1.4: Word problems
- Page ID
- 45030
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Now, let’s apply the techniques from this chapter to some common word problems. Word problems can be tricky. The goal is becoming proficient in translating an English sentence into a mathematical sentence. In this section, we focus on word problems modeled by a linear equation and solve. We discuss geometry problems including perimeter and triangles, number, and distance problems.
Number Problems
If \(28\) less than five times a number is \(232\), what is the number?
Solution
First, let \(n\) be the number. Now, translate the key words in the sentence: \[\underset{\color{blue}{5n-28}}{\color{blue}{\underbrace{\color{black}{...28\text{ less than }\underset{5n}{\underbrace{\text{five times a number}}}}}}}\underset{=}{\underbrace{\text{is}}}\underset{232}{\underbrace{232}}...\nonumber\]
Notice, after translating, we obtain the equation \[5n-28=232\nonumber\]
Let's solve: \[\begin{array}{rl}5n-8=232&\text{Isolate the variable term }5n \\ 5n-28+\color{blue}{28}\color{black}{}=232+\color{blue}{28}\color{black}{}&\text{Simplify} \\ 5n=260&\text{Multiply by the reciprocal of }5 \\ \color{blue}{\frac{1}{5}}\color{black}{}\cdot 5n=260\cdot\color{blue}{\frac{1}{5}}\color{black}{}&\text{Simplify} \\ n=52&\text{Solution}\end{array}\nonumber\]
Thus, the number is \(52\).
Fifteen more than three times a number is the same as ten less than six times the number. What is the number?
Solution
Notice, this sentence is a bit more challenging than Example \(\PageIndex{1}\), but we still follow the method. Let \(n\) be the number.
\[\underset{\color{blue}{3n+15}}{\color{blue}{\underbrace{\color{black}{\text{Fifteen more than }\underset{3n}{\underbrace{\text{three times a number }}} }}}}\underset{=}{\underbrace{\text{ is the same as }}}\underset{\color{blue}{6n-10}}{\color{blue}{\underbrace{\color{black}{\text{ ten less than }\underset{6n}{\underbrace{\text{ six times the number}}}}}}}\nonumber\]
Notice, after translating, we obtain the equation \[3n+15=6n-10\nonumber\]
Let's solve: \[\begin{array}{rl}3n+15=6n-10&\text{Combine like terms} \\ 3n+15+\color{blue}{(-6n)}\color{black}{}=6n-10+\color{blue}{(-6n)}\color{black}{}&\text{Simplify} \\ -3n+15=-10&\text{Isolate the variable term} \\ -3n+15+\color{blue}{(-15)}\color{black}{}=-10+\color{blue}{(-15)}\color{black}{}&\text{Simplify} \\ -3n=-25&\text{Multiply by the reciprocal of }-3 \\ \color{blue}{-\frac{1}{3}}\color{black}{}\cdot -3n=-25\cdot\color{blue}{-\frac{1}{3}}\color{black}{}&\text{Simplify} \\ n=\frac{25}{3}&\text{Solution}\end{array}\nonumber\]
Thus, the number is \(\frac{25}{3}\).
Consecutive Integers
Another type of number problem involves consecutive integers.
Consecutive integers are integers that come one after the other (such as \(3,\: 4,\: 5,\) or \(−3,\: −2,\: −1\)).
- If we are trying to find several consecutive integers, it important to identify the first integer and then assign names to the following integers. E.g., if \(x\) is the first integer, then \(x + 1\) will be the next, and \(x + 2\) will be the following, and so on.
- If we are trying to find several even or odd consecutive integers, it important to identify the first integer and then assign names to the following even or odd integers. E.g., if \(x\) is the first integer, then \(x + 2\) will be the next odd or even integer, and \(x + 4\) will be the following, and so on.
The sum of three consecutive positive integers is \(93\). What are the positive integers?
