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## "Mixture" Word Problems

Explanation Examples

## What are "mixture" word problems?

Mixture word problems are exercises which involve creating a mixture from two or more different things, and then determining some quantity (such as percentage, price, number of liters, etc) of the resulting mixture. There will always be a "rate" of some sort, such as miles per hour or cost per pound.

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Mixture Word Problems

Here is an example:

Your school is holding a family-friendly event this weekend. Students have been pre-selling tickets to the event; adult tickets are $5.00, and child tickets (for kids six years old and under) are$2.50 . From past experience, you expect about 13,000 people to attend the event.

You consult with your student ticket-sellers, and discover that they have not been keeping track of how many child tickets they have sold. The tickets are identical, until the ticket-seller punches a hole in the ticket, indicating that it is a child ticket.

But they don't remember how many holes they've punched. They only know that they've sold 548 tickets for $2460 . How much revenue from each of child and adult tickets can you expect? To solve this, we need to figure out the ratio of tickets that have already been sold. If we work methodically, we can find the answer. Let A stand for the number of adult tickets pre-sold. Since a total of 548 tickets have been sold, then the number of children tickets pre-sold thus far must be 548 − A . this construction is important! When you've got a variable that stands for part of whatever it is that you're working with, then the amount that is left for the other part of whatever you're working with is found by subtracting the variable from the total. That is, (the total amount) less (the amount that is being represented by the variable) is (the amount that is left for ther other amount). This "how much is left" construction is something you will need to understand and use. Since each adult ticket cost$5.00 , then 5A stands for the revenue brought in from the adult tickets pre-sold; likewise, 2.5(548 − A) stands for the revenue brought in from the child tickets. (Note: The per-ticket cost is the "rate" for this exercise.)

Organizing this information in a grid, we get:

From the last column, we get (total $from the adult tickets) plus (total$ from the child tickets) is (the total $so far), or, as an equation: 5A + 2.5((548 − A) = 2,460 5A + 1,370 − 2.5A = 2,460 1,370 + 2.5A = 2,460 2.5A = 1,090 A = (1,090)/(2.5) = 436 So 436 adult tickets were pre-sold, so the number of child tickets pre-sold is: C = 548 − 436 = 112 So 112 child tickets were pre-sold. Advertisement We expect about 13,000 people in total to attend this event. We have a ratio of adult pre-sold tickets to total pre-sold tickets. Assuming that the pre-sold rate (or ratio) is representative of the total number of adult tickets, we can set up a proportion (of pre-sold adult tickets to total pre-sold tickets), using a variable for the unknown total number of adult tickets expected to be sold: 436/548 = x /13,000 [(436)(13,000)]/548 = x 10,343.0656934... = x This works out to about 10,343 adult tickets. The remaining 2,657 tickets, of the expected total of 13,000 , will be child tickets. Then the expected total ticket revenue is given by: 5(10,343) + 2.5(2,657) = 58,357.5 So the expected total revenues from ticket sales is$58,357.50 .

Let's try another one. This time, suppose you work in a lab. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?

Let w stand for the number of liters of the weaker 10% solution. Since the total number of liters is going to be 10 , then the number of liters left, after the first w liters have been poured, will be 10 −  w liters needed of the 30% solution.

(The clear labeling of the variable is important. While I picked w to stand for the w eaker acid, I might not remember this by the end of the exercise. If I don't label, I might not be able correctly to interpret my answer in the end.)

For mixture problems, it is often very helpful to do a grid, so let's do that here:

When the problem is set up like this, we can usually use the last column to write our equation. In this case, the liters of acid within the 10% solution, plus the liters of acid within the 30% solution, must add up to the liters of acid within the 15% solution. So, adding down the right-hand column, setting the inputs equal to the mixed output, we get:

0.10 w + 0.30(10 − w ) = 1.5

0.10 w + 3 − 0.30 w = 1.5

3 − 0.2 w = 1.5

1.5 = 0.2 w

(1.5)/(0.2) = w = 7.5

Looking back and confirming that the variable stands for the number of liters of the weaker acid, we see that we would need 7.5 liters of 10% acid . This means that we would need another 10 − 7.5 = 2.5 liters of the stronger 30% acid .

(If you think about it, this makes sense. Fifteen percent is closer to 10% than it is to 30% , so we ought to need more 10% solution in our mix.)

