Find practice problems and solutions for these concepts at Formulas and Mixtures Word Problems Practice Problems.
Some word problems can be solved using a specific formula. We must use the context and keywords of the problem to figure out which formula to use, and then substitute the values given in the problem into the formula to solve for the missing value. We'll look at some simple formulas, and then a more complex type of a word problem—a mixture problem.
Formulas: Subject Review
The formula for distance is D = rt, where D, distance, is equal to the product of r, rate, and t, time.
If an object travels at 30 miles per hour for three hours, it will travel 30 × 3 = 90 miles.
Sometimes we need to rearrange a formula to solve for a certain value. For instance, D = rt, but if we are given distance and rate and are looking for time, we must rewrite the formula to solve for time. Dividing both sides of the equation by r gives us t = . To find time given distance and rate, we must divide the distance by the rate.
In the same way, I = prt, but if we are looking for the principal given the interest, rate, and time, we must divide both sides of the formula by r and t to get p alone on one side of the equation. Principal is equal to . The formula for interest is I = prt, where I, interest, is equal to the product of p, principal, r, rate, and t, time. The rate is given as a percent and should be converted to a decimal.
A principal of $500, at a rate of 4% per year for six years, will make ($500)(.04)(6) = $120.
Given a temperature in Fahrenheit, F, we can convert to Celsius, C, or given a temperature in Celsius, we can convert to Fahrenheit: C = (F – 32) and F = C + 32.
If the temperature is 86° Fahrenheit, then the temperature in Celsius is (86 – 32) = (54) = 30°. If the temperature is 20° Celsius, then the temperature in Fahrenheit is (20) + 32 = 36 + 32 = 68°.
The formula for velocity is V = at, where V, velocity, is equal to the product of a, acceleration, and t, time.
If an object at rest accelerates at 5 meters per second for ten seconds, then its velocity is (5)(10) = 50 meters per second.
There are many formulas for converting one unit of measure to another. Here are just a few:
 To convert from grams, meters, or liters to milligrams, millimeters, or milliliters, multiply by 1,000.
 To convert from milligrams, millimeters, or milliliters to grams, meters, or liters divide by 1,000.
 To convert from inches to feet, divide the number of inches by 12.
 To convert from feet to inches, multiply the number of feet by 12.
 To convert from feet to yards, divide the number of feet by 3.
 To convert from yards to feet, multiply the number of yards by 3.
 To convert from ounces to pounds, divide the number of ounces by 16.
 To convert from pounds to ounces, multiply the number of pounds by 16.
Example
David's model car travels at 2 feet per second. If the car travels for ten seconds, how far will it travel?
Read the entire word problem.
We are given the rate at which a model car travels and the time in which it travels.
Identify the question being asked.
We are looking for the distance it travels.
Underline the keywords.
The keyword per can signal multiplication or division.
Cross out extra information and translate words into numbers.
There is no extra information in this problem.
List the possible operations.
To find the distance an object travels, we must multiply the rate by the time. The keyword per signals multiplication, since the rate gives us the distance traveled in one second, and we are looking for the distance traveled in ten seconds.
Write number sentences for each operation.
Our number sentence is the distance formula, with 2 substituted for r and 10 substituted for t: D = (2)(10)
Solve the number sentences and decide which answer is reasonable.
20 feet = (2)(10)
Check your work.
Since D = rt, we can check our distance answer by dividing the distance by the rate to see if that quotient equals the given time, 10 seconds: = 10 seconds. We also could have divided distance by the time to see if that quotient equals the given rate, 2 feet per second: = 2 feet per second.
At the same time we study a word problem for keywords, we should also search for words that tell u to use a specific formula. The word per can mean "rate," and the word travel often means that we are working with a distance. The words Celsius and Fahrenheit may indicate that we need to convert from one unit to the other.
Example
Kimberly hears that the temperature in Sydney, Australia, is 22° Celsius. What is the temperature in Sydney in degrees Fahrenheit?
Read the entire word problem.
We are given a temperature in degrees Celsius.
Identify the question being asked.
We are looking for a temperature in degrees Fahrenheit.