Solution
Since we want to obtain three consecutive positive integers, then we can assign each integer as the following: \[\begin{array}{rl}x&\text{is the first integer} \\ x+1&\text{is the second integer} \\ x+2&\text{is the third integer}\end{array}\nonumber\]
The sum of these three integers is given to be \(93\). Translating this into an equation, we get \[x+(x+1)+(x+2)=93\nonumber\]
Let’s solve this equation for \(x\). Then we can obtain the other two integers.
\[\begin{array}{rl}x+(x+1)+(x+2)=93&\text{Rewrite without the parenthesis} \\ x+x+1+x+2=93&\text{Combine like terms} \\ 3x+3=93&\text{Isolate the variable term} \\ 3x+3+\color{blue}{(-3)}\color{black}{}=93+\color{blue}{(-3)}\color{black}{}&\text{Simplify} \\ 3x=90&\text{Multiply by the reciprocal of }3 \\ \color{blue}{\frac{1}{3}}\color{black}{}\cdot 3x=90\cdot\color{blue}{\frac{1}{3}}\color{black}{}&\text{Simplify} \\ x=30&\text{First integer}\end{array}\nonumber\]
Since the first integer is \(30\), the next two integers would be \[\begin{array}{rl}30+1=31&\text{is the second even integer} \\ 30+2=32&\text{is the third even integer}\end{array}\nonumber\]
Thus, the integers are \(30,\: 31,\) and \(32\).
The sum of three consecutive even positive integers is \(246\). What are the numbers?
Solution
Since we want to obtain three consecutive even positive integers, then we can assign each integer as the following: \[\begin{array}{rl}x&\text{is the first odd integer} \\ x+2&\text{is the second odd integer} \\ x+4&\text{is the third odd integer}\end{array}\nonumber\]
The sum of these three even integers is given to be \(246\). Translating this into an equation, we get \[x+(x+2)+(x+4)=246\nonumber\]
Let’s solve this equation for \(x\). Then we can obtain the other two integers.
\[\begin{array}{rl}x+(x+2)+(x+4)=246&\text{Rewrite without the parenthesis} \\ x+x+2+x+4=246&\text{Combine like terms} \\ 3x+6=246&\text{Isolate the variable term} \\ 3x+6+\color{blue}{(-6)}\color{black}{}=246+\color{blue}{(-6)}\color{black}{}&\text{Simplify} \\ 3x=240&\text{Multiply by the reciprocal of }3 \\ \color{blue}{\frac{1}{3}}\color{black}{}\cdot 3x=240\cdot\color{blue}{\frac{1}{3}}\color{black}{}&\text{Simplify} \\ x=80&\text{First integer}\end{array}\nonumber\]
Since the first integer is 80, the next two even integers would be \[\begin{array}{rl}80+2=82&\text{is the second even integer} \\ 80+4=32&\text{is the third even integer}\end{array}\nonumber\]
Thus, the integers are \(80,\: 82,\) and \(84\).
Find three consecutive odd positive integers so that the sum of twice the first integer, the second integer, and three times the third integer is \(152\).
Solution
Since we want to obtain three consecutive odd positive integers, then we can assign each integer as the following: \[\begin{array}{rl}x&\text{is the first odd integer} \\ x+2&\text{is the second integer} \\ x+4&\text{is the third odd integer}\end{array}\nonumber\]
The sum of twice the first integer, the second integer, and three times the third integer is given to be \(152\). Translating this into an equation, we get \[2\cdot x+(x+2)+3\cdot (x+4)=152\nonumber\]
Let’s solve this equation for \(x\). Then we can obtain the other two integers.
\[\begin{array}{rl}2\cdot x+(x+2)+3\cdot (x+4)=152&\text{Rewrite without the parenthesis} \\ 2x+x+2+3x+12=152&\text{Combine like terms} \\ 6x+14=152&\text{Isolate the variable term} \\ 6x+14+\color{blue}{(-14)}\color{black}{}=152+\color{blue}{(-14)}\color{black}{}&\text{Simplify} \\ 6x=138&\text{Multiply by the reciprocal of }6 \\ \color{blue}{\frac{1}{6}}\color{black}{}\cdot 6x=138\cdot\color{blue}{\frac{1}{6}}\color{black}{}&\text{Simplify} \\ x=23&\text{First integer}\end{array}\nonumber\]
Since the first integer is \(23\), the next two odd integers would be \[\begin{array}{rl}23+2=25&\text{is the second odd integer} \\ 23+4=27&\text{is the third odd integer}\end{array}\nonumber\]
Thus, the integers are \(23,\: 25,\) and \(27\).