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## How to Solve Mixture Word Problems

Last Updated: March 29, 2019

This article was co-authored by Daron Cam . Daron Cam is an Academic Tutor and the Founder of Bay Area Tutors, Inc., a San Francisco Bay Area-based tutoring service that provides tutoring in mathematics, science, and overall academic confidence building. Daron has over eight years of teaching math in classrooms and over nine years of one-on-one tutoring experience. He teaches all levels of math including calculus, pre-algebra, algebra I, geometry, and SAT/ACT math prep. Daron holds a BA from the University of California, Berkeley and a math teaching credential from St. Mary's College. This article has been viewed 102,405 times.

Mixture word problems involve creating a mixture from two ingredients. A common type of problem is creating a solution of a certain strength, such as a 20% saline solution, from two solutions of varying strengths. Since these are multi-step problems involving a bit of logic, they can sometimes be confusing to solve. It is helpful to begin these types of problems by setting up a table that can help you keep track of variables. From there you can use algebra to find the missing information.

## Setting Up a Table

• For example, you might have a 20% saline solution, and a 15% saline solution. If you need to make 5 liters of an 18% saline solution, how many liters of each solution do you need to combine?
• For this problem you would label the three rows “20% solution,” “15% solution,” and “18% Mixture.”

• For example, you would label the second column “Percent Saline.” Since the first ingredient is 20% saline, in the first row you will write .20. Since the second solution is 15% saline, in the second row you will write .15. Since the final mixture needs to be 18% saline, in the third row you will write .18.

## Applying the Concept to Price Problems

• For example, you might be trying to solve the following problem: The student council is selling 100 cups of punch at a school dance. The punch is made from a combination of fruit juice and lemon-lime soda. They want to sell each cup of punch for $1.00. Normally they would sell a cup of fruit juice for$1.15 and a cup of lemon lime soda for $0.75. How many cups of each ingredient should the student council use to make the punch? • In this problem, fruit juice and lemon-lime soda are the two ingredients. • For example, since you know the student council plans on making 100 cups of punch, you would write 100 in the third row of the first column. • For example, you know that the punch will be sold for$1.00 per cup, so write a 1 in the second column for the mixture. The fruit juice sells for $1.15 per cup, so write 1.15 in the second column for this ingredient. The soda sells for$0.75 per cup, so write 0.75 in the second column for lemon-lime soda.

## You Might Also Like

• ↑ http://www.algebralab.org/Word/Word.aspx?file=Algebra_Mixtures.xml
• ↑ http://purplemath.com/modules/mixture.htm

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Chapter 6: Polynomials

## 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. The basic structure of this table is shown below:

The first column in the table (Name) is used to identify the fluids or objects being mixed in the problem. The second column (Amount) identifies the amounts of each of the fluids or objects. The third column (Value) is used for the value of each object or the percentage of concentration of each fluid. The last column (Equation) contains the product of the Amount times the Value or Concentration.

Example 6.8.1

Jasnah has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane? Find the equation.

• The solution names are 50% (S 50 ), 60% (S 60 ), and 80% (S 80 ).
• The amounts are S 50 = 70 mL, S 80 , and S 60 = 70 mL + S 80 .
• The concentrations are S 50 = 0.50, S 60 = 0.60, and S 80 = 0.80.

The equation derived from this data is 0.50 (70 mL) + 0.80 (S 80 ) = 0.60 (70 mL + S 80 ).

Example 6.8.2

Sally and Terry blended a coffee mix that sells for $\2.50$ by mixing two types of coffee. If they used 40 mL of a coffee that costs $\3.00,$ how much of another coffee costing $\1.50$ did they mix with the first?

The equation derived from this data is:

$\begin{array}{ccccccc} 1.50(C_{1.50})&+&3.00(40)&=&2.50(40&+&C_{1.50}) \\ 1.50(C_{1.50})&+&120&=&100&+&2.50(C_{1.50}) \\ -2.50(C_{1.50})&-&120&=&-120&-&2.50(C_{1.50}) \\ \hline &&-1.00(C_{1.50})&=&-20&& \\ \\ &&(C_{1.50})&=&\dfrac{-20}{-1}&& \\ \\ &&C_{1.50}&=&20&& \end{array}$

This means 20 mL of the coffee selling for $\1.50$ is needed for the mix.

Example 6.8.3

Nick and Chloe have two grades of milk from their small dairy herd: one that is 24% butterfat and another that is 18% butterfat. How much of each should they use to end up with 42 litres of 20% butterfat?

