% and $4000 at 2%.