Perimeter Problems
Another problem from geometry involves perimeter or the distance around an object.
The formula for the perimeter of a rectangle is given by \[P=2w+2\ell ,\nonumber\] where \(w\) is the width and \(ℓ\) is the length of the rectangle.
The perimeter of a rectangle is \(44\) cm. The length is \(5\) less than double the width. Find the dimensions.
Solution
Let \(w\) be the width of the rectangle. Then the length is \(2w − 5\). Since the perimeter is \(44\text{ cm}\), the we can use the perimeter formula to obtain the dimensions.
\[\begin{array}{rl}P=2w+2\ell &\text{Substitute in the width, length, and perimeter} \\ 44=2(w)+2(2w-5)&\text{Rewrite with no parenthesis} \\ 44=2w+4w-10&\text{Combine like terms} \\ 44=6w-10&\text{Isolate the variable term} \\ 54=6w&\text{Multiply by the reciprocal of }6 \\ 9=w&\text{Length of the rectangle}\end{array}\nonumber\]
Since the width is \(9\text{ cm}\), then the length is \((2(9) − 5) = 13\text{ cm}\).
Triangles
Given a triangle, the sum of the three angles is \(180^{\circ}\). I.e., if the angles in a triangle are \(a^{\circ},\: b^{\circ},\) and \(c^{\circ}\), then \[a^{\circ}+b^{\circ}+c^{\circ}=180^{\circ}\nonumber\]
German mathematician Bernhart Thibaut in 1809 tried to prove that the angles of a triangle add to \(180\) without using Euclid’s parallel postulate (a point of much debate in math history). He created a proof, but it was later shown to have an error in the proof.
The second angle of a triangle is double the first. The third angle is \(40\) less than the first. Find the three angles.
Solution
Let \(x\) be the measure of the first angle. Then \[\begin{array}{rl}2x&\text{is the measure of the second angle} \\ x-40&\text{is the measure of the third angle}\end{array}\nonumber\]
Since the sum of these three angles is \(180^{\circ}\), then we can write the equation \[x+2x+(x-40)=180\nonumber\]
Let’s solve for the first angle \(x\):
\[\begin{array}{rl}x + 2x + (x − 40) = 180 &\text{Rewrite without parenthesis} \\ x + 2x + x − 40 = 180 &\text{Combine like terms} \\ 4x − 40 = 180 &\text{Isolate the variable term} \\ 4x=220&\text{Multiply by the reciprocal of }4 \\ x=55&\text{Measure of the first angle}\end{array}\nonumber\]
Since the measure of the first angle is \(55^{\circ}\), then the measures of the second and third angle are
\[\begin{array}{rl} 2(55)=110^{\circ}&\text{is the measure of the second angle} \\ 55-40=15^{\circ}&\text{is the measure of the third angle}\end{array}\nonumber\]
Uniform Motion Problems
Another common application of linear equations is uniform motion problems. When solving uniform motion problems, we use the relationship \(rt = d\) or \[\text{rate (speed)}\cdot\text{time}=\text{distance}\nonumber\]
For example, if a person were to travel \(30\) miles per hour (mph) for \(4\) hours, to find the total distance we would multiply rate and the time: \((30)(4) = 120\). Hence, this person traveled a distance of \(120\) miles. The problems we solve in this section are just a few more steps than described. To keep the information in the problem organized, we use tables.
Opposite Directions
Two joggers start from opposite ends of an \(8\) mile course running towards each other. One jogger is running at a rate of \(4\) miles per hour, and the other is running at a rate of \(6\) miles per hour. After how long will the joggers meet?
Solution
First, we can make a table to organize the given information and then create an equation. Let \(t\) represent the length of time until the joggers meet.