The equation derived from this data is:

$\begin{array}{rrrrrrr} 0.24(B_{24})&+&0.18(42&- &B_{24})&=&0.20(42) \\ 0.24(B_{24})&+&7.56&-&0.18(B_{24})&=&8.4 \\ &-&7.56&&&&-7.56 \\ \hline &&&&0.06(B_{24})&=&0.84 \\ \\ &&&&B_{24}&=&\dfrac{0.84}{0.06} \\ \\ &&&&B_{24}&=&14 \end{array}$

This means 14 litres of the 24% buttermilk, and 28 litres of the 18% buttermilk is needed.

Example 6.8.4

In Natasha’s candy shop, chocolate, which sells for $\4$ a kilogram, is mixed with nuts, which are sold for $\2.50$ a kilogram. Chocolate and nuts are combined to form a chocolate-nut candy, which sells for $\3.50$ a kilogram. How much of each are used to make 30 kilograms of the mixture?

$\begin{array}{rrrrrrl} 4.00(C)&+&2.50(30&-&C)&=&3.50(30) \\ 4.00(C)&+&75&-&2.50(C)&=&105 \\ &-&75&&&&-75 \\ \hline &&&&1.50(C)&=&30 \\ \\ &&&&C&=&\dfrac{30}{1.50} \\ \\ &&&&C&=&20 \end{array}$

Therefore, 20 kg of chocolate is needed for the mixture.

With mixture problems, there is often mixing with a pure solution or using water, which contains none of the chemical of interest. For pure solutions, the concentration is 100%. For water, the concentration is 0%. This is shown in the following example.

Example 6.8.5

Joey is making a a 65% antifreeze solution using pure antifreeze mixed with water. How much of each should be used to make 70 litres?

$\begin{array}{rrrrl} 1.00(A)&+&0.00(70-A)&=&0.65(0.70) \\ &&1.00A&=&45.5 \\ &&A&=&45.5 \\ \end{array}$

This means the amount of water added is 70 L − 45.5 L = 24.5 L.

For questions 1 to 9, write the equations that define the relationship.

• A tank contains 8000 litres of a solution that is 40% acid. How much water should be added to make a solution that is 30% acid?
• How much pure antifreeze should be added to 5 litres of a 30% mixture of antifreeze to make a solution that is 50% antifreeze?
• You have 12 kilograms of 10% saline solution and another solution of 3% strength. How many kilograms of the second should be added to the first in order to get a 5% solution?
• How much pure alcohol must be added to 24 litres of a 14% solution of alcohol in order to produce a 20% solution?
• How many litres of a blue dye that costs $\1.60$ per litre must be mixed with 18 litres of magenta dye that costs $\2.50$ per litre to make a mixture that costs $\1.90$ per litre?
• How many grams of pure acid must be added to 40 grams of a 20% acid solution to make a solution which is 36% acid?
• A 100-kg bag of animal feed is 40% oats. How many kilograms of pure oats must be added to this feed to produce a blend of 50% oats?
• A 20-gram alloy of platinum that costs $\220$ per gram is mixed with an alloy that costs $\400$ per gram. How many grams of the $\400$ alloy should be used to make an alloy that costs $\300$ per gram?
• How many kilograms of tea that cost $\4.20$ per kilogram must be mixed with 12 kilograms of tea that cost $\2.25$ per kilogram to make a mixture that costs $\3.40$ per kilogram?

Solve questions 10 to 21.

