This section covers:
 English to Math Translation
 Unit Rate Problem
 “Find the Numbers” Word Problem
 Percent Word Problem
 Percent Increase Word Problem
 Ratio/Proportion Word Problems
 Weighted Average Word Problem
 Consecutive Integer Word Problem
 Age Word Problem, Money (Coins) Word Problem
 Mixture Word Problem, Percent Mixture Word Problem
 Rate/Distance Word Problem
 Profit Word Problem
 Converting Repeating Decimal to Fraction Word Problem
 Inequality Word Problems
 Integer Function Problem
 Direct, Inverse, Joint and Combined Variation Word Problems
 Work Word Problems (in Rational Functions and Equations)
 Systems of Equations Word Problems
 More Word Problems using Rational Functions
 Absolute Value Word Problems
Note that Using Systems to Solve Algebra Word Problems can be found here in the Systems of Linear Equations and Word Problems section.
Now that you can do these difficult algebra problems, you can trick your friends by doing some fancy word problems; these are a lot of fun. The problems here only involve one variable; later we’ll work on some that involve more than one.
English to Math Translation for Word Problems
Doing word problems is almost like learning a new language like Spanish or French; you can basically translate wordforword from English to Math, and here are some translations:
English 
Math 
is, yields, will be 
\(=\) 
what number, how much (look at question)  “\(x\)” (or any variable) 
in addition to, added to, increased by  \(+\) 
sum of \(x\) and \(y\)  \(x+y\) 
difference of \(x\) and \(y\)  \(xy\) 
product of \(x\) and \(y\)  \(x\times y\) 
quotient of \(x\) and \(y\)  \(\displaystyle x\div y\,\,\,\,\,\text{or }\,\,\,\frac{x}{y}\) 
opposite of \(x\)  \(–x\) 
ratio of \(x\) to \(y\)  \(\displaystyle x\div y\,\,\,\,\,\text{or}\,\,\,\,\frac{x}{y}\) 
a number \(n\) less 3  \(n3\) 
a number \(n\) less than 3  \(3n\) 
a number \(n\) reduced by 3  \(n3\) 
of  times 
\(p\) percent 
\(\displaystyle \frac{p}{{100}}\), or move decimal left 2 places 
half, twice  \(\displaystyle \frac{n}{2},\,\,\,2n\) 
consecutive numbers 
Let \(n=\) first number, \(n+1=\) second number, \(n+2=\) third number… 
odd/even consecutive numbers 
Let \(n=\) first number, \(n+2=\) second number, \(n+4=\) third number… (Note: Even if you are looking for odd consecutive numbers, use \(n, n+2, n+4, …\)). 
average of \(x,y\) and \(z\) (and so on)  \(\displaystyle \frac{{x+y+z+…}}{{\text{(how many}\,\,\text{numbers}\,\,\text{on}\,\,\text{top)}}}\) 
\(x\) per \(y\), \(x\) to \(y\), \(x\) over \(y\), \(x\) part of \(y\)  \(x\div y\) or \(\displaystyle \frac{x}{y}\)
Example: number of girls to total people can be represented by \(\displaystyle \frac{{\text{girls}}}{{\text{total}}}\). 
\(x\) per \(y\), as in \(x\) “for every” \(y\)  Multiplication, or \(x\times y\).
Example: if you drive 50 miles per hour, how many miles will you drive in 5 hours: 250 miles. 
\(y\) increased by \(x\%\)  \(\displaystyle y+\left( {y\times \frac{x}{{100}}} \right)\) 
\(y\) decreased by \(x\%\)  \(\displaystyle y\left( {y\times \frac{x}{{100}}} \right)\) 
\(y\) is at least (or no less than) \(x\)  \(y\ge x\) 
\(y\) is at most (or no more than) \(x\)  \(y\le x\) 
\(y\) is between \(x\) and \(z\) 
\(x\le y\le z\) (inclusive) \(x<y<z\) (exclusive)

Remember these important things:
 If you’re wondering what the variable (or unknown) should be when working on a word problem, look at what the problem is asking. This is usually what your variable is!
 If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!
 If the problem asks for a Unit Rate, you want the ratio of the \(y\)value (typically the dollar amount) to the \(x\)value, when the \(x\)value is 1.This is basically the slope of the linear functions. You may see feet per second, miles per hour, or amount per unit; these are all unit rates.
Note that most of these word problems can also be solved with Algebraic Linear Systems, here in the Systems of Linear Equations section.
Now let’s do some problems that use some of the translations above. We’ll get to more difficult algebra word problems later. Trick your friends with these problems!
Unit Rate Problem:
You buy 5 pounds of apples for $3.75. What is the unit rate of a pound of apples?
Solution:
To get the unit rate, we want the amount for one pound of apples; this is when “\(x\)” (apples) is 1. We can set up a ratio: \(\displaystyle \frac{5}{{3.75}}=\frac{1}{x};\,\,\,x=\$.75\).