Table \(\PageIndex{1}\)
| rate | time | distance | |
|---|---|---|---|
| Jogger 1 | \(4\) | \(t\) | \(4t\) |
| Jogger 2 | \(6\) | \(t\) | \(6t\) |
Now we can set up the equation. If the total distance is \(8\) miles, then \[4t=6t+8,\nonumber\] i.e., the sum of Jogger 1’s distance and Jogger 2’s distance is \(8\) miles. Let’s solve.
\[\begin{array}{rl}4t+6t=8&\text{Combine like terms} \\ 10t=8&\text{Multiply by the reciprocal of }10 \\ t=\frac{4}{5}&\text{Hours until they meet}\end{array}\nonumber\]
It will be \(\frac{4}{5}\) hours (or \(48\) minutes) until they meet.
Bob and Fred start from the same point and walk in opposite directions. Bob walks \(2\) miles per hour faster than Fred. After \(3\) hours they are \(30\) miles apart. How fast did each walk?
Solution
First, we can make a table to organize the given information and then create an equation. Let \(r\) represent the rate of Fred.
Table \(\PageIndex{2}\)
| rate | time | distance | |
|---|---|---|---|
| Bob | \(r+2\) | \(3\) | \(3(r+2)\) |
| Fred | \(r\) | \(3\) | \(3r\) |
Now we can set up the equation. If the total distance is \(30\) miles, then \[3(r+2)+3r=30,\nonumber\] i.e., the sum of Bob’s distance and Fred’s distance is \(30\) miles. Let’s solve.
\[\begin{array}{rl}3(r+2)+3r=30&\text{Distribute} \\ 3r+6+3r=30&\text{Combine like terms} \\ 6r+6=30&\text{Isolate the variable term} \\ 6r=24&\text{Multiply by the reciprocal of }6 \\ r=4&\text{Rate of Fred}\end{array}\nonumber\]
Since the rate of Fred is \(4\) mph, then Bob’s rate is \(6\) mph \((4 + 2 = 6)\).
Two campers left their campsite by canoe and paddled downstream at an average speed of \(12\) miles per hour. They turned around and paddled back upstream at an average rate of \(4\) miles per hour. The total trip took \(1\) hour. After how much time did the campers turn around downstream?
Solution
First, we can make a table to organize the given information and then create an equation. Let \(t\) represent the time it took to travel upstream.
Table \(\PageIndex{3}\)
| rate | time | distance | |
|---|---|---|---|
| upstream | \(4\) | \(t\) | \(4t\) |
| downstream | \(12\) | \(1-t\) | \(12(1-t)\) |
Now we can set up the equation. If the upstream and downstream routes’ distances are the same, then \[4t=12(1-t)\nonumber\]
Let’s solve.
\[\begin{array}{rl}4t=12(1-t)&\text{Distribute} \\ 4t=12-12t&\text{Combine like terms} \\ 16t=12&\text{Multiply by the reciprocal of }16 \\ t=\frac{12}{16}&\text{Reduce} \\ t=\frac{3}{4}&\text{Time going upstream}\end{array}\nonumber\]
Since the time going upstream is \(\frac{3}{4}\) hours, then downstream’s time is \(\frac{1}{4}\) hours \(\left( 1 −\frac{3}{4} = \frac{1}{4}\right)\). Thus, the campers spent \(15\) minutes going downstream.
Catch-Up
Mike leaves his house traveling \(2\) miles per hour. Joy leaves \(6\) hours later to catch up with him traveling \(8\) miles per hour. How long will it take her to catch up with him?
Solution
First, we can make a table to organize the given information and then create an equation. Let \(t\) represent the time Joy traveled.
Table \(\PageIndex{4}\)
| rate | time | distance | |
|---|---|---|---|
| Mike | \(2\) | \(t+6\) | \(2(t+6)\) |
| Joy | \(8\) | \(t\) | \(8t\) |
Now we can set up the equation. If Joy catches up to Mike, then Mike and Joy would have traveled the same distance. Hence, giving the equation \[2(t+6)=8t,\nonumber\] i.e., Mike’s distance and Joy’s distance are the same. Let’s solve.
\[\begin{array}{rl}2(t+6)=8t&\text{Distribute} \\ 2t+12=8t&\text{Combine like terms} \\ 12=6t&\text{Multiply by the reciprocal of }6 \\ 2=t&\text{Time Joy traveled}\end{array}\nonumber\]
Since the time Joy traveled was \(2\) hours, then Mike traveled \(8\) hours \((2 + 6 = 8)\). Thus, it took \(2\) hours for Joy to catch up with Mike.
