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Solving Mixture Problems

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Mixture problems involve combining two or more things and determining some characteristic of either the ingredients or the resulting mixture. For example, we might want to know how much water to add to dilute a saline solution, or we might want to determine the percentage of concentrate in a jug of orange juice.

Introduction to Mixtures

Using a table to problem solve, practice problems.

We can use fractions, ratios, or percentages to describe quantities in mixtures.

If 200 grams of a saline solution has 40 grams of salt, what percentage of the solution is salt? Answer $\frac{40}{200} = 0.20 = 20\%$ of the solution is salt.

If each can of orange juice concentrate contains the same amount of concentrate, which recipe will make the drink that is most orangey?

There is a general strategy for solving these mixture problems that uses simple algebra organized with a chart.

How much 40% rubbing alcohol do we need to add to 90% rubbing alcohol to make a 50% solution of rubbing alcohol? We could organize the data we are given in the following chart: Solution Type Concentration Amount of Solution Amount of Pure Alcohol 40% rubbing alcohol 0.4 ? liters ? liters 90% rubbing alcohol 0.9 ? liters ? liters 50% rubbing alcohol 0.5 10 liters $0.5(10) = 5$ liters In general, the rows of the chart are the mixture types that you have. The columns describe the amount of each compound you have and the concentration of that component in each mixture (represented as a decimal). When you don't have some of the information needed to fill in the cart, use a variable instead. Solution Type Concentration Amount of Solution Amount of Pure Compound 40% rubbing alcohol 0.4 $x$ $0.4(x)$ 90% rubbing alcohol 0.9 $10-x$ $0.9(10-x)$ 50% rubbing alcohol 0.5 10 $0.5(10) = 5$ The amount of 40% solution that we'll need is unknown (so make it $x$). The amount of 90% solution that we'll need is also unknown, but must be $10-x$ liters so that we'll, in total, make 10 liters of the final solution. The amount of alcohol that each part of the mixture adds to the final result is equal to the amount of each solution mixed in, times the fraction of alcohol that that solution is made from. To use this chart to solve the problem, we will use the fourth column as an equation to solve for $x.$ The 10 liters of our final mixture must have a total volume of 5 liters of alcohol in it in order to be 50% alcohol. Those 5 liters must come from a combination of the amount of 40% solution we mix in and the amount of 90% solution. If the volume (in liters) of 40% solution that we mix in is $x,$ then $0.4x$ will be the volume (in liters) of the amount of alcohol contributed. Similarly, $0.9(10-x)$ will be the amount of alcohol contributed by the $10-x$ liters of 90% alcohol solution that we add. Therefore, in total, $0.4x + 0.9(10-x)$ must be equal to the 5 total liters we'll need in the final solution to make it 50% alcohol. Solving the equation, \[\begin{align} 0.4x + (9 - 0.9x) &= 5\\ -0.5x + 9 &= 5\\ 0.5x &= 4\\ \Rightarrow x &= \frac{4}{0.5}= 8. \end{align}\] Therefore, we need $x = 8$ liters of the 40% solution.

How many grams of pure water must be added to 40 grams of a 10% saline solution to make a saline solution that is 5% salt? Answer Let's set up a table to solve this problem, using $x$ to represent the number of grams of water that must be added. Solution Type Concentration Amount of Solution Total Amount of Salt water 0 $x$ 0 10% solution 0.1 40 $(0.1)(40)$ 5% solution 0.05 40+$x$ $(0.05)(40+x)$ Because the total amount of salt remains the same after we add the water, we can set up and solve the following equation: \[\begin{align} (0.1)(40)&=(0.05)(40+x) \\ 4 &= 2 + 0.05x \\ x&=40.\end{align}\] We need to add $40 $ grams of pure water to create the solution that is $5\%$ salt.

I have a 100 ml mixture that is 20% isopropyl alcohol (80 ml of water and 20 ml of isopropyl alcohol).

How much more alcohol do I need to add to make the mixture 25% alcohol?

There is a 40 litre solution of milk and water in which the concentration of milk is 72%. How much water must be added to this solution to make it a solution in which the concentration of milk is 60% ?

Let's practice using the strategies from above on a variety of problems.

Jill mixes 100 liters of A-beverage that contains 45% juice with 200 liters of B-beverage. The resulting C-beverage is 30% fruit juice. What is the percent of fruit juice in the 200 liters of the B-beverage? Answer Let's begin by making a table to show what we know. Beverage Liters of Beverage Concentration Liters of Juice Beverage A 100 0.45 $(0.45)(100) = 45$ Beverage B 200 $x$ $200x$ Beverage C 300 0.30 $(0.30)(300)=90$ The total number of liters of juice in Beverages A and B must equal 90, so \[\begin{align} 45 + 200x &= 90 \\ 200x &= 45 \\ 0.225 &= x.\end{align}\] The concentration of juice in Beverage B is 22.5%.

Unequal amounts of 40% and 10% acid solutions were mixed and the resulting mixture was 30% concentrated. However, the required concentration is 25%, so the Chemist added 300 cubic meters of 20% acid solution in order to get the required concentration. What was the original amount of 40% acid solution?

Write only the quantity without the units (cubic meters).

Legend: wt = weight

Strawberries contain about 15 wt% solids and 85 wt% water. To make strawberry jam, crushed strawberries and sugar are mixed in a 45:55 mass ratio, and the mixture is heated to evaporate water until the residue contains one-third water by mass.

Question: 1.) Calculate how many pounds of strawberries are needed to make a pound of jam.

In a mixture of 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:

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Examples of Setting Up "Mixture" Word Problems

Explanation Examples

The solutions to mixture problems tend to be generated by linear equations; that is, solving the set-up for mixture exercises tends to be pretty easy. It's the set-up that can be a bear.

What follows are some examples showing that set-up.

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Mixture Word Problems

These examples demonstrate how to take the information you've been given, and how to use that to pick variables, create a table to organize everything, and then create the equations. In the exercises below are a few more problems, together with their grids (but not their solutions; I've left that part for you to do).

How many liters of a 70% alcohol solution must be added to 50 liters of a 40% alcohol solution to produce a 50% alcohol solution?

You will be mixing a stronger solution with a weaker solution. The number of liters of the stronger solution is currently unknown; I'll use s to stand for the number of liters of the stronger solution. Then the number of liters of the mixture you're making will be s + 50 .

Make a table, with the columns showing the numbers of liters of solutions, the percentage of alcohol in each solution (this is your "rate"), and the number of liters of actual alcohol in each solution.

From the last column, you get the equation:

0.7 s + 20 = 0.50( s + 50)

Solve for the value of the variable. Remember to put appropriate units (in this case, "liters") on your hand-in answer .

Forty percent is closer to 50% than is 70% . So you would logically expect that you'll end up using fewer liters of 70% solution than the 50 liters of 40% solution.

How many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt?

I'll use w to stand for the number of ounces of water that are needed. And, since pure water has no salt in it, then the percentage of water that is salt is zero, and the number of ounces of salt is also zero.

7.5 = 0.10(50 + w )

Solve for the value of w . Remember to put the appropriate units (in this case, "ounces") on your hand-in answer .

Note: This exercise is typical of algebra courses, and is used to teach problem-solving skills in algebra classes. But, in real life, the actual process is different. ( Example ) If you're in a chemistry lab, expect the measuring and mixing to work in a different way.

Find the selling price per pound of a coffee mixture made from 8 pounds of coffee that sells for $9.20 per pound and 12 pounds of coffee that costs $5.50 per pound.

This exercise asks us to make the implicit assumption that the selling price of the mix is based only on the selling price (and amounts) of the inputs. Of course, in real life, the selling price of the mix would be a markup on the cost of the mix, and the cost of the mix would be related to the costs of the inputs, plus the extra costs involved in mixing and re-bagging. But this is algebra, not real life.

The price per pound is the "rate" for this exercise. The sum of the prices of the inputs is assumed to be equal to the total price for the mixture.

From the last row, you see that you have 20 pounds of coffee mixture. This mixture will sell for $139.60 . To find the selling price per pound of the mixture, divide ( $139.60 ) by ( 20 pounds). Simplify the division to find the unit rate.

Remember to put appropriate units (in this case, "dollars per pound") on your hand-in answer.

Note that, in this case, no variable was actually necessary. But if you'd picked a variable to replace the query-mark (that is, the " ? " in the bottom row), this would also have been fine.

How many pounds of lima beans that cost $0.90 per pound must be mixed with 16 pounds of corn that costs $0.50 per pound to make a mixture of vegetables that costs $0.65 per pound?

The cost per pound is the "rate" for this exercise.

You are given the number of pounds of corn, but not the number of pounds of beans. I'll use b to stand for this amount.

The cost of the inputs add to the cost of the mix, which (from the far right column) gives the equation:

0.90 b + 8 = 0.65( b + 16)

Solve for the value of the variable. Remember to put the appropriate units (in this case, "pounds") on your hand-in answer .

Two hundred liters of a punch that contains 35% fruit juice is mixed with 300 liters (L) of another punch. The resulting fruit punch is 20% fruit juice. Find the percent of fruit juice in the 300 liters of punch.

The percentage of the punch that comes from actual fruit is the "rate" for this exercise. Since the exercise is asking for a percentage, I will use the variable f .