“Find the Numbers” Word Problems:
The sum of two numbers is 18. Twice the smaller number decreased by 3 equals the larger number. What are the two numbers?
Solution:
We always have to define a variable, and we can look at what they are asking. The problem is asking for both the numbers, so we can make “\(n\)” the smaller number, and “\(18n\)” the larger.
Do you see why we did this? The way I figured this out is to pretend the smaller is 10. (This isn’t necessarily the answer to the problem!) But I knew the sum of the two numbers had to be 18, so do you see how you’d take 10 and subtract it from 18 to get the other number? See how much easier it is to think of real numbers, instead of variables when you’re coming up with the expressions?
We don’t need to worry about “\(n\)” being the smaller number (instead of “\(18n\)”); the problem will just work out this way!
Let’s translate the English into math and solve:
Problem/Math  Notes 
The sum of two numbers is 18. Twice the smaller number decreased by 3 equals the larger number. What are the two numbers?
\(\begin{array}{l}2n3\,\,\,=\,\,18n\\\underline{{+n\,\,\,\,\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,+n}}\\3n\,3\,\,=\,\,\,18\\\underline{{\,\,\,\,\,\,\,\,\,+3\,\,=\,+3}}\\\,\,3n\,\,\,\,\,\,\,\,\,\,=\,\,\,21\\\,\frac{{3n}}{3}\,\,\,\,\,\,\,\,\,\,\,=\,\,\frac{{21}}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n=7\,\,\,\,\,\,\text{(smaller number)}\\\,\,\,187=11\,\,\,\,\text{(larger number)}\end{array}\) 
Let \(n=\) the smaller number, and \(18n=\) the larger number.
The translation is pretty straight forward; note that we didn’t have distribute the 2 since the problem only calls for twice the smaller number, and then we subtract 3.
Let’s check our work:
The sum of 7 and 11 is 18. √ Twice the smaller (\(2\times 7\)) decreased by 3 would be \(143=11\). √ 
Another Problem:
If the product of a number and –7 is reduced by 3, the resulting number is 33 less than twice the opposite of that number. What is the number?
Solution:
We always have to define a variable, and we can look at what they are asking. The problem is asking for a number, so let’s make that \(n\).
Now let’s try to translate wordforword, and remember that the “opposite” of a number just means to make it negative if it’s positive or positive if it’s negative. We can just put a negative sign in front of the variable.
If you’re not sure if you should multiply, add, or subtract, try “real numbers” to see what you should do. For example, “8 reduced by 3” is 5, so for the “reduce by 3” part, we need to subtract 3. Also, “33 less than 133” is 100, so for the “33 less than”, we need to subtract 33 at the end:
Problem/Math 
Notes 
If the product of a number and –7 is reduced by 3, the resulting number is 33 less than twice the opposite of that number. What is the number?
\(\displaystyle \begin{array}{l}\left( {7} \right)n3=2\left( {n} \right)33\\\,\,\,\,\,7n3=2n33\\\,\,\,\,\,\,\underline{{+7n\,\,\,\,\,\,\,\,=\,+7n}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,=\,\,\,5n33\\\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,+33\,=\,\,\,\,\,\,\,\,\,\,\,+33}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,30\,\,=\,\,\,5n\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{30}}{5}\,\,=\,\,\frac{{5n}}{5}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n=6\end{array}\) 
Note again the opposite of a number means we basically just multiply the number by –1 (in our case, put a negative in front of it). We translate English to math, and, after solving, get 6 as our answer.
Let’s check our work: If we take the product of 6 and –7 (–42) and reduce it by 3, we get –45. Is this number 33 less than twice the opposite of 6?
Twice the opposite of 6 is –12, and 33 less than –12 is \(1233=45\). We got it! √ 
Percent Word Problem:
60 is 20% of what number? (We saw similar problems in the Percents, Ratios, and Proportions section!)
Solution:
The translation is pretty straight forward; note that we had to turn 20% into a decimal (Remember: we need to get rid of the % – we’re afraid of it – so we move the decimal 2 places away from it).
Percent Problem/Math 
Notes 
60 is 20% of what number?
\(\displaystyle \begin{align}60&=.20\times n\\\frac{{60}}{{.2}}&=\frac{{.2n}}{{.2}}\\300&=\,n\\n&=300\end{align}\) 
Remember again that of = times.
Let’s check our work: \(20\%\,\,\,\text{of}\,\,\,300=.2\times 300=60\) √ 
Percent Increase Word Problem:
The price of a pair of shoes has increased by 15%. The original price of the shoes was $20. What is the new price?
Solution:
Percent Increase Problem/Math  Notes 
The price of a pair of shoes has increased by 15%. The original price of the shoes was $20. What is the new price?