The \(10,000\)-meter race is the longest standard track event. Ten-thousand meters is approximately \(6.2\) miles. The current (at the time of printing) world record for this race is held by Ethiopian Kenenisa Bekele with a time of \(26\) minutes, \(17.53\) seconds. That is a rate of \(12.7\) miles per hour.
Total Time
On a \(130\)-mile trip, a car traveled at an average speed of \(55\) mph and then reduced its speed to \(40\) mph for the remainder of the trip. The trip took \(2.5\) hours. For how long did the car travel \(40\) mph?
Solution
First, we can make a table to organize the given information and then create an equation. Let \(t\) represent the time the car traveled at the faster speed.
Table \(\PageIndex{5}\)
| rate | time | distance | |
|---|---|---|---|
| First part | \(55\) | \(t\) | \(55t\) |
| Second part | \(40\) | \(2.5-t\) | \(40(2.5-t)\) |
Now we can set up the equation. Since the total distance of the trip was \(130\) miles, then \[55t+40(2.5-t)=130,\nonumber\] i.e., the sum of the first part’s distance and the second part’s distance is \(130\) miles. Let’s solve.
\[\begin{array}{rl} 55t + 40(2.5 − t) = 130 &\text{Distribute} \\ 55t + 100 − 40t = 130 &\text{Combine like terms} \\ 15t+100=130&\text{Isolate the variable term} \\ 15t=30&\text{Multiply by the reciprocal of }15 \\ t=2&\text{First part's travel time}\end{array}\nonumber\]
Since the first part of the trip took \(2\) hours, then the car traveled \(0.5\) hours (or \(30\) minutes) at \(40\) mph.
Word Problems Homework
When five is added to three more than a certain number, the result is \(19\). What is the number?
If five is subtracted from three times a certain number, the result is \(10\). What is the number?
When \(18\) is subtracted from six times a certain number, the result is \(−42\). What is the number?
A certain number added twice to itself equals \(96\). What is the number?
A number plus itself, plus twice itself, plus \(4\) times itself, is equal to \(−104\). What is the number?
Sixty more than nine times a number is the same as two less than ten times the number. What is the number?
Eleven less than seven times a number is five more than six times the number. Find the number.
Fourteen less than eight times a number is three more than four times the number. What is the number?
The sum of three consecutive integers is \(108\). What are the integers?
The sum of three consecutive integers is \(−126\). What are the integers?
Find three consecutive integers such that the sum of the first, twice the second, and three times the third is \(−76\).
The sum of two consecutive even integers is \(106\). What are the integers?
The sum of three consecutive odd integers is \(189\). What are the integers?
The sum of three consecutive odd integers is \(255\). What are the integers?
Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is \(70\).
The second angle of a triangle is the same size as the first angle. The third angle is \(12\) degrees larger than the first angle. How large are the angles?
Two angles of a triangle are the same size. The third angle is \(12\) degrees smaller than the first angle. Find the measure the angles.
Two angles of a triangle are the same size. The third angle is \(3\) times as large as the first. How large are the angles?
The third angle of a triangle is the same size as the first. The second angle is \(4\) times the third. Find the measure of the angles.
The second angle of a triangle is \(3\) times as large as the first angle. The third angle is \(30\) degrees more than the first angle. Find the measure of the angles.
The second angle of a triangle is twice as large as the first. The measure of the third angle is \(20\) degrees greater than the first. How large are the angles?
The second angle of a triangle is three times as large as the first. The measure of the third angle is \(40\) degrees greater than that of the first angle. How large are the three angles?
The second angle of a triangle is five times as large as the first. The measure of the third angle is \(12\) degrees greater than that of the first angle. How large are the angles?