The sum of the input amounts of juice will equal the total amount of juice in the mixture. You can use the last column to create the equation:

70 + 300 p = 100

Or you can just eyeball the amounts and notice that 300 p must equal 30 . Either way, do the division to find the value of the variable. Remember that you're looking for a percentage, so you'll need to convert the decimal solution into percentage form.

Ten grams of sugar are added to a 40 -g serving of a breakfast cereal that is 30% sugar. What is the percent concentration of sugar in the resulting mixture?

Note that, because sugar is 100% sugar, the percentage of sugar in what is added to the bowl, in decimal form, is 1.00 .

I will use the variable s to stand for the percentage of sugar in the mixture.

From the bottom row, you see that there are 22 grams of sugar in the 50 grams in the bowl, or 22 / 50 . Convert the decimal value to percentage form .

As it turns out in this case, the variable wasn't actually necessary. But there's no harm in developing the habit of defining one.

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Mixture Word Problems - Different Costs or Concentrations

Related Topics: More Lessons for Algebra Math Worksheets Mixture Problem (Adding to the Solution) Mixture Problem (Removing from the Solution) Mixture Problem (Replacing the Solution) Mixture Problem (Mixing Quantities of Different Costs)

Mixture problems are word problems where items or quantities of different values are mixed together.

We recommend using a table to organize your information for mixture problems. Using a table allows you to think of one number at a time instead of trying to handle the whole mixture problem at once.

Mixing Quantities Of Different Costs or Different Concentrations

How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to produce a mixture worth $1.00 a pound?

Set up a table for different types of chocolate.

Mixture Problems Some word problems using systems of equations involve mixing two quantities with different prices. To solve mixture problems, knowledge of solving systems of equations. is necessary. Most often, these problems will have two variables, but more advanced problems have systems of equations with three variables. Other types of word problems using systems of equations include rate word problems and work word problems. Mixing Quantities Of Different Costs Example: At the Grind House, Honduran coffee sells for $8.00 per pound and Costa Rican coffee sells for $11.50 per pound. If one pound of the Central American Blend coffee contains 0.4lb Costa Rican coffee and 0.6lb Honduran coffee, how much should one pound sell for?

Mixing Quantities Of Different Concentration Example: How many liters of a 10% salt solution should be added to 80 liters of a 35% salt solution to obtain a mixture that is a 30% salt solution?

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3.3: Solve Mixture Applications

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Learning Objectives

By the end of this section, you will be able to:

Solve coin word problems
Solve ticket and stamp word problems
Solve mixture word problems
Use the mixture model to solve investment problems using simple interest

Before you get started, take this readiness quiz.

Multiply: $14(0.25)$. If you missed this problem, review Exercise 1.8.19 .
Solve: $0.25x+0.10(x+4)=2.5$. If you missed this problem, review Exercise 2.4.22 .
The number of dimes is three more than the number of quarters. Let q represent the number of quarters. Write an expression for the number of dimes. If you missed this problem, review Exercise 1.3.43 .

Solve Coin Word Problems

In mixture problems , we will have two or more items with different values to combine together. The mixture model is used by grocers and bartenders to make sure they set fair prices for the products they sell. Many other professionals, like chemists, investment bankers, and landscapers also use the mixture model.

Doing the Manipulative Mathematics activity Coin Lab will help you develop a better understanding of mixture word problems.

We will start by looking at an application everyone is familiar with—money!

Imagine that we take a handful of coins from a pocket or purse and place them on a desk. How would we determine the value of that pile of coins? If we can form a step-by-step plan for finding the total value of the coins, it will help us as we begin solving coin word problems.

So what would we do? To get some order to the mess of coins, we could separate the coins into piles according to their value. Quarters would go with quarters, dimes with dimes, nickels with nickels, and so on. To get the total value of all the coins, we would add the total value of each pile.

$Piles of pennies, nickels, dimes, and quarters$

How would we determine the value of each pile? Think about the dime pile—how much is it worth? If we count the number of dimes, we’ll know how many we have—the number of dimes.

But this does not tell us the value of all the dimes. Say we counted 17 dimes, how much are they worth? Each dime is worth $0.10—that is the value of one dime. To find the total value of the pile of 17 dimes, multiply 17 by $0.10 to get $1.70. This is the total value of all 17 dimes. This method leads to the following model.

TOTAL VALUE OF COINS

For the same type of coin, the total value of a number of coins is found by using the model

\[number\cdot value = total\space value\]

where number is the number of coins

value is the value of each coin

total value is the total value of all the coins

The number of dimes times the value of each dime equals the total value of the dimes.

\[\begin{aligned} \text {number.} \cdot \text { value } &=\text { total value } \\ 17 \cdot \$ 0.10 &=\$ 1.70 \end{aligned}\]

We could continue this process for each type of coin, and then we would know the total value of each type of coin. To get the total value of all the coins, add the total value of each type of coin.

Let’s look at a specific case. Suppose there are 14 quarters, 17 dimes, 21 nickels, and 39 pennies.

$This table has five rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Quarters, 14, 0.25, and 3.50. The third row reads Dimes, 17, 0.10, and 1.70. The fourth row reads Nickels, 21, 0.05, and 1.05. The fifth row reads Pennies, 39, 0.01, and 0.39. The extra cell reads 6.64.$

Table $\PageIndex{1}$

The total value of all the coins is $6.64.

Notice how the chart helps organize all the information! Let’s see how we use this method to solve a coin word problem.

Example $\PageIndex{1}$

Adalberto has $2.25 in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?

Step 1. Read the problem. Make sure all the words and ideas are understood.

Label the columns “type,” “number,” “value,” “total value.”
List the types of coins.
Write in the value of each type of coin.
Write in the total value of all the coins.

$This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Dimes, blank, 0.10, and blank. The third row reads Nickels, blank, 0.05, and blank. The extra cell reads 2.25.$

Step 2. Identify what we are looking for.

We are asked to find the number of dimes and nickels Adalberto has.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

Multiply the number times the value to get the total value of each type of coin.

Next we counted the number of each type of coin. In this problem we cannot count each type of coin—that is what you are looking for—but we have a clue. There are nine more nickels than dimes. The number of nickels is nine more than the number of dimes.

\[\begin{aligned} \text { Let } d &=\text { number of dimes. } \\ d+9 &=\text { number of nickels } \end{aligned}\]

Fill in the “number” column in the table to help get everything organized.

$This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Dimes, d, 0.10, and blank. The third row reads Nickels, d plus 9, 0.05, and blank. The extra cell reads 2.25.$

Now we have all the information we need from the problem!

We multiply the number times the value to get the total value of each type of coin. While we do not know the actual number, we do have an expression to represent it.

And so now multiply $\text{number}\cdot\text{value}=\text{total value}$. See how this is done in the table below.

$This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Dimes, d, 0.10, and 0.10d. The third row reads Nickels, d plus 9, 0.05, and 0.05 times the quantity (d plus 9). The extra cell reads 2.25.$

Notice that we made the heading of the table show the model.

Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence. Translate the English sentence into an algebraic equation.

Write the equation by adding the total values of all the types of coins.

$The sentence, “value of dimes plus value of nickels equals total value of coins,” can be translated to an equation. Translate “value of dimes” to 0.10d, translate “value of nickles” to 0.05d, and translate “total value of coins” to 2.25. The full equation is 0.10d plus 0.05 times the quantity d plus 9 equals 2.25.$

Step 5. Solve the equation using good algebra techniques.

Step 6. Check the answer in the problem and make sure it makes sense.

Does this check?

\[\begin{array}{llll}{12 \text { dimes }} & {12(0.10)} &{=} &{1.20} \\ {21 \text { nickels }} & {21(0.05)} & {=} &{\underline{1.05}} \\ {} &{} &{}&{$ 2.25\checkmark} \end{array}\]

Step 7. Answer the question with a complete sentence.

If this were a homework exercise, our work might look like the following.

$A homework assignment written on lined loose leaf paper. The assignment reads: “Adalberto has 2 dollars and 25 cents in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type does he have?” Below this is a table. The first row of the table is a header row, and each cell names the column or columns below it. The first cell from the left is named “Type.” The second cell contains the equation “Number” times “Value” equals “Total Value,” with one column corresponding to “Number,” one column corresponding to “Value,” and one column corresponding to total value. Hence the content of the “Number” column times the content of the “Value” column equals the content of the “Total Value” column. In the second row of the table, the “Type” column contains “Dimes,” the “Number” column contains d, the “Value” column contains 0.10, and the “Total Value” column contains 0.10d. In the third row of the table, the “Type” column contains “Nickels,” the “Number” column contains d plus 9, the “Value column contains 0.05, and the “Total Value” column contains 0.05 times the quantity d plus 9. One row down, the “Total Value” column contains one more cell, which contains 2.25. Below the table is the equation 0.10d plus 0.05d plus 0.45 equals 2.25. Below this is 0.15d plus 0.45 equals 2.25. Below this is 0.15d equals 1.80. To the right is d plus 9, which translates to 12 plus 9, or 21 nickels. To the right of this is the checking stage, where we see if 12 dimes and 21 nickels amount to 2 dollars and 25 cents. 12 times 0.10 equals 1.20, and 21 times (0.05) equals 1.05. 1.20 plus 1.05 equals 2.25.$

Try It $\PageIndex{1}$

Michaela has $2.05 in dimes and nickels in her change purse. She has seven more dimes than nickels. How many coins of each type does she have?

