3.8: Problem Solving Strategies: Use a Formula
Learning Objectives
- Read and understand given problem situations.
- Develop and apply the strategy: use a formula.
- Plan and compare alternative approaches to solving problems.
Introduction
In this chapter, we have been solving problems in which quantities vary directly with one another other. In this section, we will look at few examples of ratios and percents occurring in real-world problems. We will follow the Problem Solving Plan.
Step 1 Understand the problem
Read the problem carefully. Once you have read the problem, list all the components and data that are involved. This is where you will be assigning your variables.
Step 2 Devise a plan – Translate
Come up with a way to solve the problem. Set up an equation or formula.
Step 3 Carry out the plan – Solve
This is where you solve the formula you came up with in Step 2.
Step 4 Look – Check and Interpret
Check to see if you used all your information and that the answer makes sense.
It is important that you first know what you are looking for when solving problems in mathematics. Math problems often require that you extract information and use it in a definite procedure. You must collect the appropriate information and use it (using a strategy or strategies) to solve the problem Many times, you will be writing out an equation which will enable you to find the answer.
Example 1
An architect is designing a room that is going to be twice as long as it is wide. The total square footage of the room is going to be 722 square feet. What are the dimensions in feet of the room?
Step 1 Collect Relevant Information.
\begin{align*}\text{Width of room} & = \text{unknown} =x\\
\text{Length of room} & = 2 \times \text{width}\\
\text{Area of room} & = 722 \ \text{square feet}\end{align*}
Step 2 Make an Equation
\begin{align*}\text{Length of room} & = 2x\\
\text{Area of room} & = x \times 2x=2x^2\\
2x^2 & = 722\end{align*}
Step 3 Solve
\begin{align*}2x^2 & = 722 & & \text{Divide both sides by} \ 2\\
x^2 & = 361 & & \text{Take the ‘square root’ of both sides}.\\
x & = \sqrt{361} = 19 \\
2x & = 2 \times 19 = 38\end{align*}
Solution
The dimensions of the room are 19 feet by 38 feet.
Step 4 Check Your Answer
Is 38 twice 19?
\begin{align*}2 \times 19 =38 & & \text{TRUE} \ - \ \text{This checks out}.\end{align*}
Is 38 times 19 equal to 722?
\begin{align*}38 \times 19= 722 & & \text{TRUE} - \text{This checks out}\end{align*}
The answer checks out.
Example 2
A passenger jet initially climbs at 2000 feet per minute after take-off from an airport at sea level. At the four minute mark this rate slows to 500 feet per minute. How many minutes pass before the jet is at 20000 feet?
Step 1
\begin{align*}\text{Initial climb rate} & = \frac{2000\ feet}{1\ minute}\\
\text{Initial climb time} & = 4 \ minutes\\
\text{Final climb rate} & = \frac{500\ feet}{1\ minute}\\
\text{Final climb time} & = \text{unknown} = x\\
\text{Final altitude} & = 20000 \ feet\end{align*}
The first two pieces on information can be combined. Here is the result.
Height at four minute mark \begin{align*}= 4 \ minutes \ \cdot \frac{2000\ feet}{1\ minute} = 8000\ feet.\end{align*}
Step 2 Write an equation.
Since we know that the height at four minutes is 8000 feet, we need to find the time taken to climb the final \begin{align*}(20000 -8000) = 12000 \ feet\end{align*}
We will use \begin{align*}\text{distance} = \text{speed} \times \text{time}\end{align*}
\begin{align*}\text{time} = \frac{\text{distance}} {\text{speed}} = \text{distance} \ \cdot \left ( \frac{1} {\text{speed}}\right )\end{align*}
Step 3 Solve.
\begin{align*}x & = 12,0\cancel{0}\cancel{0} \ feet \cdot \left (\frac{1 \ minute}{5\cancel{0}\cancel{0} \ feet}\right ) & & \text{Note that the units of feet will also cancel}\\
x & = 24\ minutes\\
\text{total time} & = x+4\end{align*}
Solution
The time taken to reach 20000 feet is 28 minutes.
Step 4 check your answer
What is 4 times 2000?
\begin{align*}4 \times 2000 = 8000 & & \text{The initial climb is through} \ 8000 \ feet.\end{align*}
What is 24 times 500?
\begin{align*}24 \times 500 = 12000 & & \text{The second part of the climb is through} \ 8000 \ feet.\end{align*}
The total climb = initial climb + secondary climb = \begin{align*}(8000 + 12000) = 20000 \ feet\end{align*}
The answer checks out.
Example 3
The time taken for a moving body to travel a given distance is given by \begin{align*}time=\frac{distance}{speed}\end{align*}
Step 1 We will write out the most important information.
\begin{align*}\text{Distance} & = 10,000 \ meters\\ \text{Speed in air} & = \frac{340\ meters} {1\ second}\\ \text{Speed in water} & = \frac{1500\ meters} {1\ second}\\ \text{Time through air} & = \text{unknown} \ x\\ \text{Time through water} & = \text{unknown} \ y\\ \text{Delay} & = x-y\end{align*}
Step 2 We will convert this information into equations.
