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Formulas and Mixtures Word Problems Practice Problems

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Updated on Aug 24, 2011

To review these concepts, go to Formulas and Mixtures Word Problems Study Guide.

Practice

Directions: Solve the following word problems using the eight step process.

  1. Bettyann puts $2,000 in the bank and gains interest at a rate of 4.5% per year. How much interest has she earned after five years?
  2. Carlos throws a paper airplane across a room. If the plane accelerates from rest at a rate of 3.2 feet per second squared, what is the velocity of the plane after three seconds?
  3. Dimitri is riding a train that is traveling at 82 miles per hour. If he rides the train for two and a quarter hours, how far will he have traveled?
  4. Shelly's thermometer shows that it is 65° Fahrenheit today. If it was 5° Fahrenheit cooler yesterday, what was the temperature in degrees Celsius yesterday?
  5. If a jug weighs 86.4 ounces, how many pounds does it weigh?

Directions: Try to solve these word problems by identifying which formula to use. If you have trouble, use the eight-step process.

  1. How many feet are in 40 yards?
  2. A rock falling from a cliff accelerates from rest at a rate of 9.8 meters per second squared. How many seconds will it take for the rock to reach a velocity of 31.36 meters per second?
  3. If a plane travels 455 miles per hour, how long will it take to travel 2,375 miles? Round your answer to the nearest hundredth of an hour.
  4. Juan puts $5,000 in a bank account that gains interest at a rate of 4.7% per year. How much interest will Juan have earned after six months?
  5. One lap at the Atlanta Motor Speedway is 1.54 miles. If Travis completes a lap in 30.8 seconds, how fast, in miles per hour, was he traveling?

Directions: Round your answers to the nearest hundredth.

  1. Raymar's 25-ounce salt solution is 25% salt. How much salt must he add to the solution to make it 30% salt?
  2. Rosaria has a 40-ounce salt solution that is 10% salt. How much salt must she add to the solution to make it 25% salt?
  3. How many liters of water must be evaporated from 30 liters of a 45% alcohol solution to make the concentration 55% alcohol?
  4. Yao has 5 liters of a solution that is 55% alcohol. How many liters of water must be evaporated to make the concentration 80% alcohol?
  5. Libby has 22 liters of a 40% alcohol solution. How many liters of 70% alcohol solution must be added to create a 60% alcohol solution?
  6. Jana has 16 liters of a 5% alcohol solution. How many liters of a 45% alcohol solution must be added to create an 18% alcohol solution?

Solutions

  1. Read the entire word problem.
  2. We are given the principal, rate, and time.

    Identify the question being asked.

    We are looking for the interest earned by that principal at that rate over that time.

    Underline the keywords and words that indicate formulas.

    The keyword per can signal multiplication or division. The words interest and rate tell us that we need to use the formula for interest.

    Cross out extra information and translate words into numbers.

    There is no extra information in this problem.

    List the possible operations.

    To find interest earned given the principal, rate, and time, we must multiply those three values.

    Write number sentences for each operation.

    Convert 4.5% to a decimal. Our number sentence is the interest formula, with 2,000 substituted for p, 0.045 substituted for r, and 5 substituted for t: I = (2,000)(0.045)(5).

    Solve the number sentences and decide which answer is reasonable.

    I = (2,000)(0.045)(5) = $450

    Check your work.

    Since I = prt, we can check our interest answer by dividing the interest by the rate and time to see if that quotient equals the given principal, $2,000: = $2,000.

  3. Read the entire word problem.
  4. We are given the acceleration and hang time of a paper airplane.

    Identify the question being asked.

    We are looking for the velocity of the plane after three seconds.

    Underline the keywords and words that indicate formulas.

    The keyword per can signal multiplication or division. The words accelerates and rate tell us that we need to use the velocity formula.

    Cross out extra information and translate words into numbers.

    There is no extra information in this problem.

    List the possible operations.

    To find velocity given the acceleration and time, we must multiply the acceleration by the time.

    Write number sentences for each operation.

    Our number sentence is the velocity formula, with 3.2 substituted for a and 3 substituted for t: V = (3.2)(3).

    Solve the number sentences and decide which answer is reasonable.

    V = (3.2)(3) = 9.6 feet per second

    Check your work.

    Since V = at, we can check our velocity answer by dividing the velocity by the acceleration to see if that quotient equals the given time, 3 seconds: = 3 seconds.

  5. Read the entire word problem
  6. We are given the rate at which a train is traveling and the time the train travels at that rate.

    Identify the question being asked.

    We are looking for the distance traveled by Dimitri.