9 nickels, 16 dimes

Try It $\PageIndex{2}$

Liliana has $2.10 in nickels and quarters in her backpack. She has 12 more nickels than quarters. How many coins of each type does she have?

17 nickels, 5 quarters

SOLVE COIN WORD PROBLEMS.

Determine the types of coins involved.

Create a table to organize the information.

Identify what we are looking for.
Use variable expressions to represent the number of each type of coin and write them in the table.
Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the sentence into an equation. Write the equation by adding the total values of all the types of coins.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

Example $\PageIndex{2}$

Maria has $2.43 in quarters and pennies in her wallet. She has twice as many pennies as quarters. How many coins of each type does she have?

Step 1. Read the problem.

We know that Maria has quarters and pennies.

We are looking for the number of quarters and pennies.

Step 3. Name. Represent the number of quarters and pennies using variables.

$No Alt Text$

Step 4. Translate. Write the equation by adding the ‘total value’ of all the types of coins.

$\begin{array} {ll} {\textbf{Step5. Solve} \text{ the equation.}} &{0.25q + 0.01(2q) = 2.43} \\{\text{Multiply.}} &{0.25q + 0.02q = 2.43} \\ {\text{Combine like terms.}} &{0.27q = 2.43} \\ {\text{Divide by 0.27}} &{q = 9 \text{ quarters}} \\ {\text{The number of pennies is 2q.}} &{2q} \\ {} &{2\cdot 9} \\ {} &{18 \text{ pennies}} \\ {\textbf{Step 6. Check} \text{ the answer in the problem.}} &{} \\\\ {\text{Maria has 9 quarters and 18 pennies. Dies this}} &{} \\ {\text{make }$2.43?} &{} \end{array}$ $\begin{array} {llll}\\ {9\text{ quarters }} &{ 9(0.25)} &{=} &{2.25} \\ {18\text{ pennies }} &{18(0.01)} &{=} &{\underline{0.18}} &{}\\ {\text{Total}} &{} &{} &{$2.43\checkmark} \end{array}$ $\begin{array} {ll} \\ {\textbf{Step 7. Answer}\text{ the question.}} &{\text{Maria has nine quarters and eighteen pennies.}} \end{array}$

Try It $\PageIndex{3}$

Sumanta has $4.20 in nickels and dimes in her piggy bank. She has twice as many nickels as dimes. How many coins of each type does she have?

42 nickels, 21 dimes

Try It $\PageIndex{4}$

Alison has three times as many dimes as quarters in her purse. She has $9.35 altogether. How many coins of each type does she have?

51 dimes, 17 quarters

In the next example, we’ll show only the completed table—remember the steps we take to fill in the table.

Example $\PageIndex{3}$

Danny has $2.14 worth of pennies and nickels in his piggy bank. The number of nickels is two more than ten times the number of pennies. How many nickels and how many pennies does Danny have?

Try It $\PageIndex{5}$

Jesse has $6.55 worth of quarters and nickels in his pocket. The number of nickels is five more than two times the number of quarters. How many nickels and how many quarters does Jesse have?

41 nickels, 18 quarters

Try It $\PageIndex{6}$

Elane has $7.00 total in dimes and nickels in her coin jar. The number of dimes that Elane has is seven less than three times the number of nickels. How many of each coin does Elane have?

22 nickels, 59 dimes

Solve Ticket and Stamp Word Problems

Problems involving tickets or stamps are very much like coin problems. Each type of ticket and stamp has a value, just like each type of coin does. So to solve these problems, we will follow the same steps we used to solve coin problems.

Example $\PageIndex{4}$

At a school concert, the total value of tickets sold was $1,506. Student tickets sold for $6 each and adult tickets sold for $9 each. The number of adult tickets sold was five less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?

Determine the types of tickets involved. There are student tickets and adult tickets.

$This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Student, blank, 6, and blank. The third row reads Adult, blank, 9, and blank. The extra cell reads 1506.$

We are looking for the number of student and adult tickets.

Step 3. Name. Represent the number of each type of ticket using variables.

We know the number of adult tickets sold was five less than three times the number of student tickets sold.

$This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Student, s, 6, and 6s. The third row reads Adult, 3s minus 5, 9, and 9 times the quantity (3s minus 5). The extra cell reads 1506.$

Step 4. Translate. Write the equation by adding the total values of each type of ticket.

\[6 s+9(3 s-5)=1506 \nonumber\]

Step 5. Solve the equation.

\[\begin{array}{rcl}{6 s+27 s-45} &{=} &{1506} \\ {33 s-45} &{=} &{1506} \\ {33 s} &{=} &{1551} \\ {s} & {=} &{47 \text { student tickets }} \\ {\text{Number of adult tickets}} &{=} &{3s-5} \\ {} &{=} &{3(47)-5} \\ {\text{So there were}} &{136} &{\text{adult tickets}}\end{array} \nonumber\]

Step 6. Check the answer.

There were 47 student tickets at $6 each and 136 adult tickets at $9 each. Is the total value $1,506? We find the total value of each type of ticket by multiplying the number of tickets times its value then add to get the total value of all the tickets sold.

\[\begin{array}{lll} {47\cdot 6} &{=} &{282} \\ {136\cdot 9} &{=} &{\underline{1224}} \\ {} &{} &{1506\checkmark} \\\end{array} \nonumber\]

Step 7. Answer the question. They sold 47 student tickets and 136 adult tickets.

Try It $\PageIndex{7}$

The first day of a water polo tournament the total value of tickets sold was $17,610. One-day passes sold for $20 and tournament passes sold for $30. The number of tournament passes sold was 37 more than the number of day passes sold. How many day passes and how many tournament passes were sold?

330 day passes, 367 tournament passes

Try It $\PageIndex{8}$

At the movie theater, the total value of tickets sold was $2,612.50. Adult tickets sold for $10 each and senior/child tickets sold for $7.50 each. The number of senior/child tickets sold was 25 less than twice the number of adult tickets sold. How many senior/child tickets and how many adult tickets were sold?

112 adult tickets, 199 senior/child tickets

We have learned how to find the total number of tickets when the number of one type of ticket is based on the number of the other type. Next, we’ll look at an example where we know the total number of tickets and have to figure out how the two types of tickets relate.

Suppose Bianca sold a total of 100 tickets. Each ticket was either an adult ticket or a child ticket. If she sold 20 child tickets, how many adult tickets did she sell?

Did you say ‘80’? How did you figure that out? Did you subtract 20 from 100?

If she sold 45 child tickets, how many adult tickets did she sell?

Did you say ‘55’? How did you find it? By subtracting 45 from 100?

What if she sold 75 child tickets? How many adult tickets did she sell?

The number of adult tickets must be 100−75. She sold 25 adult tickets.

Now, suppose Bianca sold x child tickets. Then how many adult tickets did she sell? To find out, we would follow the same logic we used above. In each case, we subtracted the number of child tickets from 100 to get the number of adult tickets. We now do the same with x .

We have summarized this below.

$This table has five rows and two columns. The top row is a header row that reads from left to right Child tickets and Adult tickets. The second row reads 20 and 80. The third row reads 45 and 55. The fourth row reads 75 and 25. The fifth row reads x and 100 plus x.$

Table $\PageIndex{2}$

We can apply these techniques to other examples

Example $\PageIndex{5}$

Galen sold 810 tickets for his church’s carnival for a total of $2,820. Children’s tickets cost $3 each and adult tickets cost $5 each. How many children’s tickets and how many adult tickets did he sell?

Determine the types of tickets involved. There are children tickets and adult tickets.

$This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Children, blank, 3, and blank. The third row reads Adult, blank, 5, and blank. The extra cell reads 2820.$

We are looking for the number of children and adult tickets.
We know the total number of tickets sold was 810.
This means the number of children’s tickets plus the number of adult tickets must add up to 810.
Let $c$ be the number of children tickets.

$This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Children, c, 3, and 3c. The third row reads Adult, 810 minus c, 5, and 5 times the quantity (810 minus c). The extra cell reads 2820.$

Then $810−c$ is the number of adult tickets.
Multiply the number times the value to get the total value of each type of ticket.

Step 4. Translate.

Write the equation by adding the total values of each type of ticket.

\[\begin{align*} 3 c+5(810-c) &=2,820 \\ 3 c+4,050-5 c &=2,820 \\-2 c &=-1,230 \\ c &=615 \text { children tickets } \end{align*}\]

How many adults?

\[\begin{array}{c}{810-c} \\ {810-615} \\ {195 \text { adult tickets }}\end{array} \nonumber\]

Step 6. Check the answer. There were 615 children’s tickets at $3 each and 195 adult tickets at $5 each. Is the total value $2,820?

\[\begin{array}{rrl}{615 \cdot 3} &{=} & {1845} \\ {195 \cdot 5} &{=} & {\underline{975}} \\ {} &{} &{2,820\checkmark} \end{array} \nonumber\]

Step 7. Answer the question. Galen sold 615 children’s tickets and 195 adult tickets.

Try It $\PageIndex{9}$

During her shift at the museum ticket booth, Leah sold 115 tickets for a total of $1,163. Adult tickets cost $12 and student tickets cost $5. How many adult tickets and how many student tickets did Leah sell?

