# Algebra Word Problems

This section covers:

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 word-for-word 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 $$y$$ $$x-y$$ 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 $$n-3$$ a number $$n$$ less than 3 $$3-n$$ a number $$n$$ reduced by 3 $$n-3$$ 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 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:

### 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 cross-multiply 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?

Solution:

 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\\\\24-12&=3M-2M\\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(10-Q)&=1.45\\.25Q+1-.1Q=1.45\\.15Q+1&=1.45\\.15Q&=.45\\\frac{{.15Q}}{{.15}}&=\frac{{.45}}{{.15}}\\Q&=3\\D&=10-3=7\end{align} Let $$Q=$$ the number of quarters that Briley has. Then we know that she has $$10-Q$$ 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: $$25Q+.10\left( {10-Q} \right)=1.45$$;  then solve to get $$Q=3$$.

# Mixture Word Problem:

Note: Mixture Word Problems are also done using Systems of Equations, like here.

### 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=100-20$80, so the number sold would be $$x-20$$.

 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): $$3\left( {x-20} \right)-1.5x=15$$; solve to get $$x=50$$.

Here’s the math:

### 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!