Mia has a jar containing nickels and dimes worth $4.80 in total. If she has 1212 more dimes than nickels, how many of each coin does she have?

Nickels=24;Dimes=36Nickels=24;Dimes=36

Nickels=36;Dimes=24Nickels=36;Dimes=24

Nickels=30;Dimes=42Nickels=30;Dimes=42

Nickels=42;Dimes=30Nickels=42;Dimes=30

A technician needs to prepare a disinfectant by mixing a 70%70\% isopropyl alcohol solution with some 50%50\% solution to obtain 65%65\% alcohol. If the technician uses 500mL⁡500\operatorname{\mathrm{mL}} of the 50%50\% solution, how many mLmL of the 70%70\% solution must be added?

1500mL⁡1500\operatorname{\mathrm{m}\mathrm{L}}1500mL

600mL⁡600\operatorname{\mathrm{m}\mathrm{L}}600mL

667mL⁡667\operatorname{\mathrm{m}\mathrm{L}}667mL

67mL⁡67\operatorname{\mathrm{m}\mathrm{L}}67mL

3. Solving Word Problems

Mixture Problem Solving

3. Solving Word Problems

Mixture Problem Solving: Videos & Practice Problems

Video Lessons Worksheet

Topic summary

Mixture Problem Solving focuses on combining two or more quantities to form a total and modeling that situation with algebra. The key idea is to write each part of the mixture as a product of amount and value, rate, or percent, then add those parts to match the total. A common setup is $ \text{part}_1 + \text{part}_2 = \text{total} $, along with a second equation relating the amounts, such as the sum of the quantities.

For percent mixtures, convert each percent to decimal form and use $ \text{percent} \cdot \text{whole} = \text{partial amount} $. This is used for acid solutions, sugar syrup, and average cost situations. For simple interest mixtures, the total interest comes from adding the interest from each investment, using $ I = Prt $.

Across mixture problems, the process is consistent: identify the total, define variables for the unknown amounts, build an equation by summing the parts, rewrite in one variable using a second relationship, and solve. This same structure applies to coins, ticket sales, investments, chemical solutions, and tile cost mixtures.

Concept

Intro to Mixture Problems

Video duration:

Intro to Mixture Problems Video Summary

Mixture problems involve combining two or more quantities to form a single mixture, and they can be solved using systematic problem-solving steps similar to other word problems. These problems often appear in various contexts such as money and coins, tile laying, ticket sales, or mixing chemical solutions. A common example is determining the number of different coins in a collection when given the total value and a relationship between the quantities.

Consider a scenario where the total amount of money is \$2.20, composed of dimes and nickels. If there are eight more nickels than dimes, we can represent the number of dimes as d and the number of nickels as n. The total value equation is constructed by multiplying the number of each coin by its value and summing these amounts:

\[ 0.10d + 0.05n = 2.20 \]

Since the number of nickels is eight more than the number of dimes, we express this relationship as:

\[ n = d + 8 \]

Substituting this into the total value equation gives:

\[ 0.10d + 0.05(d + 8) = 2.20 \]

Distributing and simplifying leads to:

\[ 0.10d + 0.05d + 0.40 = 2.20 \] \[ 0.15d + 0.40 = 2.20 \]

Isolating the variable term by subtracting 0.40 from both sides:

\[ 0.15d = 1.80 \]

Dividing both sides by 0.15 to solve for d:

\[ d = \frac{1.80}{0.15} = 12 \]

Knowing the number of dimes, the number of nickels is:

\[ n = 12 + 8 = 20 \]

This approach highlights the key steps in solving mixture problems: first, identify the total quantity and the individual components; second, translate the problem into an algebraic equation by expressing one variable in terms of another; third, substitute and simplify to solve for a single variable; and finally, interpret the solution in the context of the problem.

Understanding how to set up and solve these equations is essential for tackling a wide range of mixture problems, whether involving money, materials, or chemical solutions. Mastery of this method enhances problem-solving skills and builds a strong foundation for more complex applications.