84 adult tickets, 31 student tickets

Try It $\PageIndex{10}$

A whale-watching ship had 40 paying passengers on board. The total collected from tickets was $1,196. Full-fare passengers paid $32 each and reduced-fare passengers paid $26 each. How many full-fare passengers and how many reduced-fare passengers were on the ship?

26 full-fare, 14 reduced fare

Now, we’ll do one where we fill in the table all at once.

Example $\PageIndex{6}$

Monica paid $8.36 for stamps. The number of 41-cent stamps was four more than twice the number of two-cent stamps. How many 41-cent stamps and how many two-cent stamps did Monica buy?

The types of stamps are 41-cent stamps and two-cent stamps. Their names also give the value!

“The number of 41-cent stamps was four more than twice the number of two-cent stamps.”

\[\begin{array}{l}{\text { Let } x=\text { number of } 2 \text { -cent stamps. }} \\ {2 x+4=\text { number of } 41-\text { cent stamps }}\end{array} \nonumber\]

$This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads 41 cent stamps, 2x plus 4, 0.41, and 0.41 times the quantity (2x plus 4). The third row reads 2 cent stamps, x, 0.02, and 0.02x. The extra cell reads 8.36.$

\[\begin{array}{lr} {\text{Write the equation from the total values.}} &{0.41(2x + 4) + 0.02x = 8.36} \\ {} &{0.82x + 1.64 + 0.02x = 8.36} \\ {} &{0.84x + 1.64 = 8.36} \\ {\text{Solve the equation.}} &{0.84x = 6.72} \\ {} &{x = 8} \\\\ {\text{Monica bought eight two-cent stamps.}} &{} \\{\text{Find the number of 41-cent stamps she bought}} &{2x + 4 \text{ for } x = 8} \\{\text{by evaluating}} &{2x + 4} \\{} &{2(8) + 4} \\ {} &{20} \end{array} \nonumber\]

\[\begin{array} {rll} {8(0.02) + 20(0.41)} &{\stackrel{?}{=}} &{8.36} \\ {0.16 + 8.20} &{\stackrel{?}{=}} &{8.36} \\{8.36} &{=} &{8.46\checkmark} \end{array}\] \[\begin{array} {ll} \\ {} &{\text{Monica bought eight two-cent stamps and 20}} \\ {} &{\text{41-cent stamps}} \end{array} \nonumber\]

Try It $\PageIndex{11}$

Eric paid $13.36 for stamps. The number of 41-cent stamps was eight more than twice the number of two-cent stamps. How many 41-cent stamps and how many two-cent stamps did Eric buy?

32 at $0.41, 12 at $0.02

Try It $\PageIndex{12}$

Kailee paid $12.66 for stamps. The number of 41-cent stamps was four less than three times the number of 20-cent stamps. How many 41-cent stamps and how many 20-cent stamps did Kailee buy?

26 at $0.41, 10 at $0.20

Solve Mixture Word Problems

Now we’ll solve some more general applications of the mixture model. Grocers and bartenders use the mixture model to set a fair price for a product made from mixing two or more ingredients. Financial planners use the mixture model when they invest money in a variety of accounts and want to find the overall interest rate. Landscape designers use the mixture model when they have an assortment of plants and a fixed budget, and event coordinators do the same when choosing appetizers and entrees for a banquet.

Our first mixture word problem will be making trail mix from raisins and nuts.

Example $\PageIndex{7}$

Henning is mixing raisins and nuts to make 10 pounds of trail mix. Raisins cost $2 a pound and nuts cost $6 a pound. If Henning wants his cost for the trail mix to be $5.20 a pound, how many pounds of raisins and how many pounds of nuts should he use?

As before, we fill in a chart to organize our information.

The 10 pounds of trail mix will come from mixing raisins and nuts.

\[\begin{array}{l}{\text { Let } x=\text { number of pounds of raisins. }} \\ {10-x=\text { number of pounds of nuts }}\end{array} \nonumber\]

We enter the price per pound for each item.

We multiply the number times the value to get the total value.

$This table has four rows and four columns. The top row is a header row that reads from left to right Type, Number of pounds, Price per pound ($), and Total Value ($). The second row reads raisins, x, 2, and 2x. The third row reads nuts, 10 minus x, 6, and 6 times the quantity (10 minus x). The fourth row reads trail mix, 10, 5.20, and 10 times 5.20.$

Notice that the last line in the table gives the information for the total amount of the mixture.

We know the value of the raisins plus the value of the nuts will be the value of the trail mix.

Try It $\PageIndex{13}$

Orlando is mixing nuts and cereal squares to make a party mix. Nuts sell for $7 a pound and cereal squares sell for $4 a pound. Orlando wants to make 30 pounds of party mix at a cost of $6.50 a pound, how many pounds of nuts and how many pounds of cereal squares should he use?

5 pounds cereal squares, 25 pounds nuts

Try It $\PageIndex{14}$

Becca wants to mix fruit juice and soda to make a punch. She can buy fruit juice for $3 a gallon and soda for $4 a gallon. If she wants to make 28 gallons of punch at a cost of $3.25 a gallon, how many gallons of fruit juice and how many gallons of soda should she buy?

21 gallons of fruit punch, 7 gallons of soda

We can also use the mixture model to solve investment problems using simple interest . We have used the simple interest formula, $I=Prt$, where $t$ represented the number of years. When we just need to find the interest for one year, $t=1$, so then $I=Pr$.

Example $\PageIndex{8}$

Stacey has $20,000 to invest in two different bank accounts. One account pays interest at 3% per year and the other account pays interest at 5% per year. How much should she invest in each account if she wants to earn 4.5% interest per year on the total amount?

We will fill in a chart to organize our information. We will use the simple interest formula to find the interest earned in the different accounts.

The interest on the mixed investment will come from adding the interest from the account earning 3% and the interest from the account earning 5% to get the total interest on the $20,000.

\[\begin{aligned} \text { Let } x &=\text { amount invested at } 3 \% \\ 20,000-x &=\text { amount invested at } 5 \% \end{aligned}\]

The amount invested is the principal for each account.

We enter the interest rate for each account.

We multiply the amount invested times the rate to get the interest.

$This table has four rows and four columns. The top row is a header row that reads from left to right Type, Amount invested, Rate, and Interest. The second row reads 3%, x, 0.03, and 0.03x. The third row reads 5%, 20,000 minus x, 0.05, and 0.05 times the quantity (20,000 minus x). The fourth row reads 4.5%, 20,000, 0.045, and 0.045 times 20,000.$

Notice that the total amount invested, 20,000, is the sum of the amount invested at 3% and the amount invested at 5%. And the total interest, $0.045(20,000)$, is the sum of the interest earned in the 3% account and the interest earned in the 5% account.

As with the other mixture applications, the last column in the table gives us the equation to solve.

Try It $\PageIndex{15}$

Remy has $14,000 to invest in two mutual funds. One fund pays interest at 4% per year and the other fund pays interest at 7% per year. How much should she invest in each fund if she wants to earn 6.1% interest on the total amount?

$4,200 at 4%, $9,800 at 7%

Try It $\PageIndex{16}$

Marco has $8,000 to save for his daughter’s college education. He wants to divide it between one account that pays 3.2% interest per year and another account that pays 8% interest per year. How much should he invest in each account if he wants the interest on the total investment to be 6.5%?

$2,500 at 3.2%, $5,500 at 8%

Key Concepts

Total Value of Coins For the same type of coin, the total value of a number of coins is found by using the model. number·value=total value where number is the number of coins and value is the value of each coin; total value is the total value of all the coins
Label the columns type, number, value, total value.
Name what we are looking for. Choose a variable to represent that quantity. Use variable expressions to represent the number of each type of coin and write them in the table. Multiply the number times the value to get the total value of each type of coin.

Anatomy & Physiology
Astrophysics
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Precalculus
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English Grammar
U.S. History
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... and beyond

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Featured Answers

Mixture Problems

Key Questions

Let's try to think about the general form of a word problem involving mixtures.

In general, we have the following scenario:

a merchant sells two kinds of products (coffee, sweets, etc).

we know the unit prices for both kinds of products and for the final mixture #p_1# US dollars per pound for the first kind of product, #p_2# US dollars per pound for the second kind of product #p_m# US dollars per pound for the mixture

we know the total quantity formed by the mixture of the two products ( #q# pounds)

we have to find out the quantities of each product needed to form the mixture (here we have the variables : #x# denoting the quantity of the first kind of product and #y# denoting the quantity of the second kind of product)

Now, we have sufficient information to work out the equations.

First, we know that the sum of the two quantities is #q# pounds, which gives us the first equation : #x+y=q#

Second, we know that the sale price is the product of quantity and unit price, which gives us the second equation : #p_1 x + p_2 y = p_m*q#

Now, we have a system of two linear equations that can be easily solved by substitution.

Consider the following question:

How many pints of 20% acid solution and 70% acid solution must be mixed to obtain 40 pints of 50% acid solution?

First, I'll set up my table. I'll fill in the unknowns with variables #x# and #y# .

From this, we can easily set up the two equations.