• How many litres of a solvent that costs $\80$ per litre must be mixed with 6 litres of a solvent that costs $\25$ per litre to make a solvent that costs $\36$ per litre?
• How many kilograms of hard candy that cost $\7.50$ per kg must be mixed with 24 kg of jelly beans that cost $\3.25$ per kg to make a mixture that sells for $\4.50$ per kg?
• How many kilograms of soil supplement that costs $\7.00$ per kg must be mixed with 20 kg of aluminum nitrate that costs $\3.50$ per kg to make a fertilizer that costs $\4.50$ per kg?
• A candy mix sells for $\2.20$ per kg. It contains chocolates worth $\1.80$ per kg and other candy worth $\3.00$ per kg. How much of each are in 15 kg of the mixture?
• A certain grade of milk contains 10% butterfat and a certain grade of cream 60% butterfat. How many litres of each must be taken so as to obtain a mixture of 100 litres that will be 45% butterfat?
• Solution A is 50% acid and solution B is 80% acid. How much of each should be used to make 100 cc of a solution that is 68% acid?
• A paint that contains 21% green dye is mixed with a paint that contains 15% green dye. How many litres of each must be used to make 600 litres of paint that is 19% green dye?
• How many kilograms of coffee that is 40% java beans must be mixed with coffee that is 30% java beans to make an 80-kg coffee blend that is 32% java beans?
• A caterer needs to make a slightly alcoholic fruit punch that has a strength of 6% alcohol. How many litres of fruit juice must be added to 3.75 litres of 40% alcohol?
• A mechanic needs to dilute a 70% antifreeze solution to make 20 litres of 18% strength. How many litres of water must be added?
• How many millilitres of water must be added to 50 millilitres of 100% acid to make a 40% solution?
• How many litres of water need to be evaporated from 50 litres of a 12% salt solution to produce a 15% salt solution?

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## Mixture word problems

More mixture word problems.

Example 2 and 3 are more challenging mixture word problems.

Suppose a car can run on ethanol and gas and you have a 15 gallons tank to fill. You can buy fuel that is either 30 percent ethanol or 80 percent ethanol.

How much of each type of fuel should you mix so that the mixture is 40 percent ethanol?

Solution #1: Use of one variable leading to a linear equation

Distance rate time word problems

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## Mixture Word Problems Lesson

• Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
• Demonstrate an understanding of the six-step method used for solving applications of linear equations
• Learn how to solve mixture word problems

## How to Solve Mixture Word Problems

Six-step method for solving word problems with linear equations in one variable.

• Read the problem and determine what you are asked to find
• If more than one unknown exists, we express the other unknowns in terms of this variable
• Write out an equation that describes the given situation
• Solve the equation
• State the answer using a nice clear sentence
• We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.

## Mixture Word Problems

Skills check:.

Solve each word problem.

A metallurgist needs to make 12 oz. of an alloy containing 71% copper. She is going to melt and combine one metal that is 78% copper with another metal that is 66% copper. How much of each should she use?

3 ounces of Ryan's Premium Molasses was made by combining 2 ounces of cane molasses which costs $1 per ounce with 1 ounce of beet molasses which costs$4 per ounce. Find the cost per ounce of the mixture.

How many gallons of an 80% sugar solution must be mixed with 7 gallons of a 35% sugar solution to make a 59% sugar solution?

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## Solving Mixture Problems

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:

## Mixture Word Problems and Solutions

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.

Mixture Problems: Example:

John has 20 ounces of a 20% of salt solution, How much salt should he add to make it a 25% solution?

Fill in the table with information given in the question.

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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.

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. Table $$\PageIndex{1}$$ The total value of all the coins is6.64.

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

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. 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. 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. 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. 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. ## Try It $$\PageIndex{1}$$ Michaela has2.05 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

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?

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.

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

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

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.

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

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.

• 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.

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.

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

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

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. 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.

• 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.

• 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*}

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

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

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.”

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

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

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

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.

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. 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