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Problem

Mia has a jar containing nickels and dimes worth \$4.80 in total. If she has $12$ more dimes than nickels, how many of each coin does she have?

$N i c k e l s = 36; D i m e s = 24$

$N i c k e l s = 24; D i m e s = 36$

$N i c k e l s = 30; D i m e s = 42$

$N i c k e l s = 42; D i m e s = 30$

Example

Intro to Mixture Problems Example 1

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Intro to Mixture Problems Example 1 Video Summary

To determine the number of student and adult tickets Market purchased for a basketball game, we start by defining variables and setting up equations based on the given information. Student tickets cost \$12 each, adult tickets cost \$20 each, and Market bought a total of 28 tickets, spending \$464 in total.

Let s represent the number of student tickets and a represent the number of adult tickets. The total cost equation can be expressed as:

\[ 12s + 20a = 464 \]

Additionally, since the total number of tickets is 28, we have the relationship:

\[ s + a = 28 \]

Solving for s in terms of a gives:

\[ s = 28 - a \]

Substituting this expression for s into the total cost equation yields:

\[ 12(28 - a) + 20a = 464 \]

Distributing 12 across the terms inside the parentheses results in:

\[ 336 - 12a + 20a = 464 \]

Combining like terms simplifies the equation to:

\[ 336 + 8a = 464 \]

Subtracting 336 from both sides gives:

\[ 8a = 128 \]

Dividing both sides by 8 isolates a:

\[ a = 16 \]

Using the value of a to find s:

\[ s = 28 - 16 = 12 \]

Therefore, Market purchased 12 student tickets and 16 adult tickets. This problem demonstrates how to apply systems of linear equations to real-world scenarios, using substitution to solve for unknown variables. Understanding how to translate word problems into algebraic expressions and manipulate equations is essential for solving similar problems involving cost, quantity, and total values.

Problem

Maya needs $48$ sq ft of tile for a backsplash. Basic tiles cost \$9 per sq ft and designer tiles cost \(25 per sq ft. She wants the overall average cost to be $dollar sign 14$ per sq ft. How many square feet of each tile should she use?

Basic tiles $equals 18 sq of ft of$ ; Designer tiles $equals 30 sq of ft of$

Basic tiles $equals 30 sq of ft of$ ; designer tiles $equals 18 sq of ft of$

Basic tiles $equals 15 sq of ft of$ ; Designer tiles $equals 33 sq of ft of$

Basic tiles $equals 33 sq of ft of$ ; Designer tiles $equals 15 sq of ft of$

Concept

Mixture Problems Involving Percents

Video duration:

Mixture Problems Involving Percents Video Summary

Mixture problems involving percents require applying the same fundamental methods used in standard mixture problems, with the added step of converting percentages to decimals. When solving these problems, the key is to express each component of the mixture as a product of its quantity and concentration, then sum these products to equal the total desired concentration.

For example, consider a chemist who needs 14 liters of a 50% acid solution but only has access to 40% and 70% acid solutions. To determine how much of each solution to mix, define variables for the unknown quantities: let x represent liters of the 40% solution and y represent liters of the 70% solution. The total volume equation is:

\[x + y = 14\]

Next, express the acid content from each solution as the product of the volume and its decimal concentration (percent converted to decimal). The total acid content in the final mixture is 50% of 14 liters, or:

\[0.4x + 0.7y = 0.5 \times 14\]

Substituting the total acid content:

\[0.4x + 0.7y = 7\]

Using the total volume equation, solve for one variable, for example:

\[y = 14 - x\]

Substitute this into the acid content equation:

\[0.4x + 0.7(14 - x) = 7\]

Distribute and simplify:

\[0.4x + 9.8 - 0.7x = 7\] \[-0.3x + 9.8 = 7\]

Isolate x:

\[-0.3x = 7 - 9.8\] \[-0.3x = -2.8\] \[x = \frac{-2.8}{-0.3} = 9.\overline{3}\]

This means approximately 9.33 liters of the 40% acid solution are needed. To find y, substitute back:

\[y = 14 - 9.33 = 4.67\]

Thus, 4.67 liters of the 70% acid solution are required. This approach highlights that solving percent mixture problems involves setting up equations based on total volume and total concentration, converting percentages to decimals, and solving for unknown quantities. Understanding how to translate percent concentrations into decimal form and apply algebraic methods is essential for accurately solving these types of problems.