\(\displaystyle \begin{array}{l}x=\$20+\left( {15\%\,\times 20} \right)\\x=\$20+\left( {.15\times 20} \right)\\x=\$20+\$3=\$23\\x=\$23\,\end{array}\) or \(\displaystyle \begin{array}{l}x=\$20\times \left( {1+15\%} \right)\\x=\$20\times \left( {1+.15} \right)\\x=\$20\times \left( {1.15} \right)\\x=\$23\,\end{array}\)

15% of the original amount \(=15%\times 20\), since of = times. We then need to turn the 15% back into a decimal and add to the original amount.
The second way we did it was to multiply the original amount ($20) by 1.15 (100% + 15%), which added 15% to the original amount before we multiplied. 
Ratio/Proportion Word Problems
Relating Two Things Together: a Rate
It takes 2 minutes to print out 3 color photos on Erin’s printer. Write an equation relating the number of color photos \(p\) to the number of minutes \(m\).
Solution:
This problem seems easy, but you have to think about what the problem is asking. When we are asked to relate something to something else, typically we use the last thing (the “to the” part) as the \(y\), or the dependent variable.
I like to set up these types of problems as proportions, but what we’re looking for is actually a rate of minutes to photos, or how many minutes to print 1 photo. Remember that rate is “how many \(y\)” to “one \(x\)”, or in our case, how many “\(m\)” to one “\(p\)”. We will see later that this is like a Slope that we’ll learn about in the Coordinate System and Graphing Lines including Inequalities section. Here’s the math:
Ratio Problem/Math  Notes 
It takes 2 minutes to print out 3 color photos on Erin’s printer. Write an equation relating the number of color photos \(p\) to the number of minutes \(m\).
\(\displaystyle \begin{align}\frac{{\text{2 minutes}}}{{\text{3 color photos}}}&=\frac{{\text{how many minutes}}}{{\text{1 color photo}}}\\\frac{\text{2}}{\text{3}}&=\frac{m}{{1p}}\\3m&=2p\\m&=\frac{2}{3}p\end{align}\) 
To get the rate of minutes to photos, we can set up a proportion with the minutes on the top and the photos on the bottom, and then cross multiply.
So the equation relating the number of color photos \(p\) to the number of minutes \(m\) is \(\displaystyle m=\frac{2}{3}p\). √ 
Ratio/Proportion Problem:
The ratio of boys to girls in your new class is 5 : 2. The sum of the kids in the class is 28. How many boys are in the class?
Solution:
This is a ratio problem; we learned about ratios in the Percents, Ratios, and Proportions section. A ratio is a comparison of two numbers; a ratio of 5 to 2 (also written 5:2 or \(\displaystyle \frac{5}{2}\)) means you have 5 boys for every 2 girls in your class. So if you had only 7 in your class, you’d have 5 boys and 2 girls. But what if you had 14? You’d have 10 boys and 4 girls, since 10 is 5 times 2, and 4 is 2 times 2. Let’s see how we can set this up in an equation, though, so we can do the algebra!
There are actually a couple of different ways to do this type of problem. Probably the most common is to set up a proportion like we did here earlier. Let \(x=\) the number of boys in the class.
Ratio Problem/Math  Notes 
The ratio of boys to girls in your new class is 5:2. The sum of the kids in the class is 28. How many boys are in the class?
\(\require{cancel} \displaystyle \begin{align}\frac{{\text{boys}}}{{\text{total in class}}}&=\frac{x}{{28}}=\frac{5}{7}\\\\7x&=5\times 28=140\\\frac{{\cancel{7}x}}{{\cancel{7}}}&=\frac{{140}}{7}\\x&=20\text{ boys}\end{align}\) 
We know the sum of the numbers in the ratio is 7 (boys and girls: 5 + 2), and the sum of the kids in the class is 28. And we also know the “boys” part of the ratio is 5.
We need to set up a proportion with the same things on top or on bottom; our ratios will have “boys” on top and “total in class” on bottom. In other words, we need to see how many boys out of 28 will keep a ratio of 5 boys to 7 total in the class. We can crossmultiply and get \(x=20\).
There are 20 boys and 8 girls (28 – 20) in the new class. Let’s check: the ratio of 20 to 8 is the same as the ratio of 5 to 2. And the number of boys and girls add up to 28! √ 
There’s another common way to handle these types of problems, but this way can be a little trickier since the variable in the equation is not what the problem is asking for; we will make the variable a “multiplier” for the ratio. The advantage to this way is we don’t have to use fractions.
Ratio Word Problem/Math  Notes 
The ratio of boys to girls in your new class is 5:2. The sum of the kids in the class is 28. How many boys are in the class?
\(\begin{align}5x+2x&=28\\7x=28\\\frac{{7x}}{7}&=\frac{{28}}{7}\\x&=4\\\\5\times 4&=20\,\,\,\,\text{boys}\\2\times 4&=8\,\,\,\,\text{girls}\end{align}\) 
We have to multiply both numbers by the same thing to keep the ratio the same – try this with some numbers to see this.