The second angle of a triangle is three times the first, and the third is \(12\) degrees less than twice the first. Find the measures of the angles.
The second angle of a triangle is four times the first and the third is \(5\) degrees more than twice the first. Find the measures of the angles.
The perimeter of a rectangle is \(150\) cm. The length is \(15\) cm greater than the width. Find the dimensions.
The perimeter of a rectangle is \(304\) cm. The length is \(40\) cm longer than the width. Find the length and width.
The perimeter of a rectangle is \(152\) meters. The width is \(22\) meters less than the length. Find the length and width.
The perimeter of a rectangle is \(280\) meters. The width is \(26\) meters less than the length. Find the length and width.
The perimeter of a college basketball court is \(96\) meters and the length is \(14\) meters more than the width. What are the dimensions?
A is \(60\) miles from B. An automobile at A starts for B at the rate of \(20\) miles per hour at the same time that an automobile at B starts for A at the rate of \(25\) miles an hour. How long will it be before the automobiles meet?
Two automobiles are \(276\) miles apart and start at the same time to travel toward each other. They travel at rates differing by \(5\) miles per hour. If they meet after \(6\) hours, find each rate.
Two trains travel toward each other from points which are \(195\) miles apart. They travel at rate of \(25\) and \(40\) miles an hour, respectively. If they start traveling at the same time, how long before the trains will meet?
Car A and Car B start traveling towards each other at the same time from points \(150\) miles apart. If Car A went at the rate of \(20\) miles an hour, at what rate must B travel if they meet in \(5\) hours?
A passenger and a freight train start toward each other at the same time from two points \(300\) miles apart. If the rate of the passenger train exceeds the rate of the freight train by \(15\) miles per hour, and they meet after \(4\) hours, what are the rates of the passenger and train?
Two automobiles started at the same time from a point, but traveled in opposite directions. Their rates were \(25\) and \(35\) miles per hour, respectively. After how many hours were they \(180\) miles apart?
A man having ten hours at his disposal made an excursion, riding out at the rate of \(10\) miles an hour and returning on foot at the rate of \(3\) miles an hour. Find the distance he rode.
A man walks at the rate of \(4\) miles per hour. How far can he walk into the country and ride back on a trolley that travels at the rate of \(20\) miles per hour if he must be back home \(3\) hours from the time he started?
A boy rides away from home in an automobile at the rate of \(28\) miles an hour and walks back at the rate of \(4\) miles an hour. The round trip requires \(2\) hours. How far does he ride in the automobile?
A motorboat leaves a harbor and travels at an average speed of \(15\) mph toward an island. The average speed on the return trip was \(10\) mph. How far was the island from the harbor if the total trip took \(5\) hours?
A family drove to a resort at an average speed of \(30\) mph and later returned over the same road at an average speed of \(50\) mph. Find the distance to the resort if the total driving time was \(8\) hours.
As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was \(90\) mph, and the average speed returning was \(120\) mph. Find the distance between the two airports if the total flying time was \(7\) hours.
Annie, who travels \(4\) miles an hour starts from a certain place \(2\) hours in advance of Brandie, who travels \(5\) miles an hour in the same direction. How many hours must Brandie travel to overtake Annie?
A man travels \(5\) miles an hour. After traveling for \(6\) hours another man starts at the same place following the first man at the rate of \(8\) miles an hour. When will the second man overtake the first man?
A motorboat leaves a harbor and travels at an average speed of \(8\) mph toward a small island. Two hours later a cabin cruiser leaves the same harbor and travels at an average speed of \(16\) mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cruiser be alongside the motorboat?
A long distance runner started on a course running at an average speed of \(6\) mph. One hour later, a second runner began the same course at an average speed of \(8\) mph. How long after the second runner started will the second runner overtake the first runner?
A car traveling at \(48\) mph overtakes a cyclist who, riding at \(12\) mph, has had a \(3\)-hour head start. How far from the starting point does the car overtake the cyclist?