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. 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 has14,000 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. ## Free Mathematics Tutorials ## Mixture Problems With Solutions Mixture problems and their solutions are presented along with their solutions. Percentages are also used to solve these types of problems. Problem 1: How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution? Solution to Problem 1: Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hence x + 40 = y We shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term. 20% x + 50% * 40 = 30% y Substitute y by x + 40 in the last equation to obtain. 20% x + 50% * 40 = 30% (x + 40) Change percentages into fractions. 20 x / 100 + 50 * 40 / 100= 30 x / 100 + 30 * 40 / 100 Multiply all terms by 100 to simplify. 20 x + 50 * 40 = 30 x + 30 * 40 Solve for x. x = 80 liters 80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution. Problem 2: John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a 2% alcohol solution with a 7% alcohol solution. What are the quantities of each of the two solutions (2% and 7%) he has to use? Solution to Problem 2: Let x and y be the quantities of the 2% and 7% alcohol solutions to be used to make 100 ml. Hence x + y = 100 We now write mathematically that the quantity of alcohol in x ml plus the quantity of alcohol in y ml is equal to the quantity of alcohol in 100 ml. 2% x + 7% y = 5% 100 The first equation gives y = 100 - x. Substitute in the last equation to obtain 2% x + 7% (100 - x) = 5% 100 Multiply by 100 and simplify 2 x + 700 - 7 x = 5 * 100 Solve for x x = 40 ml Substitute x by 40 in the first equation to find y y = 100 - x = 60 ml Problem 3: Sterling Silver is 92.5% pure silver. How many grams of Sterling Silver must be mixed to a 90% Silver alloy to obtain a 500g of a 91% Silver alloy? Solution to Problem 3: Let x and y be the weights, in grams, of sterling silver and of the 90% alloy to make the 500 grams at 91%. Hence x + y =500 The number of grams of pure silver in x plus the number of grams of pure silver in y is equal to the number of grams of pure silver in the 500 grams. The pure silver is given in percentage forms. Hence 92.5% x + 90% y = 91% 500 Substitute y by 500 - x in the last equation to write 92.5% x + 90% (500 - x) = 91% 500 Simplify and solve 92.5 x + 45000 - 90 x = 45500 x = 200 grams. 200 grams of Sterling Silver is needed to make the 91% alloy. Problem 4: How many Kilograms of Pure water is to be added to 100 Kilograms of a 30% saline solution to make it a 10% saline solution. Solution to Problem 4: Let x be the weights, in Kilograms, of pure water to be added. Let y be the weight, in Kilograms, of the 10% solution. Hence x + 100 = y Let us now express the fact that the amount of salt in the pure water (which 0) plus the amount of salt in the 30% solution is equal to the amount of salt in the final saline solution at 10%. 0 + 30% 100 = 10% y Substitute y by x + 100 in the last equation and solve. 30% 100 = 10% (x + 100) Solve for x. x = 200 Kilograms. Problem 5: A 50 ml after-shave lotion at 30% alcohol is mixed with 30 ml of pure water. What is the percentage of alcohol in the new solution? Solution to Problem 5: The amount of the final mixture is given by 50 ml + 30 ml = 80 ml The amount of alcohol is equal to the amount of alcohol in pure water ( which is 0) plus the amount of alcohol in the 30% solution. Let x be the percentage of alcohol in the final solution. Hence 0 + 30% 50 ml = x (80) Solve for x x = 0.1817 = 18.75% Problem 6: You add x ml of a 25% alcohol solution to a 200 ml of a 10% alcohol solution to obtain another solution. Find the amount of alcohol in the final solution in terms of x. Find the ratio, in terms of x, of the alcohol in the final solution to the total amount of the solution. What do you think will happen if x is very large? Find x so that the final solution has a percentage of 15%. Solution to Problem 6: Let us first find the amount of alcohol in the 10% solution of 200 ml. 200 * 10% = 20 ml The amount of alcohol in the x ml of 25% solution is given by 25% x = 0.25 x The total amount of alcohol in the final solution is given by 20 + 0.25 x The ratio of alcohol in the final solution to the total amount of the solution is given by [ ( 20 + 0.25 x ) / (x + 200)] If x becomes very large in the above formula for the ratio, then the ratio becomes close to 0.25 or 25% (The above function is a rational function and 0.25 is its horizontal asymptote). This means that if you increase the amount x of the 25% solution, this will dominate and the final solution will be very close to a 25% solution. To have a percentage of 15%, we need to have [ ( 20 + 0.25 x ) / (x + 200)] = 15% = 0.15 Solve the above equation for x 20 + 0.25 x = 0.15 * (x + 200) x = 100 ml ## Popular Pages • Math Problems, Questions and Online Self Tests • Work Rate Problems with Solutions • Geometry Problems with Answers and Solutions - Grade 10 • Percent Math Problems • Free Algebra Questions and Problems with Answers ## Solver Title ## Generating PDF... • Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Expanded Form Mean, Median & Mode • Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems • Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. 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Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer. • How do you identify word problems in math? • Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem. • Is there a calculator that can solve word problems? • Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems. • What is an age problem? • An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age. word-problems-calculator • Middle School Math Solutions – Inequalities Calculator Next up in our Getting Started maths solutions series is help with another middle school algebra topic - solving... Please add a message. Message received. Thanks for the feedback. #### IMAGES 1. How to Solve Mixture Word Problems (with Pictures) 2. Mixture Word Problems (video lessons, examples and solutions) 3. Mixture Word Problems (video lessons, examples and solutions) 4. Mixture Word Problems (video lessons, examples and solutions) 5. Mixture Word Problems (video lessons, examples and solutions) 6. Algebra #### VIDEO 1. Mixture Word Problem 2. Intro Algebra Word Problems #37c 3. (LA13) Applications of Linear Systems 4. Systems of Equations: Word Problems 5. HOW TO SOLVE TRICKY MIXTURE WORD PROBLEMS EASY ALGEBRA 6. How to Increase Math Understanding with Mixture Problems Worksheets #### COMMENTS 1. How do you set up and solve mixture word problems? 0.30 (10 − w) mixture. 10. 0.15. 0.15 (10) = 1.5. When the problem is set up like this, we can usually use the last column to write our equation. In this case, the liters of acid within the 10% solution, plus the liters of acid within the 30% solution, must add up to the liters of acid within the 15% solution. 2. How to Solve Mixture Word Problems (with Pictures) For this problem you would label the three rows "20% solution," "15% solution," and "18% Mixture.". 2. Label and fill in the first column. The first column will include values that represent the part of the total mixture or solution each ingredient is. Label the column "Amount" and fill in the cell for each ingredient. 3. 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. The basic structure of this table is shown below: Example ... 4. 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. 5. Mixture Word Problems Solution #1: Use of one variable leading to a linear equation. Let x be the number of pounds of cashews. So, 150 - x will represent the number of almonds. Since each pound of the mixture costs 3 dollars, 150 pounds will cost 3 × 150 = 450 dollars. Cost of cashews + cost of almonds = 450. 2 × x + (150 - x) × 5 = 450. 2x + 150 × 5 - x × 5 = 450. 6. Mixture Word Problems (solutions, examples, videos) We set up and solve a mixture problem using a system of equations with two variables. Before solving the problem, a short introduction to what a -solution- with talking about a chemical mixture. Example: A chemist mixes a 12% acid solution with a 20% acid solution to make 300 milliliters of an 18% acid solution. 7. Mixture Problems This math video tutorial explains how to solve mixture problems that can be found in a typical algebra or a chemistry course. This video contains plenty of ... 8. Mixture Problems Here are some examples for solving mixture problems. Example 1. Coffee worth$1.05 per pound is mixed with coffee worth 85¢ per pound to obtain 20 pounds of a mixture worth 90¢ per pound. How many pounds of each type are used?