5 times a number, and 2 times that same number must equal 28. Let \(x\) be the multiplier – not the number of boys or girls.
We get 4 as the multiplier, but we’re looking for the number of boys in the class (5 times 4 = 20) and the number of girls in the class (2 times 4 = 8).
There are 20 boys and 8 girls. Let’s check: the ratio of 20 to 8 is the same as the ratio of 5 to 2 (each is divided by 4 – the multiplier!) And the number of boys and girls add up to 28! √

Here’s a ratio problem that’s pretty tricky; we have to do it in a lot of steps:
Problem: One ounce of solution X contains ingredients a and b in a ratio of 2:3. One ounce of solution Y contains ingredients a and b in a ratio of 1:2. If solution Z is made by mixing solutions X and Y in a ratio of 3:11, then 1260 ounces of solution Z contains how many ounces of ingredient a?
Ratio Problem/Math  Notes 
One ounce of solution X contains ingredients a and b in a ratio of 2:3. One ounce of solution Y contains ingredients a and b in a ratio of 1:2.
If solution Z is made by mixing solutions X and Y in a ratio of 3:11, then 1260 ounces of solution Z contains how many ounces of ingredient a?
Solution Z: \(\begin{array}{c}3x+11x=1260;\,\,\,\,\,\,\,x=90\\3\times 90=270\,\,\,\text{oz}\text{. solution X}\\11\times 90=990\,\,\,\text{oz}\text{. solution Y}\end{array}\) 
Let’s work backwards on this problem, and first work with solution Z, since we know there are 1260 ounces of it. It’s good to start with the parts of the problems with numbers first!
Since we know the ratio of X and Y is 3:11 in solution Z, we can find the ratio multiplier, and find how much of solutions X and Y are in Z. We see that there are 270 ounces of X and 990 of Y in solution Z. 
Solution X: \(\begin{array}{c}2x+3x=270;\,\,\,\,\,\,x=54\\2\times 54=108\,\,\,\text{oz}\text{. ingredient a}\\3\times 54=162\,\,\,\,\text{oz}\text{. ingredient b}\end{array}\)
Solution Y: \(\begin{array}{c}1x+2x=990;\,\,\,\,\,\,x=330\\1\times 330=330\,\,\,\text{oz}\text{. ingredient a}\\\,2\times 330=660\,\,\,\text{oz}\text{. ingredient b}\end{array}\) 
Now we know that there are 270 ounces of solution X in solution Z. We can find out how much of ingredients a and b are in solution X by using a ratio multiplier again (one ounce of solution X contains ingredients a and b in a ratio of 2:3). We see that there are 108 ounces of ingredient a in solution X.
We can do the same for solution Y, which contains ingredients a and b in a ratio of 1:2. We see that there are 330 ounces of ingredient a in solution Y. 
\(108+330=438\)  The problem asks for the amount of ingredient a in solution Z, so add the amounts of ingredient a in Solutions X and Y to get 438.
1260 ounces of solution Z contains 438 ounces of ingredient a. √ 
Weighted Average Word Problem:
You’ve taken four tests in your Algebra II class and made an 89, 92, 78, and 83. The final is worth two test grades. What do you need to make on the final to make an A in the class for the semester?
Solution:
We always have to define a variable, and we can look at what they are asking.
Let \(x=\) what you need to make on the final. Now we have 6 test grades that will count towards our semester grade: 4 regular tests and 2 test grades that will be what you get on the final (since it counts twice, we need to add it 2 times). This is called a weighted average, since we “weighted” the final test grade twice.
Let’s use the equation for an average:
Weighted Average Problem/Math  Notes 
You’ve taken four tests in your Algebra II class and made an 89, 92, 78, and 83. The final is worth two test grades. What do you need to make on the final to make an A in the class for the semester?
\(\require{cancel} \displaystyle \begin{align}\frac{{89+92+78+83+x+x}}{6}&=90\\\frac{{89+92+78+83+2x}}{{\cancel{6}}}\times \frac{{\cancel{6}}}{1}&=90\times \frac{6}{1}\\342+2x&=540\\2x&=198\\\frac{{2x}}{2}&=\frac{{198}}{2}\\\\x&=99\end{align}\) 
The average or mean equation is just adding up all the values, and then dividing by the number of values that we just added up. We have to divide by 6, since we have 4 tests given, and then the final is worth 2 test grades.
You have to make a 99 on the final to make an A in the class! Yikes! Good luck – you can do it!
