Lesson More Mixture problems

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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 x+%2B+y

liters.
According to the problem condition, this volume should be equal to 8 liters. This gives us the first equation
x+%2B+y+=+8

.

Further,

liters of the 12% acid solution contains 0.12%2Ax

liters of the pure acid, and

liters of the 32% acid solution contains 0.32%2Ay

liters of the pure acid.
Therefore, the volume of the pure acid in the x+%2B+y

liters of the mixture is equal to 0.12%2Ax+%2B+0.32%2Ay

liters.
According to the problem condition, this volume is equal to 20% of 8 liters. This gives us the second equation
0.12%2Ax+%2B+0.32%2Ay+=+0.2%2A8+=+1.6

.

Thus, we have the system of two linear equations in two unknowns
system%28x+%2B++++++y+=+8%2C%0D%0A+++++0.12%2Ax+%2B+0.32%2Ay+=+1.6%0D%0A%29

.

To simplify the solution, multiply the second equation by 100. You get
system%28x+%2B++++y+=+8%2C%0D%0A+++++++12%2Ax+%2B+32%2Ay+=+160%0D%0A%29

.

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
system%2812%2Ax+%2B+12%2Ay+=+96%2C%0D%0A++++++++++12%2Ax+%2B+32%2Ay+=+160%0D%0A%29

.

Now subtract the first equation from the second one. You get
20%2Ay+=+160-96+=+64

.

Thus, the solution for

liters.

To calculate

, simply substitute this value of

into the very first equation. You get
x+=+8+-+3.2+=+4.8

.

Thus, the solution for

liters.

To check this solution, substitute the found values of

and

into the second equation of the system. You get
0.12%2Ax+%2B+0.32%2Ay+=+0.12%2A4.8+%2B+0.32%2A3.2+=+1.6

, 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 x+%2B+y

lbs.
According to the problem condition, it should be equal to 2 lbs. This gives us the first equation
x+%2B+y+=+2

.

Further,

lbs of the first metal contains 0.6%2Ax

lbs of silver, and

lbs of the second metal contains 0.4%2Ay

lbs of silver.
Therefore, the new alloy contains 0.6%2Ax+%2B+0.4%2Ay

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
0.6%2Ax+%2B+0.4%2Ay+=+1

.

Thus, we have the system of two linear equations in two unknowns
system%28x+%2B+++++y+=+2%2C%0D%0A++++++0.6%2Ax+%2B+0.4%2Ay+=+1%0D%0A%29

.

To solve the system, multiply the first equation by 0.6. You get
system%280.6%2Ax+%2B+0.6%2Ay+=+1.2%2C%0D%0A++++++++++0.6%2Ax+%2B+0.4%2Ay+=+1%0D%0A%29

.

Now subtract the second equation from the first one. You get
0.2%2Ay+=+1.2-1+=+0.2

.

Subdivide both sides of this equation by 0.2. You get
y+=+0.2%2F0.2+=+1

.

Thus, the solution for

lbs.

To calculate

, simply substitute this value of

into the very first equation. You get
x+=+2+-+1+=+1

.

To check this solution, substitute the found values of

and

into the first and the second equations of the system. You get
x++++%2B++++++y+=+1+++%2B+1+=+2

for the first equation, and
0.6%2Ax+%2B+0.4%2Ay+=+0.6+%2B+0.4+=+1

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
x+%2B+y+=+285

.

The second equation is for the amount of collected money:
5x+%2B+2y+=+1065

.

Thus, we have the system of two linear equations in two unknowns
system%28x+%2B+++y+=+285%2C%0D%0A++++++++5%2Ax+%2B+2%2Ay+=+1065%0D%0A%29

.

To solve the system, multiply the first equation by 2. You get
system%282%2Ax+%2B+2%2Ay+=+570%2C%0D%0A++++++++++5%2Ax+%2B+2%2Ay+=+1065%0D%0A%29

.

Now subtract the first equation from the second one. You get
3%2Ax+=+1065-570+=+495

.

Subdivide both sides of this equation by 3. You get
x+=+495%2F3+=+165

.