A jet plane traveling at \(600\) mph overtakes a propeller-driven plane which has had a \(2\)-hour head start. The propeller-driven plane is traveling at \(200\) mph. How far from the starting point does the jet overtake the propeller-driven plane?
Two men are traveling in opposite directions at the rate of \(20\) and \(30\) miles per hour at the same time and from the same place. In how many hours will they be \(300\) miles apart?
Running at an average rate of \(8\) meters per second, a sprinter ran to the end of a track and then jogged back to the starting point at an average rate of \(3\) meters per second. The sprinter took \(55\) seconds to run to the end of the track and jog back. Find the length of the track.
A motorboat leaves a harbor and travels at an average speed of \(18\) mph to an island. The average speed on the return trip was \(12\) mph. How far was the island from the harbor if the total trip took \(5\) hours?
A motorboat leaves a harbor and travels at an average speed of \(9\) mph toward a small island. Two hours later a cabin cruiser leaves the same harbor and travels at an average speed of \(18\) mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cruiser be alongside the motorboat?
A jet plane traveling at \(570\) mph overtakes a propeller-driven plane that has had a \(2\)-hour head start. The propeller-driven plane is traveling at \(190\) mph. How far from the starting point does the jet overtake the propeller-driven plane?
Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is \(6\) miles per hour more than the rate of the other and they are \(168\) miles apart at the end of \(4\) hours, what is each rate?
As part of flight training, a student pilot was required to fly to an airport and then return. The average speed on the way to the airport was \(100\) mph, and the average speed returning was \(150\) mph. Find the distance between the two airports if the total flight time was \(5\) hours.
Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours they are \(72\) miles apart. Find the rate of each cyclist.
A car traveling at \(56\) mph overtakes a cyclist who, riding at \(14\) mph, has had a \(3\)-hour head start. How far from the starting point does the car overtake the cyclist?
Two small planes start from the same point and fly in opposite directions. The first plane is flying \(25\) mph slower than the second plane. In two hours, the planes are \(430\) miles apart. Find the rate of each plane.
A bus traveling at a rate of \(60\) mph overtakes a car traveling at a rate of \(45\) mph. If the car had a \(1\)-hour head start, how far from the starting point does the bus overtake the car?
Two small planes start from the same point and fly in opposite directions. The first plane is flying \(25\) mph slower than the second plane. In \(2\) hours, the planes are \(470\) mi apart. Find the rate of each plane.
A truck leaves a depot at \(11\) a.m. and travels at a speed of \(45\) mph. At noon, a van leaves the same place and travels the same route at a speed of \(65\) mph. At what time does the van overtake the truck?
A family drove to a resort at an average speed of \(25\) mph and later returned over the same road at an average speed of \(40\) mph. Find the distance to the resort if the total driving time was \(13\) hours.
Three campers left their campsite by canoe and paddled downstream at an average rate of \(10\) mph. They then turned around and paddled back upstream at an average rate of \(5\) mph to return to their campsite. How long did it take the campers to canoe downstream if the total trip took \(1\) hour?
A motorcycle breaks down and the rider has to walk the rest of the way to work. The motorcycle was being driven at \(45\) mph, and the rider walks at a speed of \(6\) mph. The distance from home to work is \(25\) miles, and the total time for the trip was \(2\) hours. How far did the motorcycle go before it broke down?
A student walks and jogs to college each day. The student averages \(5\) kilometers per hour walking and \(9\) kilometers per hour jogging. The distance from home to college is \(8\) kilometers, and the student makes the trip in one hour. How far does the student jog?
On a \(130\)-mile trip, a car traveled at an average speed of \(55\) mph and then reduced its speed to \(40\) mph for the remainder of the trip. The trip took a total of \(2.5\) hours. For how long did the car travel at \(40\) mph?
On a \(220\)-mile trip, a car traveled at an average speed of \(50\) mph and then reduced its average speed to \(35\) mph for the remainder of the trip. The trip took a total of \(5\) hours. How long did the car travel at each speed?
An executive drove from home at an average speed of \(40\) mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at and average speed of \(60\) mph. The entire distance was \(150\) miles. The entire trip took \(3\) hours. Find the distance from the airport to the corporate offices.