Let’s see if it works:
\(\displaystyle \frac{{86+92+78+83+99+99}}{6}=\frac{{540}}{6}\,=90\,\,\,\,\,\surd \) 
HINT: For any problem with weighted averages, you can multiply each value by the weight in the numerator, and then divide by the sum of all the weights that you’ve used. For example, if you had test 1 (say, an 89) counting 20% of your grade, test 2 (say, an 80) counting 40% of your grade, and test 3 (say, a 78) counting 40% of your grade, you will take the weighted average as in the formula below. Don’t forget to turn percentages into decimals and make sure that all the percentages that you use (the “weights”) add up to 100 (all the decimals you use as weights should add up to 1). When using decimals, your denominator should be 1:
\(\displaystyle \frac{{\left( {89\times .2} \right)\,+\,\left( {80\times .4} \right)\,+\,\left( {78\times .4} \right)}}{{.2+.4+.4}}=\frac{{17.8+32+31.2}}{1}=81\)
Consecutive Integer Word Problem:
The sum of the least and greatest of 3 consecutive integers (numbers in a row) is 60. What are the values of the 3 integers?
Solution:
You’ll see these “consecutive integer” problems a lot in algebra. When you see these, you always have to assign “\(n\)” to the first number, “\(n+1\)” to the second, “\(n+2\)” to the third, and so on. This makes sense, since consecutive means “in a row” and we’re always adding 1 to get to the next number.
Consecutive Integer Problem/Math  Notes 
The sum of the least and greatest of 3 consecutive integers (numbers in a row) is 60. What are the values of the 3 integers?
\(\displaystyle \begin{align}n+n+2&=60\\2n+2&=60\\2n&=58\\\frac{{2n}}{2}&=\frac{{58}}{2}\end{align}\)
\(\displaystyle n=29\,\,\,\,\,\,n+1=30\,\,\,\,\,\,n+2=31\) 
Again, we assign “\(n\)” to the first number, “\(n+1\)” to the second, and “\(n+2\)” to the third, since they are consecutive numbers.
Let’s translate the English into math. The least of the 3 consecutive numbers is “\(n\)“, and the greatest is “\(n+2\)”. We just need to add the least number and the greatest to get 60.
The three consecutive numbers are 29, 30, and 31. Let’s see if it works: \(29+31=60\,\,\,\, \surd \) 
Note: If the problem asks for even or odd consecutive numbers, use “\(n\)”, “\(n+2\)”, “\(n+4\)”, and so on – for both even and odd numbers! It will work; trust me!
Age Word Problem:
Your little sister Molly is one third the age of your mom. In 12 years, Molly will be half the age of your mom. How old is Molly and your mom now?
Solution:
Doesn’t this one sound complicated? It’s a great one to try on your friends! It’s not that bad though – let’s first define a variable by looking at what the problem is asking.
Age Word Problem/Math  Notes 
Your little sister Molly is one third the age of your mom. In 12 years, Molly will be half the age of your mom. How old is Molly and your mom now?
\(\begin{align}M+12&=\frac{1}{2}\left( {3M+12} \right)\\2\times \left( {M+12} \right)&=3M+12\\2M+24&=3M+12\\\\2412&=3M2M\\12&=M\end{align}\)
\(M=12\,\,\,\,\,\,\,\,\,3M=36\) 
Let \(M=\) the age of sister Molly now. Then we know that your mom is \(3M\) (make it into an easier problem – if Molly is 10, your mom is 30).
Turn English into math (second sentence). Remember that we have to add 12 years to both ages (\(M+12\) for Molly and \(3M+12\) for your mom), since we’re talking about 12 years from now (unfortunately, moms have to age, too).
Multiply both sides by 2 to get rid of the fraction, and then “push” the 2 through the parentheses. Then get the variables to one side, and the constants to the other.
Molly is 12, and your mother is 36.
Let’s see if it works: In 12 years, Molly will be 24, and her mom will be 48. Aha! Molly will be half of her mom’s age in 12 years. \(\surd \) 
Money (Coins) Word Problem:
Suppose Briley has 10 coins in quarters and dimes and has a total of $1.45. How many of each coin does she have?
Solution:
We always have to define a variable, and we can look at what they are asking.
Money Word Problem/Math  Notes 
Suppose Briley has 10 coins in quarters and dimes and has a total of $1.45. How many of each coin does she have?
\(\displaystyle \begin{align}.25Q+.10(10Q)&=1.45\\.25Q+1.1Q=1.45\\.15Q+1&=1.45\\.15Q&=.45\\\frac{{.15Q}}{{.15}}&=\frac{{.45}}{{.15}}\\Q&=3\\D&=103=7\end{align}\) 
Let \(Q=\) the number of quarters that Briley has. Then we know that she has \(10Q\) dimes (turn into easier problem – if she has 2 quarters, she has 10 minus 2, or 8 dimes).
Since she has a total of $1.45, and each quarter is worth $.25 and each dime is worth $.10, then the number of quarters times .25 plus the number of dimes times .10 must equal her total, or $1.45. (Again, turn into easier problem: if you have 4 quarters, you have .25 times 4 = $1.00 total). “Push” the .10 through the parentheses and solve.
Briley has 3 quarters, and 7 dimes.