    Underline the keywords and words that indicate formulas.

    The keyword per can signal multiplication or division. The word travel could mean that we need to use the distance formula.

    Cross out extra information and translate words into numbers.

    There is no extra information in this problem, but we must translate two and a quarter into 2.25.

    List the possible operations.

    To find distance given the rate and time, we must multiply the rate by the time.

    Write number sentences for each operation.

    Our number sentence is the distance formula, with 82 substituted for r and 2.25 substituted for t: D = (82)(2.25).

    Solve the number sentences and decide which answer is reasonable.

    D = (82)(2.25) = 184.5 miles

    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, 2.25 hours: = 2.25 hours.

  7. Read the entire word problem.
  8. We are given a temperature in degrees Fahrenheit and the number of degrees the temperature increased from the previous day.

    Identify the question being asked.

    We are looking for yesterday's temperature in degrees Celsius.

    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.

    The temperature was 5° cooler the previous day, so we must first find the temperature in degrees Fahrenheit yesterday, and then convert that to degrees Celsius.

    Write number sentences for each operation.

    The temperature was 5° cooler yesterday, so we must subtract 5 from 65 to find the temperature yesterday:

    65 – 5

    Solve the number sentences and decide which answer is reasonable.

    65 – 5 = 60°

    Write number sentences for each operation.

    Now that we have the temperature yesterday in degrees Fahrenheit, we can convert it to degrees Celsius. Substitute the temperature in degrees Fahrenheit into the formula:

    Solve the number sentences and decide which answer is reasonable.

    Check your work.

    We can use the formula for converting Celsius to Fahrenheit to check our answer. F = C + 32, F = (15.56) + 32 = 28 + 32 = 60°. Since the temperature that day was 5° cooler than today, add 5 to 60. This should equal the temperature in degrees Fahrenheit today, 65: 5 + 60 = 65° Fahrenheit.

  9. Read the entire word problem.
  10. We are given a weight in ounces.

    Identify the question being asked.

    We are looking for that weight in pounds.

    Underline the keywords and words that indicate formulas.

    The words ounces and pounds tell us that we need to use a weight conversion formula.

    Cross out extra information and translate words into numbers.

    There is no extra information in this problem.

    List the possible operations.

    Since 1 ounce weighs less than 1 pound, we must divide the number of ounces to find the number of pounds.

    Write number sentences for each operation.

    There are 16 ounces in 1 pound, so we must divide the number of ounces by 16 to find the number of pounds:

    Solve the number sentences and decide which answer is reasonable.

    = 5.4 pounds

    Check your work.

    We can check our work by converting the number of pounds back to ounces. Multiply the number of pounds by 16, and that should equal the weight of the jug in ounces, 86.4; 5.4 × 16 = 86.4 ounces.

  11. There are 3 feet in 1 yard, so we must multiply the number of yards by 3 to find the number of feet: (40)(3) = 120 feet.
  12. We are given an acceleration and a velocity, so we must use the velocity formula to find the time. Since V = at, we must divide both sides of the equation by a to find the formula for t, time: t = = 3.2 seconds.
  13. We are given a rate of speed and a distance, so we must use the distance formula to find the time. Since D = rt, we must divide both sides of the equation by r to find the formula for t, time: t = × 5.22 hours.
  14. We are given a principal, a rate, and a time, so we must use the interest formula to find the interest. However, we are given the rate in years and the time in months, so we must convert the time to years. There are 12 months in a year, so six months is = 0.5 years. Now we are ready to use the formula: I = prt = (5,000)(0.047)(0.5) = $117.50.
  15. We are given a distance and a time, so we must use the distance formula to find the rate of speed. Since D = rt, we must divide both sides of the equation by t to find the formula for r, rate: r = 0.05. However, since we were given the time in seconds, this rate is the number of miles per second, not miles per hour. There are 60 seconds in a minute and 60 minutes in an hour, so to convert miles per second to miles per hour, we must multiply 0.05 by 3,600: 0.05 × 3,600 = 180 miles per hour.
  16. Make a table with three rows and three columns. Since we are looking for how much salt will be added, our table will be about the percent and quantity of salt. The original solution is 25%, or 0.25, salt. It is 25 ounces in total, which means that 0.25 × 25 of it is salt. We are adding an unknown quantity of salt, 100% of which is salt. Our final solution will be 30% salt:
  17. The total quantity of the final solution will be 25 + x, and 0.30 of it will be salt:

    The salt quantity equals 0.30(25 + x) and also equals the sum of 0.25 × 25 and 1 × x. Set these two expressions equal to each other and solve for x:

    0.30(25 + x) = (0.25 × 25) + (1 × x)

    7.5 + 0.3x = 6.25 + x

    1.25 = 0.7x

    x = 1.79 ounces

  18. Make a table with three rows and three columns. Since we are looking for how much salt will be added, our table will be about the percent and quantity of salt. The original solution is 10%, or 0.10, salt. It is 40 ounces in total, which means that 0.10 × 40 of it is salt. We are adding an unknown quantity of salt, 100% of which is salt. Our final solution will be 25% salt:
  19. The total quantity of the final solution will be 40 + x, and 0.25 of it will be salt:

    The salt quantity is the product of the percent concentration and the total quantity: 0.25(40 + x). It is also the sum of the salt quantity in the original solution and the amount added: 0.10 × 40 and 1 × x. Set these two expressions equal to each other and solve for x:

    0.25(40 + x) = (0.10 × 40) + (1 × x)

    10 + 0.25x = 4 + x

    6 = 0.75x

    x = 8 ounces

  20. Make a table with 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. The solution is 45% alcohol and 100% – 45% = 55% water. We are looking for a final solution that is only 100% – 55% = 45% water:
  21. Remember, we are removing water, so the total quantity will be 30 – x, not 30 + x:

    The final water quantity is equal to the percent concentration of the final solution times the total quantity of the solution: 0.45(30 – x). The final quantity of water is also equal to the original quantity of water minus the amount of water removed: (0.55 × 30) – (1 × x). Set these equations equal to each other and solve for x:

    0.45(30 – x) = (0.55 × 30) – (1 × x)

    13.5 – 0.45x = 16.5 – x

    0.55x = 3

    x = 5.45 liters, to the nearest hundredth

  22. Make a table with three rows and three columns. The solution is 55% alcohol and 100% – 55% = 45% water. We are looking for a final solution that is only 100% – 80% = 20% water:
  23. Remember, we are removing water, so the total quantity will be 5 – x, not 5 + x:

    The final water quantity is equal to the percent concentration of the final solution times the total quantity of the solution: 0.20(5 – x). The final quantity of water is also equal to the original quantity of water minus the amount of water removed: (0.45 × 5) – (1 × x). Set these equations equal to each other and solve for x:

    0.20(5 – x) = (0.45 × 5) – (1 × x)

    1 – 0.20x = 2.25 – x

    0.8x = 1.25

    x = 1.56 liters, to the nearest hundredth.

  24. Libby has 22 liters of a 0.40 solution, which means that the alcohol quantity of that solution is 0.4 × 22. We are adding an unknown quantity of a 0.70 solution, which has an alcohol quantity of 0.7 × x, and our final solution will be a 0.60 alcohol solution, with an alcohol quantity of 0.60 times the total quantity, 22 + x:
  25. The variable x represents how many liters of the 70% solution we must add. The final alcohol quantity is equal to the alcohol quantity of the original solution plus the alcohol quantity of the added solution: 0.4 × 22 and 0.7 × x. The final alcohol quantity is also equal to the product of percent concentration of the final solution and the total quantity of that solution: 0.6(22 + x). Set these equations equal to each other and solve for x:

    (0.4 × 22) + (0.7 × x) = 0.6(22 + x)

    8.8 + 0.7x = 13.2 + 0.6x

    0.1x = 4.4

    x = 44 liters

    Libby must add 44 liters of 70% alcohol solution to the 40% alcohol solution to create a 60% alcohol solution.

  26. Jana has 16 liters of a 0.05 solution, which means that the alcohol quantity of that solution is 0.05 × 16. We are adding an unknown quantity of a 0.45 solution, which has an alcohol quantity of 0.45 × x, and our final solution will be an 0.18 alcohol solution, with an alcohol quantity of 0.18 times the total quantity, 16 + x:
  27. The variable x represents how many liters of the 45% solution Jana must add. The final alcohol quantity is equal to the alcohol quantity of the original solution plus the alcohol quantity of the added solution: 0.05 × 16 and 0.45 × x. The final alcohol quantity is also equal to the product of percent concentration of the final solution and the total quantity of that solution: 0.18(16 + x). Set these equations equal to each other and solve for x:

    (0.05 × 16) + (0.45 × x) = 0.18(16 + x)

    0.8 + 0.45x = 2.88 + 0.18x

    0.27x = 2.08

    x = 7.70 liters, to the nearest hundredth

    Jana must add approximately 7.70 liters of 45% alcohol solution to the 5% alcohol solution to create an 18% alcohol solution.

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