Sum of values of two acids = Value of mixture

Therefore, #0.20x + 0.70y = 20#

For convenience, we'll multiply the entire equation by 10,

#2x + 7y=200#

This is equation (1)

Setting up the second equation,

Sum of amounts of each acid = Amount of mixture

To make at least one term of this equation identical to a term of equation (1), we'll multiply the entire equation by 2,

This is equation (2)

Subtracting equation (2) from (1) ,

#(+) 2x + 7y=200# #(-)2x+2y=80# #--------# #(=)0+5y=120#

Thus, #5y=120#

#y=120/5=24#

Substituting #y=24# in (2) ,

#2x+2(24)=80# #2x+48=80# #2x=80-48=32# #x=32/2=16#

So, we have #x=16# and #y=24#

We can conclude that 16 pints of 20% acid solution must be mixed with 24 pints of 70% solution to obtain 40 pints of 50% solution.

I hope your question was answered.

Module 8: Systems of Linear Equations

Mixture problems, learning outcomes.

Solve mixture problems

Write a system of linear equations representing a mixture problem, solve the system and interpret the results

One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. There are many other disciplines that use solutions as well.

The concentration or strength of a liquid solution is often described as a percentage. This number comes from the ratio of how much mass is in a specific volume of liquid. For example if you have [latex]50[/latex] grams of salt in a [latex]100mL[/latex] of water you have a [latex]50\%[/latex] salt solution based on the following ratio:

[latex]\frac{50\text{ grams }}{100\text{ mL }}=0.50\frac{\text{ grams }}{\text{ mL }}=50\text{ % }[/latex]

Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.

We will use the following table to help us solve mixture problems:

To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is [latex]120mL[/latex] of a [latex]9\%[/latex] solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.

First, find the total mass of solids for each solution by multiplying the volume by the concentration.

Next we add the new volumes and new masses.

Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.

A chemist has [latex]70 mL[/latex] of a [latex]50\%[/latex] methane solution. How much of an [latex]80\%[/latex] solution must she add so the final solution is [latex]60\%[/latex] methane?

Let’s use the problem solving process outlined previously to help us work through a solution to the problem.

Read and Understand: We are looking for a new amount – in this case a volume – based on the words “how much”. We know two starting concentrations and the final concentration, as well as one volume.

Define and Translate: Solution 1 is the [latex]70 mL[/latex] of [latex]50\%[/latex] methane and solution 2 is the unknown amount with [latex]80\%[/latex] methane. We can call our unknown amount [latex]x[/latex].

Write and Solve: Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.

Multiply amount by concentration to get total, be sure to distribute on the last row: [latex]\left(70 + x\right)0.6[/latex]Add the entries in the amount column to get final amount. The concentration for this amount is [latex]0.6[/latex] because we want the final solution to be [latex]60\%[/latex] methane.

Add the total mass for solution 1 and solution 2 to get the total mass for the [latex]60\%[/latex] solution. This is our equation for finding the unknown volume.

[latex]35+0.8x=42+0.6x[/latex]

[latex]\begin{array}{c}35+0.8x=42+0.6x\\\underline{-0.6x}\,\,\,\,\,\,\,\underline{-0.6x}\\35+0.2x=42\\\end{array}[/latex]

Subtract [latex]35[/latex] from both sides

[latex]\begin{array}{c}35+0.2x=42\\\underline{-35}\,\,\,\,\,\,\,\underline{-35}\\0.2x=7\end{array}[/latex]

Divide both sides by [latex]0.2[/latex]

[latex]\begin{array}{c}0.2x=7\\\frac{0.2x}{0.2}=\frac{7}{0.2}\end{array}[/latex] [latex]x=35[/latex]

[latex]35mL[/latex] of the [latex]80\%[/latex] solution must be added to the original [latex]70mL[/latex] of the [latex]50\%[/latex] solution to gain a solution with a concentration of [latex]60\%[/latex].

The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.

A farmer has two types of milk, one that is [latex]24\%[/latex] butterfat and another which is [latex]18\%[/latex] butterfat. How much of each should he use to end up with [latex]42[/latex] gallons of [latex]20\%[/latex] butterfat?

Read and Understand: We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is [latex]42[/latex] gallons. There are two unknowns in this problem.

Define and Translate: We will call the unknown volume of the [latex]24\%[/latex] solution [latex]x[/latex], and the unknown volume of the [latex]18\%[/latex] solution [latex]y[/latex].

Write and Solve: Fill in the table with the information we know.

Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from

When you sum the amount column you get one equation: [latex]x+ y = 42[/latex] When you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[/latex]

Use elimination to find a value for [latex]x[/latex] and [latex]y[/latex].

Multiply the first equation by [latex]-0.18[/latex]

[latex]-0.18(x+y) = (42)(-0.18)[/latex]

[latex]-0.18x - 0.18y = -7.56[/latex]

Now our system of equations looks like this:

[latex]0.24x + 0.18y = 8.4[/latex]

Adding the two equations together to eliminate the y terms gives this equation:

[latex]0.06x = 8.4[/latex]

Divide by [latex]0.06[/latex] on each side:

[latex]x = 14[/latex]

Now substitute the value for [latex]x[/latex] into one of the equations in order to solve for [latex]y[/latex].

[latex](14) + y = 42[/latex]

[latex]y = 28[/latex]

This can be interpreted as [latex]14[/latex] gallons of [latex]24\%[/latex] butterfat milk added to [latex]28[/latex] gallons of [latex]18\%[/latex] butterfat milk will give [latex]42[/latex] gallons of [latex]20\%[/latex] butterfat milk.

In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.

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Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
Ex: System of Equations Application - Mixture Problem. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/4s5MCqphpKo . License : CC BY: Attribution
Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : . License : CC BY: Attribution
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Praxis Core Math

Course: praxis core math > unit 1.

Rational number operations | Lesson
Rational number operations | Worked example

Ratios and proportions | Lesson

Ratios and proportions | Worked example
Percentages | Lesson
Percentages | Worked example
Rates | Lesson
Rates | Worked example
Naming and ordering numbers | Lesson
Naming and ordering numbers | Worked example
Number concepts | Lesson
Number concepts | Worked example
Counterexamples | Lesson
Counterexamples | Worked example
Pre-algebra word problems | Lesson
Pre-algebra word problems | Worked example
Unit reasoning | Lesson
Unit reasoning | Worked example

What are ratios and proportions?

What skills are tested.

Identifying and writing equivalent ratios
Solving word problems involving ratios
Solving word problems using proportions

How do we write ratios?

The ratio of lemon juice to sugar is a part-to-part ratio. It compares the amount of two ingredients.
The ratio of lemon juice to lemonade is a part-to-whole ratio. It compares the amount of one ingredient to the sum of all ingredients.
Determine whether the ratio is part to part or part to whole.
Calculate the parts and the whole if needed.
Plug values into the ratio.
Simplify the ratio if needed. Integer-to-integer ratios are preferred.
1 5 ‍ of the students on the varsity soccer team are lower-level students.
1 ‍ in 5 ‍ students on the varsity soccer team are lower-level students.

How do we use proportions?

Write an equation using equivalent ratios.
Plug in known values and use a variable to represent the unknown quantity.
If the numeric part of one ratio is a multiple of the corresponding part of the other ratio, we can calculate the unknown quantity by multiplying the other part of the given ratio by the same number.
If the relationship between the two ratios is not obvious, solve for the unknown quantity by isolating the variable representing it.
(Choice A) 1 : 4 ‍ A 1 : 4 ‍
(Choice B) 1 : 2 ‍ B 1 : 2 ‍
(Choice C) 1 : 1 ‍ C 1 : 1 ‍
(Choice D) 2 : 1 ‍ D 2 : 1 ‍
(Choice E) 4 : 1 ‍ E 4 : 1 ‍
(Choice A) 1 6 ‍ A 1 6 ‍
(Choice B) 1 3 ‍ B 1 3 ‍
(Choice C) 2 5 ‍ C 2 5 ‍
(Choice D) 1 2 ‍ D 1 2 ‍
(Choice E) 2 3 ‍ E 2 3 ‍
Your answer should be
an integer, like 6 ‍
a simplified proper fraction, like 3 / 5 ‍
a simplified improper fraction, like 7 / 4 ‍
a mixed number, like 1 3 / 4 ‍
an exact decimal, like 0.75 ‍
a multiple of pi, like 12 pi ‍ or 2 / 3 pi ‍

Things to remember

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Government Exam Articles
Mixture & Alligation - Concept, Formulas, Tips & Questions

Mixture & Alligation - Concept, Formulas, Tips & Questions

Word problems based on mixture and alligation are asked in the quantitative aptitude section of the various Government exams.

Mostly the number of questions asked from this section varies between 1-3 and are a bit tricky to solve. Aspirants preparing for the upcoming Government exams must practise well to be able to answer questions from this topic.

Candidates can also check the detailed quantitative aptitude syllabus and the exams in which this section is included at the linked article.

In this article, we bring to you the detailed concept of mixture and alligation, along with some important formulas to solve questions based on this topic. We shall also discuss some tips which will make the solution simpler and explain the same with the help of some sample examples.

It must be ensured that since there is sectional cut off in almost all competitive exams, it is important that equal attention is paid on the preparation of each subject.

Given below are a few quantitative aptitude concept links for your assistance:

Mixture & Alligation Questions PDF:- Download PDF Here

Mixture and Alligation – Concept

A mixture, as the name suggests is mixing two or more things together and alligation enables us to find the ratio in which the ingredients/ things have been mixed and at what price they are sole to earn profit or face loss.

Mixture & Alligation - Quantitative Aptitude for Government Exams

To solve mixture and alligation questions, one must know that alligation is used to find the mean value of a mixture when the ratio and amount of the ingredients mixed are different and also to find the proportion in which the elements are mixed.