Let’s see if it works: 3 quarters would be $.75 and 7 dimes would be $.70. If we add $.75 and $.70 we get $1.45. \(\surd \) 
We could have also done this problem (and many problems like these) with a table:
Amount  Price  Total  
Quarters  Q  .25  .25 Q  Multiply across 
Dimes  10 – Q  .10  .10 (10 – Q)  Multiply across 
Total  10  1.45  Do Nothing Here  
Add Down  Do Nothing Here  Add Down: \(25Q+.10\left( {10Q} \right)=1.45\); then solve to get \(Q=3\). 
Mixture Word Problem:
Note: Mixture Word Problems are also done using Systems of Equations, like here.
One kind of candy (jelly) sells for $5 a pound and another (chocolate) for $10 a pound. How many pounds of each should be used to make a mixture of 10 pounds of candy (both kinds) that sells for $80?
Solution:
Let’s first define a variable, and use another table like we did before. Let \(J=\) the number of pounds of jelly candy that is used in the mixture. Then \(10J\) equals the number of pounds of the chocolate candy.
Amount  Price  Total  
Jelly Candy  J  5  5J  Multiply across 
Chocolate Candy  10 – J  10  10 (10 – J)  Multiply across 
Total  10  80  Do Nothing Here  
Add Down  Do Nothing Here  Add Down: \(5J+10\left( {10J} \right)=80\); then solve to get \(J=4\). 
Here’s the math:
Math  Notes 
\(\begin{align}5J+10(10J)&=80\\5J+10010J&=80\\5J&=20\\J&=4\end{align}\)  We would need 4 pounds of the jelly candy and \(104=6\) pounds of the chocolate candy. 
Percent Mixture Word Problem:
A 20% concentrate is to be mixed with a mixture having a concentration of 60% to obtain 80 liters of a mixture with a concentration of 30%. How much of the 20% concentrate and the 60% concentrate will be needed?
Solution:
Let’s first define a variable, and use another table like we did before. Let \(T=\) the number of liters we need from the 20% concentrate, and then \(80T\) will be the number of liters from the 60% concentrate. (Put in real numbers to check this).
Let’s put this in a chart again – it’s not too bad. This one is a little more difficult since we have to multiply across for the Total row, too, since we want a 30% solution of the total.
Amount  Price  Total  
20% Concentrate  T  .20  .2T  Multiply across 
60% Concentrate  80 – T  .60  .60 (80 – T)  Multiply across 
Total (What we want) 
80  .30  24  Multiply across 
Add Down  Do Nothing Here  Add Down: \(.2T+.6\left( {80T} \right)=24\); then solve to get \(t=60\). 
Here’s the math:
Math  Notes 
\(\begin{align}.2T+.6\left( {80T} \right)&=24\\.2T+48.6T&=24\\.4T&=24\\T&=60\end{align}\)  We would need 60 liters of the 20% solutions and \(8060=20\) liters of the 60% solution.

Don’t worry if you don’t totally get these; as you do more, they’ll get easier. We’ll do more of these when we get to the Systems of Linear Equations and Word Problems topics. Also, remember that if the problem calls for a pure solution or concentrate, use 100%.
Rate/Distance Word Problem:
A train and a car start at the same place. The train is going 40 miles per hour and a car is going in the opposite direction at 60 miles per hour. How long will it be until they are 100 miles apart?
Solution:
It’s always good to draw pictures for these types of problems:
Distance Word Problem/Math  Notes 
A train and a car start at the same place. The train is going 40 miles per hour and a car is going in the opposite direction at 60 miles per hour. How long will it be until they are 100 miles apart?
\(\text{Distance}\,\,=\,\,\text{Rate}\,\,\times \,\,\text{Time}\) Total Distance: \(100=60t+40t\)
Solve: \(\begin{align}100&=60t+40t\\100&=100t\\t&=1\end{align}\) 
Remember always that \(\text{Distance}=\text{Rate}\,\times \,\text{Time}\)
The rates of the train and car are 40 and 60, respectively. Usually a rate is “something per something”.
Let \(t\) equal the how long (in hours) it will be until the train and the car are 100 miles apart. We must figure the distance of the train and car separately, and then we can add distances together to get 100.
Again, you can always add distances; look at them separately first, and then you can put them together to equal the total distance (100).
The math was pretty easy on this one! In one hour, the train and the car will be 100 miles apart. 
Note that there’s an example of a Parametric Distance Problem here in the Parametric Equations section.
Profit Word Problem
Hannah paid $1.50 each for programs to her play. She sold all but 20 of them for $3 each and made a profit of $15 total. How many programs did Hannah buy? How many did she sell? Hint: Profit = Selling Price – Purchase Price
Solution:
Let’s make a table to store the information. Let \(x=\) the number of programs that Hannah bought. Let’s put in real numbers to see how we’d get the number that she sold: if she bought 100 programs and sold all but 20 of them, she would have sold 80 of them. We can see that $80=10020$80, so the number sold would be \(x20\).
