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

N i c k e l s = 24 ; D i m e s = 36 Nickels=24;Dimes=36 N i c k e l s = 24 ; D im es = 36

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

N i c k e l s = 30 ; D i m e s = 42 Nickels=30;Dimes=42 N i c k e l s = 30 ; D im es = 42

N i c k e l s = 42 ; D i m e s = 30 Nickels=42;Dimes=30 N i c k e l s = 42 ; D im es = 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 500 m L ⁡ 500\operatorname{\mathrm{mL}} of the 50 % 50\% solution, how many mL of the 70 % 70\% solution must be added?

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

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

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

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

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

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

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

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

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

3. Solving Word Problems

Mixture Problem Solving

3. Solving Word Problems

Mixture Problem Solving: Videos & Practice Problems

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Topic summary

Mixture problems involve combining quantities, such as coins or solutions, to form a total amount. By expressing parts as products and summing them, we build equations like $0.1d + 0.05n = 2.2$ for coins or $0.4x + 0.7y = 0.5 \times 14$ for acid solutions. Using substitution to rewrite variables in terms of one unknown allows solving for quantities. Converting percents to decimals is essential when dealing with concentration mixtures. This approach applies to various contexts, reinforcing skills in forming and solving polynomial equations and understanding terms, coefficients, and constants.

concept

Intro to Mixture Problems

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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, this relationship can be expressed 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:

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

Combining like terms:

\[0.15d + 0.40 = 2.20\]

Subtracting 0.40 from both sides:

\[0.15d = 1.80\]

Dividing both sides by 0.15 to isolate 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, represent the total mixture as the sum of its parts using multiplication and addition; second, express one variable in terms of another based on the problem’s conditions; third, substitute to form a single-variable equation; and finally, solve for the unknown and interpret the result. Mastering this method allows for effective problem-solving across diverse mixture scenarios, reinforcing skills in algebraic manipulation and logical reasoning.

Problem

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

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

$Nickels=24;Dimes=36$

$Nickels=30;Dimes=42$

$Nickels=42;Dimes=30$

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 purchased for a basketball game, we start by defining variables: let s represent the number of student tickets and a represent the number of adult tickets. Student tickets cost \$12 each, while adult tickets cost \$20 each. The total amount spent on tickets was \$464, and the total number of tickets bought was 28.

The total cost can be expressed as the sum of the cost of student tickets and adult tickets, leading to the equation:

\[464 = 12s + 20a\]

Since the total number of tickets is 28, we also have:

\[s + a = 28\]

Solving for s in terms of a gives:

\[s = 28 - a\]

Substituting this expression for s into the cost equation yields:

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

Distributing 12, we get:

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

Combining like terms results in:

\[464 = 336 + 8a\]

Subtracting 336 from both sides gives:

\[128 = 8a\]

Dividing both sides by 8 isolates a:

\[a = 16\]

Using the value of a to find s:

\[s = 28 - 16 = 12\]

Therefore, 12 student tickets and 16 adult tickets were purchased. This problem demonstrates how to set up and solve a system of linear equations by substitution, a fundamental skill in algebra that connects real-world scenarios with mathematical modeling.

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 $y = 14 - x$ into the acid content equation gives:

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

Distributing and simplifying:

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

Isolating 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. Using the total volume equation, the amount of 70% solution is:

\[y = 14 - 9.33 = 4.67 \text{ liters}\]

Thus, mixing 9.33 liters of 40% acid solution with 4.67 liters of 70% acid solution yields 14 liters of a 50% acid solution.

In summary, solving percent mixture problems involves setting up equations based on total volume and total concentration, converting percentages to decimals, and solving for unknown quantities. This approach reinforces understanding of proportional relationships and algebraic manipulation, essential skills in chemistry and applied mathematics.

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 mL 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 mL 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 simple interest rates, it is essential to apply the simple interest formula, which states that the interest earned is equal to the principal amount multiplied by 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, with a combined yearly interest of \$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, summed together:

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

Substituting y from the first equation into the second gives:

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

Distributing the 0.09 yields:

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

Combining like terms results in:

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

Subtracting 1,350 from both sides:

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

Dividing 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 total investment equation:

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

Therefore, \$3,100 was invested at 9%. This problem highlights the importance of setting up systems of linear equations based on given financial conditions and using substitution to solve for unknown variables. Understanding how to translate word problems into algebraic expressions and apply the simple interest formula is crucial for solving real-world investment problems efficiently.