It may sound a bit complex but once the candidate solves questions based on the same, the concept becomes clearer.

Aspirants can check the links given below to prepare themselves even better for the upcoming competitive exams:

Important Formulas for Mixture and Alligation

To solve any numerical ability question, candidates need to know a set of formulas for each topic which make solving the questions easier and time-saving. This, given below are a few formulas which will help candidates with the same and let them solve the mixture and alligation questions more conveniently:

The basic formula which is used to find the ratio in which the ingredients are mixed is

It is also called the rule of alligation and can also be represented as

Mixture and Alligation - Rule of Alligation

Candidates who are looking forward to applying for the upcoming Government exams and wish to solve more and more question papers and mock tests to speed up their calculation skills and enhance their preparation can refer to the links given below:

Free Online Government Exam Quiz
Previous Year Government Exam Question Papers PDF with Solutions
Bank PO Question Papers with Solutions
Free Online Mock Tests with Solutions

Tips and Tricks To Solve Questions

Time management plays a very important role in the Government exams and candidates cannot afford to lose much time while solving any of the given questions. Thus, a few simple tricks and tips will also prove to be helpful for the candidates to solve the mixture alligation questions easily:

The rule of alligation can also be used to solve questions based partnerships, time and work and wages
Read a question and try placing the values in the alligation rule mentioned above to solve the question
Questions from this topic may sound a bit tricky but are easy to solve if once a candidate get familiar with the concept and the important formulas used
You can not just find the ratio between the quantity of two elements but also the rate at which the article can be sold using the alligation rule

Any tip will only be useful if a candidate spends quality time practising the mixture and alligation questions and apply the rule of alligation to find answers for the same.

Get maths tricks and shortcut to solve the quantitative aptitude questions more efficiently and easily.

Candidates who wish to perform even better must visit the links given below to analyse the syllabus for each Government sector exam:

Mixture and Alligation Questions

Below we have discussed a few mixture and alligation based questions for your reference and to help you assist in understanding the concept and applying the formula effectively to solve the questions.

Q 1. A grocer wishes to sell a mixture of two variety of pulses worth Rs.16 per kg. In what ratio must he mix the pulses to reach this selling price, when cost of one variety of pulses is Rs.14 per kg and the other is Rs.24 per kg?

Answer: (4) 4:1

Using the rule of alligation,

⇒ 8:2 = 4:1

Q 2. Cost of two types of pulses is Rs.15 and Rs, 20 per kg, respectively. If both the pulses are mixed together in the ratio 2:3, then what should be the price of mixed variety of pulses per kg?

Rs. 22 per kg
Rs. 30 per kg
Rs. 10 per kg
Rs. 18 per kg
Rs. 24 per kg

Answer: (4) Rs. 18 per kg

Let the cost of mixed variety of pulse be Rs. x

As per the alligation rule,

2:3 = (20-x) : (x-15)

⇒ 2x+3x = 60+30

Q 3. A dealer has 1000 kg sugar and he sells a part of it at 8% profit and the rest of it at 18% profit. The overall profit he earns is 14%. What is the quantity which is sold at 18% profit?

Answer: (2) 600 kg

As per the rule of alligation,

Quantity of Dearer: Quantity of Cheaper = (18-14) : (14-8) = 4:6 = 2:3

Quantity of sugar sold at 18% profit = 3/5 × 1000 = 600kg

Q 4. How much coffee of variety A, costing Rs. 5 a kg should be added to 20 kg of Type B coffee at Rs. 12 a kg so that the cost of the two coffee variety mixture be worth Rs. 7 a kg?

Answer: (5) 50 kg

As per the rule of alligation,

Quantity of Dearer: Quantity of Cheaper = (12-7) : (7-5) = 5:2

Quantity of Variety A coffee that needs to be mixed ⇒ 5:2 = x:20

The above-given questions are just a sample mixture and alligation questions. Candidates must solve more of these to get a better understanding of the topic.

Aspirants can also check the related preparation links given below to ace the upcoming Government exams:

Candidates looking for Government exam updates, information, study material or preparation tips can refer to BYJU’S for assistance.

Refer to the following video to get the continued explanation and analysis of the Mixture and Alligation concept:

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Ratio Problem Solving

Here we will learn about ratio problem solving, including how to set up and solve problems. We will also look at real life ratio problems.

There are also ratio problem solving worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is ratio problem solving?

Ratio problem solving is a collection of word problems that link together aspects of ratio and proportion into more real life questions. This requires you to be able to take key information from a question and use your knowledge of ratios (and other areas of the curriculum) to solve the problem.

A ratio is a relationship between two or more quantities . They are usually written in the form a:b where a and b are two quantities. When problem solving with a ratio, the key facts that you need to know are,

What is the ratio involved?
What order are the quantities in the ratio?
What is the total amount / what is the part of the total amount known?
What are you trying to calculate ?

As with all problem solving, there is not one unique method to solve a problem. However, this does not mean that there aren’t similarities between different problems that we can use to help us find an answer.

The key to any problem solving is being able to draw from prior knowledge and use the correct piece of information to allow you to get to the next step and then the solution.

Let’s look at a couple of methods we can use when given certain pieces of information.

When solving ratio problems it is very important that you are able to use ratios. This includes being able to use ratio notation.

For example, Charlie and David share some sweets in the ratio of 3:5. This means that for every 3 sweets Charlie gets, David receives 5 sweets.

Charlie and David share 40 sweets, how many sweets do they each get?

We use the ratio to divide 40 sweets into 8 equal parts.

Then we multiply each part of the ratio by 5.

3 x 5:5 x 5 = 15:25

This means that Charlie will get 15 sweets and David will get 25 sweets.

Dividing ratios

Step-by-step guide: Dividing ratios (coming soon)

Ratios and fractions (proportion problems)

We also need to consider problems involving fractions. These are usually proportion questions where we are stating the proportion of the total amount as a fraction.

Simplifying and equivalent ratios

Simplifying ratios

Equivalent ratios

Units and conversions ratio questions

Units and conversions are usually equivalent ratio problems (see above).

If £1:\$1.37 and we wanted to convert £10 into dollars, we would multiply both sides of the ratio by 10 to get £10 is equivalent to \$13.70.
The scale on a map is 1:25,000. I measure 12cm on the map. How far is this in real life, in kilometres? After multiplying both parts of the ratio by 12 you must then convert 12 \times 25000=300000 \ cm to km by dividing the solution by 100 \ 000 to get 3km.

Notice that for all three of these examples, the units are important. For example if we write the mapping example as the ratio 4cm:1km, this means that 4cm on the map is 1km in real life.

Top tip: if you are converting units, always write the units in your ratio.

Usually with ratio problem solving questions, the problems are quite wordy . They can involve missing values , calculating ratios , graphs , equivalent fractions , negative numbers , decimals and percentages .

Highlight the important pieces of information from the question, know what you are trying to find or calculate , and use the steps above to help you start practising how to solve problems involving ratios.

How to do ratio problem solving

In order to solve problems including ratios:

Identify key information within the question.

Know what you are trying to calculate.

Use prior knowledge to structure a solution.

Explain how to do ratio problem solving

Ratio problem solving worksheet

Get your free ratio problem solving worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on ratio

Ratio problem solving is part of our series of lessons to support revision on ratio . You may find it helpful to start with the main ratio lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

How to work out ratio
Ratio to fraction
Ratio scale
Ratio to percentage

Ratio problem solving examples

Example 1: part:part ratio.

Within a school, the number of students who have school dinners to packed lunches is 5:7. If 465 students have a school dinner, how many students have a packed lunch?

Within a school, the number of students who have school dinners to packed lunches is \bf{5:7.} If \bf{465} students have a school dinner , how many students have a packed lunch ?

Here we can see that the ratio is 5:7 where the first part of the ratio represents school dinners (S) and the second part of the ratio represents packed lunches (P).

We could write this as

Where the letter above each part of the ratio links to the question.

We know that 465 students have school dinner.

2 Know what you are trying to calculate.

From the question, we need to calculate the number of students that have a packed lunch, so we can now write a ratio below the ratio 5:7 that shows that we have 465 students who have school dinners, and p students who have a packed lunch.

We need to find the value of p.

3 Use prior knowledge to structure a solution.

We are looking for an equivalent ratio to 5:7. So we need to calculate the multiplier. We do this by dividing the known values on the same side of the ratio by each other.

So the value of p is equal to 7 \times 93=651.

There are 651 students that have a packed lunch.

Example 2: unit conversions

The table below shows the currency conversions on one day.

Use the table above to convert £520 (GBP) to Euros € (EUR).

Use the table above to convert \bf{£520} (GBP) to Euros \bf{€} (EUR).

The two values in the table that are important are GBP and EUR. Writing this as a ratio, we can state

Ratio problem solving example 2 step 1 image 2

We know that we have £520.

We need to convert GBP to EUR and so we are looking for an equivalent ratio with GBP = £520 and EUR = E.

To get from 1 to 520, we multiply by 520 and so to calculate the number of Euros for £520, we need to multiply 1.17 by 520.

1.17 \times 520=608.4

So £520 = €608.40.