Problem

A technician needs to prepare a disinfectant by mixing a $70 percent sign$ isopropyl alcohol solution with some $50 percent sign$ solution to obtain $65 percent sign$ alcohol. If the technician uses $500 m L$ of the $50 percent sign$ solution, how many $m L$ of the $70 percent sign$ solution must be added?

$1500$\operatorname{\mathrm{m}\]\mathrm{L}$}$

$600$\operatorname{\mathrm{m}\]\mathrm{L}$}$

$667$\operatorname{\mathrm{m}\]\mathrm{L}$}$

$67$\operatorname{\mathrm{m}\]\mathrm{L}$}$

Problem

A candy maker has $300 mL$ of a $45 percent sign$ sugar syrup. She wants to dilute it with pure water to make a $30 percent sign$ syrup. How many $m L$ of water should she add?

$450 mL$

$750 mL$

$150$\operatorname{\mathrm{m}\]\mathrm{L}$}$

$250$\operatorname{\mathrm{m}\]\mathrm{L}$}$

Example

Example 2: Mixing Simple Interest

Video duration:

Example 2: Mixing Simple Interest Video Summary

When solving problems involving investments with different interest rates, the simple interest formula is essential. Simple interest is calculated as the product of the principal amount, the interest rate, and the time period. In this scenario, a total of \$15,000 is split between two investments: one at 7% annual simple interest and the other at 9% annual simple interest. The total interest earned in one year from both investments is \$1,112.

To find the amounts invested at each rate, let x represent the amount invested at 7%, and y represent the amount invested at 9%. Since the total investment is \$15,000, the relationship between x and y is:

\[x + y = 15,000\]

The total interest earned from both investments can be expressed using the simple interest formula for each part:

\[0.07x + 0.09y = 1,112\]

To solve for one variable, express y in terms of x:

\[y = 15,000 - x\]

Substitute this into the interest equation:

\[0.07x + 0.09(15,000 - x) = 1,112\]

Distribute the 0.09:

\[0.07x + 1,350 - 0.09x = 1,112\]

Combine like terms:

\[-0.02x + 1,350 = 1,112\]

Subtract 1,350 from both sides:

\[-0.02x = 1,112 - 1,350\] \[-0.02x = -238\]

Divide both sides by -0.02 to isolate x:

\[x = \frac{-238}{-0.02} = 11,900\]

This means \$11,900 was invested at 7%. To find the amount invested at 9%, substitute back into the equation for y:

\[y = 15,000 - 11,900 = 3,100\]

Therefore, \$3,100 was invested at 9%. This approach demonstrates how to apply the simple interest formula and solve systems of linear equations to determine investment distributions based on total interest earned.

Problem

A retirement fund invests some money in government bonds at $2.5 percent sign$ annual simple interest, and $5,000 more than that amount in corporate bonds at $4 percent sign$ . If the total annual interest earned is $1,450, how much was invested in each type of bond?

Government bonds = $24,230.77; Corporate bonds = $19,230.77

Government bonds = $19,230.77; Corporate bonds = $24,230.77

Government bonds = $18,500.50; Corporate bonds = $23,500.50

Government bonds = $23,500.50; Corporate bonds = $18,500.50

Problem

Elena has $18,500 to invest. She invests some of it at $8.4 percent sign$ annual simple interest for $1$ year, and the remainder at $11.6 percent sign$ annual simple interest for $8$ months. At the end of the year, her total interest earned is $1,500. How much did she invest at each rate?