Thus, the solution for

, which means that 165 tickets for adults were sold.

To calculate

, substitute this value of

into the very first equation. You get
165+%2B+y+=+285

.

To check this solution, substitute the found values of

and

into the first and the second equations of the system. You get
x+++%2B+++y+=+165+++%2B+120+=+285

for the first equation, and
5%2Ax+%2B+2%2Ay+=+5%2A165+%2B+2%2A120+=+1065

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
x+%2B+y+=+205

.

Counting the legs, you get the second equation
2x+%2B+4y+=+676

.

Thus, you have the system of two linear equations in two unknowns
system%28x+%2B+++y+=+205%2C%0D%0A++++++++2%2Ax+%2B+4%2Ay+=+676%0D%0A%29

.

To solve the system, multiply the first equation by 2. You get
system%282%2Ax+%2B+2%2Ay+=+410%2C%0D%0A++++++++++2%2Ax+%2B+4%2Ay+=+676%0D%0A%29

.

Now subtract the first equation from the second one. You get
2%2Ay+=+676-410+=+266

.

Subdivide both sides of this equation by 2. You get
y+=+266%2F2+=+133

.

Thus, the solution for

, which means that there are 133 rabbits at the farm.

To calculate

, substitute the found value of

into the very first equation. You get
x+%2B+133+=+205

, 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
x+++%2B+++y+=+72+++%2B+133+=+205

for the first equation, and
2%2Ax+%2B+4%2Ay+=+2%2A72+%2B+4%2A133+=+144+%2B+532+=+676

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
x+%2B+y+=+12000

.

The interest John received from the first investment was 0.04%2Ax

.
The interest John received from the second investment was 0.02%2Ay

.

Thus, the second equation is
0.04%2Ax+%2B+0.02%2Ay+=+12000

.

Thus, we have the system of two linear equations in two unknowns
system%28x+%2B++++++y+=+12000%2C%0D%0A+++++0.04%2Ax+%2B+0.02%2Ay+=+400%0D%0A%29

.

To solve the system, multiply the first equation by 0.02. You get
system%280.02%2Ax+%2B+0.02%2Ay+=+240%2C%0D%0A++++++++++0.04%2Ax+%2B+0.02%2Ay+=+400%0D%0A%29

.

Now subtract the first equation from the second one. You get
0.02%2Ax+=+400-240+=+160

.

Subdivide both sides of this equation by 0.02. You get
x+=+160%2F0.02+=+8000

.

So, the solution for

, which means that John invested $8000 at 4%.

To calculate

, substitute the found value of

into the very first equation. You get
8000+%2B+y+=+12000

, 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
x+++%2B+++y+=+8000+++%2B+4000+=+12000

for the first equation, and
0.04%2Ax+%2B+0.02%2Ay+=+0.04%2A8000+%2B+0.02%2A4000+=+320+%2B+80+=+4000

  for the second equation,
which exactly satisfies the problem conditions.

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

My other lessons on word problems for mixtures in this site are
    - Mixture problems
    - Solving typical word problems on mixtures for solutions
    - Word problems on mixtures for antifreeze solutions
    - Word problems on mixtures for dry substances like coffee beans, nuts, cashew and peanuts
    - Word problems on mixtures for dry substances like candies, dried fruits
    - Word problems on mixtures for dry substances like soil and sand
    - Word problems on mixtures for alloys
    - Typical word problems on mixtures from the archive
    - Advanced mixture problems
    - Advanced mixture problem for three alloys
    - Unusual word problem on mixtures
    - Check if you know the basics of mixtures from Science

    - Special type mixture problems on DILUTION adding water
    - Increasing concentration of an acid solution by adding pure acid
    - Increasing concentration of alcohol solution by adding pure alcohol
    - How many kilograms of sand must be added to a mixture of sand and cement
    - Draining-replacing mixture problems
    - How much water must be evaporated
    - Advanced problems on draining and replacing
    - Using effective methodology to solve many-steps dilution problems

    - Entertainment problems on mixtures

    - OVERVIEW of lessons on word problems for mixtures

Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.

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