Example 3: writing a ratio 1:n

Liquid plant food is sold in concentrated bottles. The instructions on the bottle state that the 500ml of concentrated plant food must be diluted into 2l of water. Express the ratio of plant food to water respectively in the ratio 1:n.

Liquid plant food is sold in concentrated bottles. The instructions on the bottle state that the \bf{500ml} of concentrated plant food must be diluted into \bf{2l} of water . Express the ratio of plant food to water respectively as a ratio in the form 1:n.

Using the information in the question, we can now state the ratio of plant food to water as 500ml:2l. As we can convert litres into millilitres, we could convert 2l into millilitres by multiplying it by 1000.

2l = 2000ml

So we can also express the ratio as 500:2000 which will help us in later steps.

We want to simplify the ratio 500:2000 into the form 1:n.

We need to find an equivalent ratio where the first part of the ratio is equal to 1. We can only do this by dividing both parts of the ratio by 500 (as 500 \div 500=1 ).

So the ratio of plant food to water in the form 1:n is 1:4.

Example 4: forming and solving an equation

Three siblings, Josh, Kieran and Luke, receive pocket money per week proportional to their age. Kieran is 3 years older than Josh. Luke is twice Josh’s age. If Josh receives £8 pocket money, how much money do the three siblings receive in total?

Three siblings, Josh, Kieran and Luke, receive pocket money per week proportional to their ages. Kieran is \bf{3} years older than Josh . Luke is twice Josh’s age. If Luke receives \bf{£8} pocket money, how much money do the three siblings receive in total ?

We can represent the ages of the three siblings as a ratio. Taking Josh as x years old, Kieran would therefore be x+3 years old, and Luke would be 2x years old. As a ratio, we have

We also know that Luke receives £8.

We want to calculate the total amount of pocket money for the three siblings.

We need to find the value of x first. As Luke receives £8, we can state the equation 2x=8 and so x=4.

Now we know the value of x, we can substitute this value into the other parts of the ratio to obtain how much money the siblings each receive.

The total amount of pocket money is therefore 4+7+8=£19.

Example 5: simplifying ratios

Below is a bar chart showing the results for the colours of counters in a bag.

Express this data as a ratio in its simplest form.

From the bar chart, we can read the frequencies to create the ratio.

We need to simplify this ratio.

To simplify a ratio, we need to find the highest common factor of all the parts of the ratio. By listing the factors of each number, you can quickly see that the highest common factor is 2.

\begin{aligned} &12 = 1, {\color{red} 2}, 3, 4, 6, 12 \\\\ &16 = 1, {\color{red} 2}, 4, 8, 16 \\\\ &10 = 1, {\color{red} 2}, 5, 10 \end{aligned}

HCF (12,16,10) = 2

Dividing all the parts of the ratio by 2 , we get

Our solution is 6:8:5 .

Example 6: combining two ratios

Glass is made from silica, lime and soda. The ratio of silica to lime is 15:2. The ratio of silica to soda is 5:1. State the ratio of silica:lime:soda.

Glass is made from silica, lime and soda. The ratio of silica to lime is \bf{15:2.} The ratio of silica to soda is \bf{5:1.} State the ratio of silica:lime:soda .

We know the two ratios

We are trying to find the ratio of all 3 components: silica, lime and soda.

Using equivalent ratios we can say that the ratio of silica:soda is equivalent to 15:3 by multiplying the ratio by 3.

Ratio problem solving example 6 step 3 image 1

We now have the same amount of silica in both ratios and so we can now combine them to get the ratio 15:2:3.

Ratio problem solving example 6 step 3 image 2

Example 7: using bar modelling

India and Beau share some popcorn in the ratio of 5:2. If India has 75g more popcorn than Beau, what was the original quantity?

India and Beau share some popcorn in the ratio of \bf{5:2.} If India has \bf{75g} more popcorn than Beau , what was the original quantity?

We know that the initial ratio is 5:2 and that India has three more parts than Beau.

We want to find the original quantity.

Drawing a bar model of this problem, we have

Where India has 5 equal shares, and Beau has 2 equal shares.

Each share is the same value and so if we can find out this value, we can then find the total quantity.

From the question, India’s share is 75g more than Beau’s share so we can write this on the bar model.

Ratio problem solving example 7 step 3 image 1

We can find the value of one share by working out 75 \div 3=25g.

Ratio problem solving example 7 step 3 image 2

We can fill in each share to be 25g.

Ratio problem solving example 7 step 3 image 3

Adding up each share, we get

India = 5 \times 25=125g

Beau = 2 \times 25=50g

The total amount of popcorn was 125+50=175g.

Common misconceptions

Mixing units

Make sure that all the units in the ratio are the same. For example, in example 6 , all the units in the ratio were in millilitres. We did not mix ml and l in the ratio.

Ratio written in the wrong order

For example the number of dogs to cats is given as the ratio 12:13 but the solution is written as 13:12.

Ratios and fractions confusion

Take care when writing ratios as fractions and vice-versa. Most ratios we come across are part:part. The ratio here of red:yellow is 1:2. So the fraction which is red is \frac{1}{3} (not \frac{1}{2} ).

Ratio problem solving common misconceptions

Counting the number of parts in the ratio, not the total number of shares

For example, the ratio 5:4 has 9 shares, and 2 parts. This is because the ratio contains 2 numbers but the sum of these parts (the number of shares) is 5+4=9. You need to find the value per share, so you need to use the 9 shares in your next line of working.

Ratios of the form \bf{1:n}

The assumption can be incorrectly made that n must be greater than 1 , but n can be any number, including a decimal.

Practice ratio problem solving questions

1. An online shop sells board games and computer games. The ratio of board games to the total number of games sold in one month is 3:8. What is the ratio of board games to computer games?

8-3=5 computer games sold for every 3 board games.

2. The volume of gas is directly proportional to the temperature (in degrees Kelvin). A balloon contains 2.75l of gas and has a temperature of 18^{\circ}K. What is the volume of gas if the temperature increases to 45^{\circ}K?

3. The ratio of prime numbers to non-prime numbers from 1-200 is 45:155. Express this as a ratio in the form 1:n.

4. The angles in a triangle are written as the ratio x:2x:3x. Calculate the size of each angle.

5. A clothing company has a sale on tops, dresses and shoes. \frac{1}{3} of sales were for tops, \frac{1}{5} of sales were for dresses, and the rest were for shoes. Write a ratio of tops to dresses to shoes sold in its simplest form.

6. During one month, the weather was recorded into 3 categories: sunshine, cloud and rain. The ratio of sunshine to cloud was 2:3 and the ratio of cloud to rain was 9:11. State the ratio that compares sunshine:cloud:rain for the month.

Ratio problem solving GCSE questions

1. One mole of water weighs 18 grams and contains 6.02 \times 10^{23} water molecules.

Write this in the form 1gram:n where n represents the number of water molecules in standard form.

2. A plank of wood is sawn into three pieces in the ratio 3:2:5. The first piece is 36cm shorter than the third piece.

Calculate the length of the plank of wood.

5-3=2 \ parts = 36cm so 1 \ part = 18cm

3. (a) Jenny is x years old. Sally is 4 years older than Jenny. Kim is twice Jenny’s age. Write their ages in a ratio J:S:K.

(b) Sally is 16 years younger than Kim. Calculate the sum of their ages.

Learning checklist

You have now learned how to:

Relate the language of ratios and the associated calculations to the arithmetic of fractions and to linear functions
Develop their mathematical knowledge, in part through solving problems and evaluating the outcomes, including multi-step problems
Make and use connections between different parts of mathematics to solve problems

The next lessons are

Compound measures
Best buy maths

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Mixture and alligation are one of the most important and intrinsic topics for SSC CGL. As it is not an individual chapter, it is a method of solving problems of different topics like ratio, profit, loss, average, etc. Almost 3-4 questions are asked in the prelims exam that can be solved using the technique of allegation. In this article we will be studying the method of the allegation and how can we implement it for solving questions of various topics in a much easier and smarter way.

Let’s start with the basic definition first:-

Mixture : It is the process of mixing two or more elements/ingredients together.

Allegation : It is a process or a rule that tells us how to mix two or more ingredients/mixtures to get the desired mixture of a certain price & concentration.

Now there are 2 ways of applying the allegation technique to questions.

Formula Method:

Quantity of cheaper/ Quantity of dearer = (Price of dearer – Mean price)/ (Mean Price- Price of cheaper)

Diagram Method:

The above-mentioned formula can be better understood using the diagram method & it is more convenient to use this method.

Note 1. Please note that allegation is a concept with the help of which we can find the mean price if 2 or more ingredients are mixed in some ratio or we can also find the ratio in which two or more items should be mixed to get a certain price or concentration of the mixture.

Note 2. (d-m)/(m-c) is nothing but the ratio of quantities of both the items, so even if we are given a ratio in which 2 items are mixed, we can find the “m”..

Now let’s try to solve a variety of problems using the allegation method which will help you in understanding the concept in a much better way.

Illustration 1. If 20 kgs of rice which cost 45 Rs/Kg is mixed with 25 kgs of rice which costs 30 Rs/Kg. What will be the price of the final mixture?

Solution: In this question, we have to find “M”

Now as we know, (45-M)/(M-30)=25/20

⇒ (45-M)/(M-30)= 5/4

⇒180-4M = 5M- 150

⇒ M= 330/9= 36.6 Rs/Kg

Illustration 2. Oranges of type A that cost 60 Rs/Kg is mixed with the orange of type B that costs 80 Rs/Kg in a ratio of 3:5. What will be the price of the resultant mixture of oranges?