\$11,000 at 8.4% and \$7,500 at 11.6%

\$7,500 at 8.4% and \$11,000 at 11.6%

\$10,399.92 at 8.4% and \$8100.08 at 11.6%

\$8100.08 at 8.4% and \$10,399.9 at 11.6%

Here's what students ask on this topic:

To solve a mixture problem involving coins, start by defining variables for the number of each type of coin. For example, let $d$ be the number of dimes and $n$ be the number of nickels. Write an equation for the total value by multiplying the number of each coin by its value and summing them: $0.10d + 0.05n = T$ , where $T$ is the total amount of money. Use any additional information, such as relationships between the number of coins (e.g., $n = d + 8$ ), to express one variable in terms of the other. Substitute this into the total value equation to get one variable, then solve for it. Finally, find the other variable and interpret the results.

When solving mixture problems with percent solutions, first convert the percentages to decimals. Define variables for the amounts of each solution, such as $x$ liters of a 40% solution and $y$ liters of a 70% solution. Write two equations: one for the total volume, $x + y = 14$ liters, and one for the total amount of acid, $0.4x + 0.7y = 0.5 imes 14$ . Solve one equation for a variable (e.g., $y = 14 - x$ ) and substitute into the other. Simplify and solve for the single variable. This method ensures you find the correct amounts to mix to achieve the desired concentration.

To set up equations for mixture problems with two quantities, identify the variables representing each part of the mixture. Then, write an equation expressing the total as the sum of the parts, where each part is the product of the quantity and its value or rate. For example, if $a$ and $b$ are amounts with values $p$ and $q$ , the total $T$ is $T = pa + qb$ . Also, write a second equation relating the quantities, such as $a + b = $\text{whole}$$ . Use substitution to rewrite in one variable, then solve and interpret the solution.

First, define variables for the unknown quantities. Next, write an equation for the total amount or value by summing the parts, each as a product of quantity and rate or value. Then, use any given relationship between the variables to express one variable in terms of the other. Substitute this expression into the total equation to get a single-variable equation. Solve for that variable, then find the other variable using the relationship. Finally, interpret the results in the context of the problem.

To convert percent concentrations to decimals, divide the percent value by 100. For example, 40% becomes 0.40 and 70% becomes 0.70. This conversion is essential when setting up equations in mixture problems involving percent solutions because it allows you to multiply the decimal concentration by the volume to find the amount of the substance (like acid) in the solution. Always convert before building your equations to ensure accuracy.

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Mixture Problem Solving: Videos & Practice Problems

Intro to Mixture Problems

Intro to Mixture Problems Video Summary

Mia has a jar containing nickels and dimes worth \$4.80 in total. If she has $12$ more dimes than nickels, how many of each coin does she have?

Intro to Mixture Problems Example 1

Intro to Mixture Problems Example 1 Video Summary

Maya needs $48$ sq ft of tile for a backsplash. Basic tiles cost \$9 per sq ft and designer tiles cost \(25 per sq ft. She wants the overall average cost to be $dollar sign 14$ per sq ft. How many square feet of each tile should she use?

Mixture Problems Involving Percents

Mixture Problems Involving Percents Video Summary

A candy maker has $300 mL$ of a $45 percent sign$ sugar syrup. She wants to dilute it with pure water to make a $30 percent sign$ syrup. How many $m L$ of water should she add?

Example 2: Mixing Simple Interest

Example 2: Mixing Simple Interest Video Summary

A retirement fund invests some money in government bonds at $2.5 percent sign$ annual simple interest, and \(5,000 more than that amount in corporate bonds at $4 percent sign$ . If the total annual interest earned is \)1,450, how much was invested in each type of bond?

Elena has \(18,500 to invest. She invests some of it at $8.4 percent sign$ annual simple interest for $1$ year, and the remainder at $11.6 percent sign$ annual simple interest for $8$ months. At the end of the year, her total interest earned is \)1,500. How much did she invest at each rate?