Underline the keywords and words that indicate formulas.
The words Celsius and Fahrenheit tell us that we need to use a temperature conversion formula.
Cross out extra information and translate words into numbers.
There is no extra information in this problem.
List the possible operations.
We will use multiplication and addition to convert from degrees Celsius to degrees Fahrenheit.
Write number sentences for each operation.
Substitute the temperature in degrees Celsius into the formula:
F = (22) + 32
Solve the number sentences and decide which answer is reasonable.
F = (22) + 32 = 39.6 + 32 = 71.6°
Check your work.
We can use the formula for converting Fahrenheit to Celsius to check our answer. C = (F – 32) = (71.6 – 32) = (39.6) = 22° Celsius
Inside Track
When taking final exams and state tests, often you are given a limited amount of time to finish the exam. You may find that using the eightstep process for every word problem takes a little too much time. If you can recognize a formula word problem after studying the keywords and reading the question carefully, you can skip the eightstep process and plug the given numbers right into the formula. The eightstep process can always help you find what operations or formulas to use, and it reminds you to check your work. It's the best way to go, but if you are pressed for time, it may be faster to use a formula.

Example
Dianne's pencil is 176 millimeters long. How long is her pencil in meters?
Let's say we need to work quickly and do not have time for the eightstep process. This problem is asking us to convert millimeters to meters. To convert from millimeters to meters, we must divide the number of millimeters by 1,000: = 0.176 meters. Dianne's pencil is 0.176 meters long.
Example
Dianne earns $228 in interest on $1,500 that she kept in a bank for four years. What was the interest rate on Dianne's principal?
We know that I = prt, so to solve for r, the interest rate, we must rearrange the equation. If we divide both sides of the equation by p and t, we have r = . Substitute the interest, principal, and time into the formula to find the interest rate: = 0.038 = 3.8%.
Caution
Read the units of each value in a word problem carefully. If a distance problem gives you a rate in meters per second and a time in seconds, you are ready to multiply, but if the rate is given in meters per second and the time is given in minutes, you must either convert the rate to meters per minutes or convert the time to seconds before multiplying. In the same way, if you are looking to find interest earned, the rate and time must also be given in the same units. You may need to convert one number from months to years.

Example
Fernando throws a baseball at a speed of 90 miles per hour. How many seconds does it take the ball to travel 60.5 feet?
We are given a rate and a distance, and we are looking for a time. However, our rate is given in miles per hour and our distance is given in feet. We must either convert our rate to feet per hour or convert the distance to miles. Since there are 5,280 feet in a mile, 60.5 feet is equal to , or approximately 0.0116 miles. The distance formula is D = rt, but we are looking for t, the time. Divide both sides of the equation by r, and t = , which equals 0.000129. Our rate was given in miles per hour, so this is the number of hours it takes the ball to travel 60.5 feet. There are 60 minutes in an hour and 60 seconds in a minute, which means that we must multiply the time in hours by 60 × 60 = 3,600. So, 0.000129 × 3,600 = 0.4644 seconds.
Pace Yourself
Use a map to find how far you live from your school. Use the time it takes you to walk to school to find your average rate of speed, in miles per hour. If you take a car or bus to school, use that time to find the average rate of speed in miles per hour of the car or bus.

Mixture Problems
A mixture problem is just like what it sounds—when two quantities are mixed together. For example, a salt solution is a mixture of salt and water. We can change the solution by adding or removing salt or water. Or we could mix a solution that has a lot of salt in it with a solution that has only a little salt in it to make a new solution.
We solve mixture problems by making tables to organize the data. Let's look at an example.
Example
Miguel has 10 ounces of a solution of salt and water that is 30% salt. He has 10 ounces of this solution, but he wants the solution to be 35% salt. How much salt must he add to the solution?