Time in air \begin{align*} x = 10,000 \ meters \cdot \frac{1\ second} {340\ meters}\end{align*}
Time in water \begin{align*} y = 10,000 \ meters \cdot \frac{1\ second} {15000\ meters}\end{align*}
Step 3 Solve for \begin{align*}x, y\end{align*} and the delay.
\begin{align*}x & = 29.41 \ seconds\\ y & = 6.67 \ seconds\\ \text{Delay} & = x - y = (29.41 -6.67) \ seconds\end{align*}
Solution
The delay between the two sound waves arriving is 22.7 seconds.
Step 4 Check that the answer works.
We need to think of a different way to explain the concept.
The actual time that the sound takes in air is 29.41 seconds. In that time, it crosses the following distance.
\begin{align*}\text{Distance} = \text{speed} \times \text{time} = 340 \times 29.41 = 9999 \ meters\end{align*}
The actual time that the sound takes in water is 6.67 seconds. In that time, it crosses the following distance of.
\begin{align*}\text{Distance} = \text{speed} \times \text{time} = 1500 \times 6.67 = 10005 \ meters\end{align*}
Both results are close to the 10000 meters that we know the sound traveled. The slight error comes from rounding our answer.
The answer checks out.
Example 4:
Deandra is looking over her paycheck. Her boss took tax from her earnings at a rate of 15%. A deduction to cover health insurance took one-twelfth of what was left. Deandra always saves one-third of what she gets paid after all the deductions. If Deandra worked 16 hours at $7.50 per hour, how much will she save this week?
Step 1 Collect relevant information.
Deductions:
\begin{align*}\text{Tax} & = 15\% = 0.15\\ \text{Health} & = \frac{1} {12}\\ \text{Savings} & = \frac{1} {3}\\ \text{Hours} & = 16\\ \text{Rate} & = \$7.50\ \text{per hour}\\ \text{Savings amount} & = \text{unknown} \ x\end{align*}
Step 2 Write an equation.
Deandra’s earnings before deductions \begin{align*} = 16 \times \$7.50 = \$120\end{align*}
Fraction remaining after tax \begin{align*}= 1 - 0.15 = 0.85\end{align*}
Fraction remaining after health \begin{align*}= 0.85 \left (1- \frac{1} {12}\right ) = 0.85 \left (\frac{11} {12}\right ) \approx 0.85 \cdot 0.91667 \approx 0.779167\end{align*}
Fraction to be saved \begin{align*}= \frac{1} {3} \cdot 0.779167 \approx 0.25972\end{align*}
Step 3 Solve
Amount to save = \begin{align*}0.25972 \cdot \$120 = \$31.1664 \end{align*} Round to two decimal places.
Solution
Deandra saves $31.17.
Step 4 Check your answer by working backwards.
If Deandra saves $31.17, then her take-home pay was \begin{align*}3 \times \$31.17 = \$93.51\end{align*}
If Deandra was paid $93.51, then before health deductions health she had \begin{align*}\$93.51 \cdot \frac{12}{11} = \$102.01\end{align*}
If Deandra had $102.01 after tax, then before tax she had \begin{align*}\$102.01 \cdot \frac{100}{85} = \$120.01\end{align*}
If Deandra earned $120.01 at $7.50 per hour, then she worked for \begin{align*}\frac{\$102.01}{\$7.50} = 16.002 \ hours\end{align*}
This is extremely close to the hours we know she worked (the difference comes from the fact we rounded to the nearest penny).
The answer checks out.
Lesson Summary
The four steps of the Problem Solving Plan are:
- Understand the problem
- Devise a plan – Translate
- Carry out the plan – Solve
- Look – Check and Interpret
Review Questions
Use the information in the problems to create and solve an equation.
- Patricia is building a sandbox for her daughter. It is to be five feet wide and eight feet long. She wants the height of the sand box to be four inches above the height of the sand. She has 30 cubic feet of sand. How high should the sand box be?
- A 500 sheet stack of copy paper is 1.75 inches high. The paper tray on a commercial copy machine holds a two foot high stack of paper. Approximately how many sheets is this?
- It was sale day in Macy’s and everything was 20% less than the regular price. Peter bought a pair of shoes, and using a coupon, got an additional 10% off the discounted price. The price he paid for the shoes was $36. How much did the shoes cost originally?
- Peter is planning to show a video file to the school at graduation, but is worried that the distance that the audience sits from the speakers will cause the sound and the picture to be out of sync. If the audience sits 20 meters from the speakers, what is the delay between the picture and the sound? (The speed of sound in air is 340 meters per second).
- Rosa has saved all year and wishes to spend the money she has on new clothes and a vacation. She will spend 30% more on the vacation than on clothes. If she saved $1000 in total, how much money (to the nearest whole dollar) can she spend on the vacation?
- On a DVD, data is stored between a radius of 2.3 cm and 5.7 cm. Calculate the total area available for data storage in square cm.
- If a Blu-ray \begin{align*}^{TM}\end{align*} DVD stores 25 gigabytes (GB), what is the storage density, in GB per square cm?
Review Answers
- 13 inches
- Approximately 6860 sheets
- $50
- 0.06 seconds
- Approximately $565
- \begin{align*}85.45 \ cm^2\end{align*}
- \begin{align*}0.293 \ GB/cm^2\end{align*}
Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9613.
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