Here's what students ask on this topic:

How do you solve a mixture problem involving coins with different values?

What is the method to solve mixture problems involving percent solutions?

How do you set up equations for mixture problems involving two quantities?

What steps should I follow to solve a mixture problem with two variables?

How do you convert percent concentrations to decimals in mixture problems?

Your Intermediate Algebra tutors

Mixture Problem Solving: Videos & Practice Problems

Intro to Mixture Problems

Intro to Mixture Problems Video Summary

Mia has a jar containing nickels and dimes worth \$4.80 in total. If she has 1212 more dimes than nickels, how many of each coin does she have?

Intro to Mixture Problems Example 1

Intro to Mixture Problems Example 1 Video Summary

Maya needs 4848 sq ft of tile for a backsplash. Basic tiles cost \$9 per sq ft and designer tiles cost \(25 per sq ft. She wants the overall average cost to be \)14\$14 per sq ft. How many square feet of each tile should she use?

Mixture Problems Involving Percents

Mixture Problems Involving Percents Video Summary

A technician needs to prepare a disinfectant by mixing a 70%70\% isopropyl alcohol solution with some 50%50\% solution to obtain 65%65\% alcohol. If the technician uses 500mL500\(\operatorname{\mathrm{mL}\)} of the 50%50\% solution, how many mLmL of the 70%70\% solution must be added?

A candy maker has 300mL300\(\operatorname{mL}\) of a 45%45\% sugar syrup. She wants to dilute it with pure water to make a 30%30\% syrup. How many mLmL of water should she add?

Example 2: Mixing Simple Interest

Example 2: Mixing Simple Interest Video Summary

A retirement fund invests some money in government bonds at 2.5%2.5\% annual simple interest, and \(5,000 more than that amount in corporate bonds at 4%4\%. If the total annual interest earned is \)1,450, how much was invested in each type of bond?

Elena has \(18,500 to invest. She invests some of it at 8.4%8.4\% annual simple interest for 11 year, and the remainder at 11.6%11.6\% annual simple interest for 88 months. At the end of the year, her total interest earned is \)1,500. How much did she invest at each rate?

Here's what students ask on this topic:

How do you solve a mixture problem involving coins with different values?

How do you solve a mixture problem involving coins with different values?

What is the method to solve mixture problems involving percent solutions?

What is the method to solve mixture problems involving percent solutions?

How do you set up equations for mixture problems involving two quantities?

How do you set up equations for mixture problems involving two quantities?

What steps should I follow to solve a mixture problem with two variables?

What steps should I follow to solve a mixture problem with two variables?

How do you convert percent concentrations to decimals in mixture problems?

How do you convert percent concentrations to decimals in mixture problems?

Your Intermediate Algebra tutors

Mia has a jar containing nickels and dimes worth \$4.80 in total. If she has $12$ more dimes than nickels, how many of each coin does she have?

Maya needs $48$ sq ft of tile for a backsplash. Basic tiles cost \$9 per sq ft and designer tiles cost \(25 per sq ft. She wants the overall average cost to be $dollar sign 14$ per sq ft. How many square feet of each tile should she use?

A candy maker has $300 mL$ of a $45 percent sign$ sugar syrup. She wants to dilute it with pure water to make a $30 percent sign$ syrup. How many $m L$ of water should she add?

A retirement fund invests some money in government bonds at $2.5 percent sign$ annual simple interest, and \(5,000 more than that amount in corporate bonds at $4 percent sign$ . If the total annual interest earned is \)1,450, how much was invested in each type of bond?

Elena has \(18,500 to invest. She invests some of it at $8.4 percent sign$ annual simple interest for $1$ year, and the remainder at $11.6 percent sign$ annual simple interest for $8$ months. At the end of the year, her total interest earned is \)1,500. How much did she invest at each rate?