Our table will have three rows and three columns. We will have a row for our original solution, a row for what we are adding to the solution, and a row for our new, final solution. We will have a column to hold the percent concentration of salt in each solution, a column to hold the total quantity of each solution, and a column to hold the salt quantity of each solution:
The original solution is 10 ounces and 30% salt. Enter these figures in the table after converting 30% to a decimal. The number of ounces of salt in the solution is equal to the total number of ounces multiplied by the percent of ounces that are salt (10 × 0.30):
To this solution, we must add salt to make a solution that is 35% salt. We don't know how many ounces of salt to add, so we represent that quantity with x. The percent concentration of salt that we are adding is 100%, since we are adding pure salt to the solution. The salt quantity of what we are adding is equal to 100% of x, or 1 × x:
The total quantity in ounces of the final solution is equal to the total quantity in ounces of the original solution plus the quantity of salt in ounces that was added: 10 + x. The salt quantity of the final solution is equal to 0.35 multiplied by the total quantity, 10 + x:
The salt quantity of the final solution is also equal to the salt quantity of the original solution plus the salt quantity that was added: (0.3 × 10) + (1 × x) = 3 + x. Since the salt quantity of the final solution is equal to 3 + x and 0.35(10 + x), we can set these two values equal to each other and solve for x, the number of ounces of salt added:
3 + x = 0.35(10 + x)
3 + x = 3.5 + 0.35x
3 – 3 + x = 3.5 – 3 + 0.35x
x = 0.5 + 0.35x
0.65x = 0.5
x =
Miguel should add ounces of salt to the solution.
We can also remove water from a solution to change the concentration of the solution.
Example
Eli has 50 ounces of an alcohol solution that is 20% alcohol. How many ounces of water must be evaporated to make the concentration 40% alcohol?
Our table again will have three rows and three columns. Since we are looking for how much water will be evaporated, our table will be about the percent and quantity of water. Since the solution is 20% alcohol, it is 100% – 20% = 80% water, and we are looking for a final solution that is only 100% – 40% = 60% water:
Remember, we are removing water, so the total quantity will be 50 – x, not 50 + x:
The final water quantity is equal to 0.60(50 – x) and (0.8 × 50) – (1 × x), which is the water quantity of the original solution minus the quantity that was removed (evaporated). We can set these equations equal to each other and solve for x:
0.60(50 – x) = (0.8 × 50) – (1 × x)
30 – 0.6x = 40 – x
0.4x = 10
x = 25
To make the concentration of the solution 40% alcohol, 25 ounces of water must be evaporated.
Let's look at one last type of mixture problem: mixing two different solutions together.
Example
Dawn has 15 liters of a 20% alcohol solution. How many liters of 50% alcohol solution must be added to create a 30% alcohol solution?
We have 15 liters of a 0.20 solution, which means that the alcohol quantity of that solution is 0.2 × 15. We are adding an unknown quantity of a 0.50 solution, which has an alcohol quantity of 0.5 × x, and our final solution will be a 0.30 alcohol solution, with an alcohol quantity of 0.30 times the total quantity, 15 + x:
The variable x represents how many liters of the 50% solution we must add. Just as in the last examples, look at the final column. The sum of the first two rows, 0.2 × 15 and 0.5 × x, is equal to the product of the two columns of the last row, 0.30 and 15 + x, since both expressions are equal to the alcohol quantity of the final solution:
(0.2 × 15) + (0.5 × x) = 0.3(15 + x)
3 + 0.5x = 4.5 + 0.3x
0.2x = 1.5
x = 7.5
Dawn must add 7.5 liters of 50% alcohol solution to the 20% alcohol solution to create a 30% alcohol solution.
Pace Yourself
Try making a mixture yourself. Fill a beaker with 8 ounces of water, and add 2 ounces of salt. The beaker is now a 20% salt solution. How much salt must you add to make a 50% salt solution?

Summary
We can solve some word problems using formulas. After reviewing some common formulas, we made a small change to our eightstep process to look for words that signal which of these formulas to use. To save time, we learned to substitute values into the formula rather than using the entire eightstep process. We also learned how to rearrange formulas to solve for each part of a formula. Mixture problems are easier to solve with a table than with the eightstep process, so we learned how to set up a specific kind of table to solve this type of problem.
Find practice problems and solutions for these concepts at Formulas and Mixtures Word Problems Practice Problems.
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