Number of Programs  Price  Total  
Sold  x – 20  3.00  3(x – 20)  Multiply across 
Bought  x  1.50  1.5x  Multiply across 
Profit  15  Total Profit Given  
Do Nothing Here  Do Nothing Here  Subtract Down: To get profit, subtract Total Bought from Total Sold, and set to Profit (15): \(3\left( {x20} \right)1.5x=15\); solve to get \(x=50\). 
Here’s the math:
Math  Notes 
\(\begin{align}3\left( {x20} \right)1.5x&=15\\3x601.5x&=15\\1.5x&=75\\x&=50\end{align}\)  Hannah bought 50 programs to make a profit of $15.
Since she sold 20 less than she bought, she sold 50 – 20 = 30 programs.

Converting Repeating Decimal to Fraction Word Problem:
Convert \(.4\overline{{25}}\,\,\,(.4252525…)\) to a fraction.
Solution:
Converting repeating decimal to fraction problems can be easily solved with a little trick; we have to set it up as a subtraction, so the repeating part of the decimal is gone. To do this, let \(x=\) the repeating fraction, and then we’ll figure out ways to multiply \(x\) by 10, 100, and so on (multiples of 10) so we can subtract two numbers and eliminate the repeating part.
The rule of thumb is to multiply the repeating decimal by a multiple of 10 so we get the repeating digit(s) just to the left of the decimal point, and then multiply the repeating digit again by a multiple of 10 so we get repeating digit(s) just to the right of the decimal. Then we subtract the two equations that we just created, and solve for \(x\):
Repeating Decimal Problem/Math  Notes 
Convert \(.4\overline{{25}}\,\,\,(.4252525…)\) to a fraction.
\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,x=4.2525252…\\\text{(original repeating decimal)}\\\\\,\,\,\,\,\,\,1000x=425.252525…\\\,\,\,\,\,\,\,\,\underline{{\,\,\,\,10x=\,\,\,\,\,\,4.252525…}}\\\,\,\,\,\,\,\,\,\,990x=421.0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{421}}{{990}}\end{array}\) 
First, we’ll let \(x=.4\overline{{25}}\). Since \(.4\overline{{25}}\) has repeating digits 25, we first have to figure out how to multiply \(.4\overline{{25}}\) to get the repeating digits just to the left of the decimal place.
We can do this by multiplying \(.4\overline{{25}}\) by 1000; we get \(1000x=4\underline{{25}}.252525…\) Also, we can see that if we multiply \(x\) by 10, we get the repeating part (25) just to the right of the decimal point; we get \(10x=4.\underline{{25}}2525…\).
Now we have to line up and subtract the two equations on the left and solve for \(x\); we get \(\displaystyle x=\frac{{421}}{{990}}\). Pretty cool! Let’s see if it works: Put \(\displaystyle \frac{{421}}{{990}}\) in your graphing calculator, and then hit Enter; you should something like .4252525253. √

Inequality Word Problems
Note that inequalities are very common in realworld situation, since we commonly hear expressions like “is less than” (\(<\)), “is more than” (\(>\)),“is no more than” (\(\le \)), “is at least” (\(\ge \)), and “is at most” (\(\le \)). Here’s an example of a Quadratic Inequality word problem. We’ll also use inequalities a lot in the Introduction to Linear Programming section.
Problem:
\(\displaystyle \frac{4}{5}\) of a number is less than 2 less than the same number. Solve the inequality and graph the results.
Solution:
This is a little tricky since we have two different meanings of the words “less than”. The words “is less than” means we should use “\(<\)” in the problem; it’s an inequality. The words “2 less than the same number” means “\(x2\)” (try it with “real” numbers).
Inequality Word Problem/Math  Notes 
\(\displaystyle \frac{4}{5}\) of a number is less than 2 less than the same number. Solve the inequality and graph the results.
\(\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\frac{4}{5}x\,\,<x2\\\,\,\,\,\,\underline{{\,\,\,\,\,\,\,\,\,x<x\,\,\,\,}}\\\,\,\,\,\frac{4}{5}xx<\,\,\,\,2\\\,\,\,\,\,\,\,\frac{1}{5}x\,\,<\,\,2\\\left( {\frac{1}{5}x} \right)\left( {5} \right)>\left( {2} \right)\left( {5} \right)\\\\\,\,\,\,\,\,x>10\,\,\,\,\,\text{(watch sign!)}\end{array}\) 
Set up and solve inequalities like we do regular equations. Let “\(x\)” be the number, and translate the problem wordforword: \(\displaystyle \frac{4}{5}x<x2\).
We need to get all the “\(x\)”s to one side and all the “numbers” to the other sign. We could have also multiplied both sides by 5 to get rid of the fraction. Remember to change the sign when we multiply both sides by –5, since we’re dealing with an inequality.