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 $dollar sign 1 comma 450$ , 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 $8.4 percent sign$ and \$7,500 at $11.6 percent sign$

\$7,500 at $8.4 percent sign$ and \$11,000 at $11.6 percent sign$

\$10,399.92 at $8.4 percent sign$ and \$8100.08 at $11.6 percent sign$

\$8100.08 at $8.4 percent sign$ and \$10,399.92 at $11.6 percent sign$

Here’s what students ask on this topic:

To solve mixture problems with coins such as dimes and nickels, start by defining variables for the number of each coin type. For example, let $d$ be the number of dimes and $n$ be the number of nickels. Use the values of the coins (dimes = \$0.10, nickels = \$0.05) to write an equation for the total amount of money: $0.10 \cdot d + 0.05 \cdot n = 2.20$ . If there is a relationship between the number of coins, such as having eight more nickels than dimes, express one variable in terms of the other: $n = d + 8$ . Substitute this into the total value equation to get one variable, then solve for that variable. Finally, find the other variable using the relationship. This method applies to many coin mixture problems.

Percent mixture problems involve combining solutions with different concentrations to get a desired concentration. First, define variables for the amounts of each solution, for example, $x$ and $y$ . Convert percentages to decimals (e.g., 40% = 0.4). Write an equation for the total amount: $x + y = 14$ liters. Then write an equation for the acid content: $0.4 \cdot x + 0.7 \cdot y = 0.5 \cdot 14$ . Substitute $y = 14 - x$ into the acid content equation to get one variable. Solve for $x$ , then find $y$ . This approach works for all percent mixture problems.

In mixture problems with two variables, start by identifying the quantities being combined, such as amounts or counts. Assign variables to these unknown quantities, for example, $x$ and $y$ . Write an equation representing the total quantity, like $x + y = total$ . Then write an equation for the combined value or concentration, such as $a \cdot x + b \cdot y = total value$ , where $a$ and $b$ are constants representing values per unit. Use any given relationships to express one variable in terms of the other, substitute, and solve for one variable. This systematic approach helps solve mixture problems efficiently.

Converting percentages to decimals is crucial in mixture problems because it allows you to perform accurate mathematical operations. Percentages represent parts per hundred, but algebraic equations require decimal form for multiplication. For example, 50% becomes 0.5, which can be multiplied by the total amount to find the actual quantity of a component, such as acid in a solution. Without converting, calculations would be incorrect, leading to wrong answers. This step ensures consistency and correctness when building and solving equations in percent mixture problems.

When a relationship between quantities is given, such as 'eight more nickels than dimes,' express one variable in terms of the other. For example, if $d$ is the number of dimes, then the number of nickels $n$ can be written as $n = d + 8$ . Substitute this expression into the main equation representing the total value or quantity. This substitution reduces the equation to one variable, making it easier to solve. After finding the value of one variable, use the relationship to find the other. This method simplifies solving mixture problems with linked quantities.

Your Beginning 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 $dollar sign 4.80$ 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 mL 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 $dollar sign 1 comma 450$ , 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 mixture problems involving coins like dimes and nickels?

What is the step-by-step method to solve percent mixture problems in algebra?

How do you write equations for mixture problems involving two variables?

Why is it important to convert percentages to decimals in mixture problems?

How do you solve mixture problems when given a relationship between quantities, like 'eight more nickels than dimes'?

Your Beginning 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\$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 mL 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 mL 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\$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 mixture problems involving coins like dimes and nickels?

How do you solve mixture problems involving coins like dimes and nickels?

What is the step-by-step method to solve percent mixture problems in algebra?

What is the step-by-step method to solve percent mixture problems in algebra?

How do you write equations for mixture problems involving two variables?

How do you write equations for mixture problems involving two variables?

Why is it important to convert percentages to decimals in mixture problems?

Why is it important to convert percentages to decimals in mixture problems?

How do you solve mixture problems when given a relationship between quantities, like 'eight more nickels than dimes'?

How do you solve mixture problems when given a relationship between quantities, like 'eight more nickels than dimes'?

Your Beginning Algebra tutors

Mia has a jar containing nickels and dimes worth $dollar sign 4.80$ 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 mL 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 $dollar sign 1 comma 450$ , 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?