9. Solving Mixture Word Problems Lesson

Example 1: Solve each word problem. An acid solution was made by mixing 5 gallons of an 80% acid solution and 7 gallons of a 50% acid solution. Find the concentration of acid in the new mixture. For this problem, we don't need any variables. Let's think for a moment about how much acid is in each mixture:

10. Solving Mixture Problems

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. x. The 10 liters of our final mixture must have a total ...

11. Mixture Word Problems (examples, solutions, videos)

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.

12. How to Solve Mixture Word Problems

The mixture word problem we solve is "Henning is mixing raisins and nuts to make 25 pounds of trai... In this video we cover how to solve mixture word problems.

13. Mixture Word Problems

For a complete lesson on mixture problems, go to https://www.MathHelp.com - 1000+ online math lessons featuring a personal math teacher inside every lesson! ...

14. 3.3: Solve Mixture Applications

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 ...

15. Mixture Problems in Algebra

To solve mixture problems in algebra, we follow these four steps. 1.) Represent the unknown quantity with a variable. 2.) Create an equation involving that variable. 3.) Solve the equation for the ...

16. Mixture Word Problems

Need a custom math course? Visit https://www.MathHelp.com.This lesson covers mixture word problems. Students learn to solve "mixture" word problems -- for ex...

17. 3 Simple Steps for Solving Mixture Problems

Check out this simple, 3-step process and video for solving mixture problems.

18. Mixture Problems With Solutions

Change percentages into fractions. Multiply all terms by 100 to simplify. Solve for x. 80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution. Problem 2: John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a 2% alcohol solution with a 7% alcohol solution.

19. Algebra Mixture Word Problem

Learn how to set up a system of equations involving a mixture problem in this free math video tutorial by Mario's Math Tutoring.0:19 Example 1 Mixture Story ...

20. Word Problems Calculator

To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to ...

21. Algebra 1 Help: Solving Mixture Word Problems 1/2

http://www.greenemath.com/In this video, we learn how to set up and solve a basic word problem that deals with a mixture. Mixture word problems are one of th...

22. Algebra

Visit http://ilectureonline.com for more math and science lectures!Everyone at one point in their lives dreaded the infamous word problems in their algebra c...