Solution: Again applying the same formula,

(60-M)/(M-80)= 5/3

180-3M = 5M-400

M= 580/8= 72.5 Rs

Illustration 3. If rice or 45 Rs/kg is mixed with rice of 35 Rs/kg and the price of the resultant mixture is 42.5 Rs/kg. What is the ratio that they were mixed in?

Solution: As we know, (45-42.5)/(42.5-35) = Y/X

2.5/7.5= Y/X

1/3 = Y/X, hence they were mixed in the ratio of 3:1.

Illustration 4. Mixture A which contains 30% water is mixed with Mixture B which has 20% water in a ratio of 2:5. What will be the % of water in the resultant mixture?

Solution: Allegation can also be applied to percentages & not just absolute values.

(30%-M)/(M-20%)= 5/2

60% – 2M= 5M – 100%

M = 160%/7 = 22.8%

Illustration 5. A shopkeeper sells one item at 20% profit and another item at 5% profit. If the overall profit of the shopkeeper is 10 per cent. What is the ratio of the CP of item A to item B?

Solution: Here, the ratio of quantities will give us ratio of the cost price

So, (20%-10%)/ (10%-5%)= CP (B)/CP (A)

10%/5%= CP (B)/ CP (A)

1:2 = CP (A)/ CP(B)

Illustration 6. A shopkeeper sells Type A rice at 40 Rs/Kg and types B rice at 60 Rs/kg. If he wants to gain a profit of 10% after selling the mixture of both types in the ratio Of 2:3. What should be the selling price of the mixture?

Solution: Here the value of M will give us no profit no loss amount. (40-M)/(M-60)= 3/2

It gives us M= 260/5 = 52

Now to get a profit of 10 %, the shopkeeper has to sell it at, (52) × (110/100) = 57.2 Rs

Illustration 7. If a shopkeeper sells an item of 2000 Rs at 15 per cent profit & another item of 3500 Rs at 5 per cent loss. Find his overall profit/loss percentage?

Solution: We will take loss % with minus sign,

(15% – M)/ (M- (-5%) = 7/4

It gives us M = 2.27%

Illustration 8. If the average marks of boys in a class is 75 and the average marks of girls in that class is 80. If the ratio of boys to girls in that class is 4:3. Find the average marks of the class?

Solution: Here also, we can find the average of the whole class easily,

(75-M)/( M-80)= 3/4

Illustration 9. If a shopkeeper sold a mixture of mangoes at 100 Rs/Kg in which he mixed 2 types of mangoes that cost him 50 Rs/Kg and 75 Rs/Kg in the ratios of 2:3 respectively. What is the profit percentage of the shopkeeper?

Solution: (50-M)/( M-75)= 3/2

M= 65 Rs/kg

Now he sold this at 100 Rs/Kg, so his profit is (35/65)× 100 = 53.8 % P

Illustration 10. If a cyclist covered 3/5th distance or the track with 15 km/hr and the rest of the distance with 10km/her. What is the average speed of the cyclist?

Solution: (15-M)/(M-10)= 2/3

M= 13 km/Hr

Illustration 11. In a village, the average age of males is 35 years & in another village, the average age of males is 30 years. If the ratio of the total number of females in both the villages is 2:3. Find the average age if both the villages are combined?

Solution: Average age will be, (35-M)/ (M-30)= 3/2

That gives us M= 36 Years

Illustration 12. A lender gives money at 8% SI for 2 years to a guy & takes money from somewhere at 5% SI for 3 years. If the amount he lent to the amount he took is in the ratio of 3:4. Find his overall profit/loss %?

Solution: 8% for 2 years SI becomes 16% & 5% for 3 years SI becomes 15%.

overall profit/loss % will be, (16%- M)/ (M+15%)= 4/3

That gives us M= -12/7% = loss of 12/7%

Illustration 13. If a racer can finish a track in 10 hours & another racer can finish the same track in 15 hours. If both are supposed to take part in the relay race & racer 1 has to run the 2/5th part of the track. How long will it take for both of them to complete the race?

Solution: If racer 1 is covering 2/5th of the track, hence racer 2 will cover 3/5th of the track.

Applying the rule of the allegation, we get

(10 Hrs – M)/ (M-15 Hrs)= 3/2

This gives us, M= 13 Hrs

Illustration 14. A shopkeeper mixes water in milk If he mixes 1-litre water in 5-litre milk & the actual cost of milk is 30Rs/Litre but he sells the milk at 28 Rs/litre. Find his actual profit or loss?

Solution: The key concept here to understand is that water costs zero rupees.

Hence the no profit no loss value of the mixture is,

(30-M)/ (M-0)= 1/5

That gives us M= 25 Rs.

Now he sells this at 28 Rs, so profit % is (3/25) × 100 = 12%

Illustration 15. A guy who works at a juice centre mixes water with juice in a ratio of 5:1 respectively. If the actual price of fruits is 50 Rs/glass & he sells 60 Rs/glass. find his actual profit/glass?

Solution: Again, repeating the same logic

(50-M)/ (M-0)= 1/5

M= 250/6 Rs, so his profit % is [60- (250/6)]/ (250/6) × 100 = 44%

This was some illustration to help you understand the application of Mixture & Allegation.

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In Example 5.5.4 5.5. 4 we'll solve a coin problem. Now that we know how to work with systems of two variables, naming the variables in the 'number' column will be easy.
Mixture Word Problems (video lessons, examples and solutions)
Mixture Problems: word problems involving items of different values being mixed together, How to solve mixture problems when we are adding to the solution, removing from the solution, replacing the solution, Mixing Quantities Of Different Costs, How to set up and solve Mixture Word Problems, How to solve acid solution problems, with video lessons, examples and step-by-step solutions.
Math Mixture Problems (examples, solutions, examples)
To solve mixture problems, knowledge of solving systems of equations. is necessary. Most often, these problems will have two variables, but more advanced problems have systems of equations with three variables.
3.3: Solve Mixture Applications
In mixture problems, we will have two or more items with different values to combine together. The mixture model is used by grocers and bartenders to make sure they set fair prices for the products …
6.8 Mixture and Solution Word Problems
6.8 Mixture and Solution Word Problems. Solving mixture problems generally involves solving systems of equations. Mixture problems are ones in which two different solutions are mixed together, resulting in a new, final solution. Using a table will help to set up and solve these problems.
Mixture Problems
Learn how to solve mixture problems in algebra with videos and questions from Socratic. Get smarter and master this topic easily.
Mixture Problems in Algebra
Understand mixture problems in algebra with various examples. Learn how to solve mixture problems and be familiar with the general solving process of these types of problems.
Mixture Problems
Mixture problems are excellent candidates for solving with systems of equations methods. These problems arise in many settings, such as when combining solutions in a chemistry lab or adding ingredients to a recipe. Mixtures (and mixture problems) are made whenever different types of items are combined to create a third, "mixed" item.
How to solve Ratio and Proportion Mixture Problems 2
Learn how to solve ratio and proportion mixture problems conceptually at lightning speed using basic concepts on mixture and alligation.
Mixture Problems
Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A solution is a mixture of two or more different substances like water and salt or vinegar and oil. Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand.
Ratios and proportions
A proportion is an equality of two ratios. We write proportions to help us establish equivalent ratios and solve for unknown quantities.
How to Solve Mixture Problems in Simple Steps 1
Learn how to solve mixture problems with two or more liquids mixed in given ratios. Basic ratio mixture concepts are explained by solving selected problems.
3 Simple Steps for Solving Mixture Problems
Struggling to understand your algebra assignments? Check out this simple, 3-step process and video for solving mixture problems.
Mixture & Alligation in Quantitative Aptitude
Mixture and Alligation Questions. Below we have discussed a few mixture and alligation based questions for your reference and to help you assist in understanding the concept and applying the formula effectively to solve the questions. Q 1. A grocer wishes to sell a mixture of two variety of pulses worth Rs.16 per kg.
Examples of Mixture Problems
To solve these types of problems, it is important to think of mixtures as a type of ratio or proportion. Any situation in which two or more different variables are combined to determine a third is a type of ratio.
Ratio Problem Solving
Ratio problem solving is a collection of word problems that link together aspects of ratio and proportion into more real life questions. This requires you to be able to take key information from a question and use your knowledge of ratios (and other areas of the curriculum) to solve the problem.
Tricks To Solve Mixture and Alligation
Mixture and alligation are one of the most important and intrinsic topics for SSC CGL. As it is not an individual chapter, it is a method of solving problems of different topics like ratio, profit, loss, average, etc. Almost 3-4 questions are asked in the prelims exam that can be solved using the technique of allegation.

Solving Mixture Problems

Introduction to Mixtures

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Examples of Setting Up "Mixture" Word Problems

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Mixture Word Problems - Different Costs or Concentrations

Mixing Quantities Of Different Costs or Different Concentrations

Margin Size

3.3: Solve Mixture Applications

Learning Objectives

Solve Coin Word Problems

TOTAL VALUE OF COINS

Example \(\PageIndex{1}\)

Try It \(\PageIndex{1}\)

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SOLVE COIN WORD PROBLEMS.

Example \(\PageIndex{2}\)

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Example \(\PageIndex{3}\)

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