Once we get the answer, we can also graph the solution. Also, try numbers close to 10, like 9 and 11, to make sure it works. 
Problem:
Erica must tutor at least 12 hour per week in order to be eligible for her workstudy program at her university. She must also study 10 more hours than what’s she’s tutoring, so she can keep up her grades in the program. What is the minimum number of hours Erica must study in order to be eligible for her workstudy program?
Solution:
First we define a variable \(h\), which will be the number of hours that Erica must study (look at what the problem is asking). We know from above that “at least” can be translated to “\(\ge\)”.
If Erica works let’s say 30 hours in her work study program, she’d have to study 40 hours (it’s easier to put in real numbers). So, the amount of time she works in her work study program would be “\(h10\)”, and this number must be at least 12. Let’s set up the equation and solve:
Inequality Word Problem/Math  Notes 
Erica must tutor at least 12 hour per week in order to be eligible for her workstudy program at her university. She must also study 10 more hours than what’s she’s tutoring, so she can keep up her grades in the program. What is the minimum number of hours Erica must study in order to be eligible for her workstudy program?
\(\begin{array}{l}h10\ge 12\\\underline{{\,\,\,\,\,\,+10\ge \,\,\,+10}}\\h\,\,\,\,\,\,\,\,\,\,\,\,\ge 22\end{array}\) 
The algebra is simple, and we don’t have to worry about changing the sign, since we’re not multiplying or dividing by a negative number.
Erica would have to tutor at least 22 hours. Notice that 22 hours works, since the problems asked for “at least”.
To check your answer, try numbers right around the answer, like 21 hours (which wouldn’t be enough), and 22 hours (which would work!)

Rational Inequality Word Problem
Technically, this next problem contains a rational function, but it’s relatively easy to solve.
Inequality Word Problem  Math/Notes 
A school group wants to rent part of a bowling alley to have a party. The bowling alley costs $500 to rent, plus an additional charge of $5 per student to bowl. The group doesn’t want any student to pay more than $15 total to attend this party.
What is an inequality that could represent this situation?
How many students would need to attend so each student would pay at most $15? 
Since there is a onetime cost in addition to a perperson cost, the cost per person will depend on the number of students attending the party: the more students, the lower the cost.
For \(x\) students attending, each would have to pay \(\displaystyle \frac{{500}}{x}\) for the bowling alley rent; try it with real numbers! In addition, each student needs to pay their $5 to bowl.
Therefore, each student will need to pay \(\displaystyle \frac{{500}}{x}+5\), and since we don’t want any student to pay more than $15, the inequality that represents this situation is \(\displaystyle \frac{{500}}{x}+5\le 15\). To see how many students would have to attend to keep the cost at $15 per person, solve for \(x\):
\(\displaystyle \frac{{500}}{x}+5\le 15;\,\,\,\,\frac{{500}}{x}\le 10;\,\,\,\,500\le 10x;\,\,\,\,x\ge 50\). At least 50 students would have to attend.

Integer Function Problem (a little bit more advanced…)
The fee for hiring a tour guide to explore Italy is $1000. One guide can only take 10 tourists and additional tour guides may be hired if needed. What is the cost of hiring tour guides, as a function of the number of tourists who go on the tour? If there are 72 tourists, what is the cost of hiring guides?
Solution:
Let’s think about this by using some real numbers. From 1–10 tourists the fee is \(1\times 1000=\$1000\), for 11–20 tourists, the fee is \(2\times 2000=\$2000\), and so on. Do you see how if we divide the number of tourists by 10, and go up to the next integer, we’ll get the number of tour guides we need? This is because any fraction of a set of ten tourists requires another tour guide.
To get the function we need, we can use the Least Integer Function, or Ceiling Function, which gives the least integer greater than or equal to a number (think of this as rounding up to the closest integer). The integer function is designated by \(y=\left\lceil x \right\rceil \). (We saw a graph of a similar function, the Greatest Integer Function, in the Parent Functions and Transformations section.)
Thus, the cost of hiring tour guides is \(\displaystyle 1000\times \left\lceil {\frac{x}{{10}}} \right\rceil \). For 72 tourists, the cost is \(\displaystyle 1000\times \left\lceil {\frac{{72}}{{10}}} \right\rceil =1000\times \left\lceil {7.2} \right\rceil =1000\times 8=\$8000\). Makes sense!
Now, all these types of problems can get much more difficult (and we will see later how to use two variables with some of them), but it’s important to take baby steps with them. Don’t worry if they seem difficult at this time, but as long as you get the general idea of how we’re doing the translations, you’re in great shape! And don’t forget:
 When assigning variables (letters), look at what the problem is asking. You’ll typically find what the variables should be there.
 If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!
Learn these rules, and practice, practice, practice!
On to Systems of Linear Equation and Word Problems – you are ready!