# Algebra Word Problems

The algebra word problems here only involve one variable; later we’ll work on some that involve more than one, such as here in the Systems of Linear Equations and Word Problems section.

## 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:

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.

Here are problems that use some of the translations above. We’ll get to more difficult algebra word problems later.

## Unit Rate Problem:

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{{3.75}}{5}=\frac{x}{1};\,\,\,x=\$.75$. We also could have just divided the amount spent by the number of pounds, which makes sense.

(Another cool way to figure out what to do is to think of a simpler problem. If we buy 20 pounds of apples for $10, we “feel” the unit rate is$.50 per pound. And the math agrees with us!)

## Simple Interest Rate Problem:

Problem: How much money would need to be invested at 2% annual simple interest for 10 years to earn $1200? Use the formula$ \text{Interest}=\text{Principle }\times \,\text{Rate}\,\times \,\text{Time}$. Solution: Using the formula$ I=Prt$, solve for$ \displaystyle P:\,P=\frac{I}{{rt}}=\frac{{1200}}{{\left( {.02} \right)10}};\,\,\,P=\$6000$. Note that we had to turn $2\%$ into a decimal: we need to get rid of the % (pretend we’re afraid of it), so we move the decimal 2 places away from it.

Note: Compound Interest problems can be found here in the Exponential Functions section.

## Percent Word Problem:

(We saw similar problems in the Percentages, Ratios, and Proportions section!)

Problem: 60 is 20% of what number?

Solution: Translate word-for-word, and remember again that of = times, and $20\%=.20=.2$:

\displaystyle \begin{align}60&=.20\times n\\\frac{{60}}{{.2}}&=\frac{{.2n}}{{.2}}\\n&=300\end{align}

Check our answer: $20\%\,\,\text{of}\,\,300=.2\times 300=60$  √

## Percent Increase Word Problem:

Problem: The price of a pair of shoes has increased by 15%. The original price of the shoes was $20. What is the new price? Solution: 15% of the original amount$ =15\%\times 20$, since of = timesTurn the 15% into a decimal and add the product to the original amount. The second way to do it is to multiply the original amount ($20) by $1.15\,\,(100\%+15\%)$, which adds $\displaystyle 15\%$ to the original amount before multiplying.

$\displaystyle \begin{array}{l}x=\$20+\left( {15\%\,\times 20} \right)\\x=\$20+\left( {.15\times 20} \right)\\x=\$20+\$3=\$23\\x=\$23\,\end{array}$     or       $\displaystyle \begin{array}{l}x=\$20\times \left( {1+15\%} \right)\\x=\$20\times \left( {1+.15} \right)\\x=\$20\times \left( {1.15} \right)\\x=\$23\,\end{array}$

## Ratio/Proportion Word Problems

Rate Problem: 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 from the Coordinate System and Graphing Lines including Inequalities section. Here’s the math:

\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}

The equation relating the number of color photos $p$ to the number of minutes $m$ is $\displaystyle m=\frac{2}{3}p$.   √

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 Percentages, 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. For example, 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.

We can 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. Cross-multiply to get $x=20$:

\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}

There are 20 boys and 8 girls (28 – 20) in the new class.

Check our answer: 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.

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: We can 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):

\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}

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: Work backwards on this problem, and first work with solution Z, since there are 1260 ounces of it. It’s good to start with the parts of the problems with numbers first! Since the ratio of X and Y is $3:11$ in solution Z, we can find the ratio multiplier ($u$), and find how much of solutions X and Y are in Z. There are 270 ounces of X and 990 of Y in solution Z:

Solution Z:   $\begin{array}{c}3u+11u=1260;\,\,\,\,\,u=90\\3\times 90=270\,\,\,\text{oz}\text{. solution X}\\11\times 90=990\,\,\,\text{oz}\text{. solution Y}\end{array}$

From above, 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 ($v$), since one ounce of solution X contains ingredients a and b in a ratio of $2:3$. Do the same for solution Y (using ratio multiplier $w$), which contains ingredients a and b in a ratio of $1:2$.

Solution X:   $\begin{array}{c}2v+3v=270;\,\,\,\,v=54\\2\times 54=108\,\,\,\text{oz}\text{. ingredient a}\\3\times 54=162\,\,\,\,\text{oz}\text{. ingredient b}\end{array}$

Solution Y:   $\begin{array}{c}1w+2w=990;\,\,\,\,w=330\\1\times 330=330\,\,\,\text{oz}\text{. ingredient a}\\\,2\times 330=660\,\,\,\text{oz}\text{. ingredient b}\end{array}$

Note that there are there are 108 ounces of ingredient a in solution X and 330 ounces of ingredient a in solution Y. 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 $108+330=438$. 1260 ounces of solution Z contains 438 ounces of ingredient a.   √

## Weighted Average Word Problem:

Weighted average problems have to do with taking averages, but assigning different weights to the elements (for example, counting them more than once).

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 (a 90) in the class for the semester?

Solution: Define a variable, and look at what is being asked. 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. Use the equation for an average; divide by 6, since we have 4 tests given, and the final is worth 2 test grades.

\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}

Check our answer: $\displaystyle \frac{{86+92+78+83+99+99}}{6}=\frac{{540}}{6}\,=90\,\,\,\,\,\surd$

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 (80) counting 40% of your grade, and test 3 (78) counting 40% of your grade, 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:

$\displaystyle \frac{{\left( {89\times .2} \right)\,+\,\left( {80\times .4} \right)\,+\,\left( {78\times .4} \right)}}{{.2+.4+.4}}=\frac{{17.8+32+31.2}}{1}=81$

## “Find the Numbers” Word Problems:

Problem: 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 $18-n$ 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!) I knew the sum of the two numbers had to be 18, so I knew to take 10 and subtract it from 18 to get the other number. It is easier to think of real numbers, instead of variables when you’re coming up with the expressions.

Translate the English into math and solve:

\displaystyle \begin{align}2n-3&=18-n\\2n-3\,\,(+\,n)&=18-n\,\,(+\,n)\\3n\,-3&=18;\,\,3n=21\\\,\frac{{3n}}{3}&=\frac{{21}}{3}\\n&=7\,\,\,\,\,\,\text{(smaller number)}\\18-7&=11\,\,\,\,\text{(larger number)}\end{align}

Check our answer: The sum of 7 and 11 is 18.  √  Twice the smaller ($2\times 7$) decreased by 3 would be $14-3=11$.  √

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: Define a variable, and look at what they are asking. The problem is asking for a number, so make that $n$. Translate word-for-word, and remember that the “opposite” of a number just means to make it negative if it’s positive or positive if it’s negative. Thus, 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 subtract 3. Also, “33 less than 133” is 100, so for the “33 less than”, we subtract 33 at the end:

\displaystyle \begin{align}\left( {-7} \right)n-3&=2\left( {-n} \right)-33\\\,\,\,\,\,-7n-3&=-2n-33\\-7n-3\,\,\left( {+7n+33} \right)&=-2n-33\,\,\left( {+7n+33} \right)\\30&=5n;\,\,\frac{{30}}{5}=\frac{{5n}}{5}\\n&=6\end{align}

Check our answer: 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 $-12-33=-45$. We got it!  √

## Consecutive Integer Word Problem:

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. The trick is 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.

Translate the English into math. The least of the three consecutive numbers is $n$, and the greatest is $n+2$. Add the least number and the greatest to get 60:

\displaystyle \begin{align}n+n+2&=60\\2n+2&=60\\2n&=58;\,\frac{{2n}}{2}=\frac{{58}}{2}\end{align}

$\displaystyle n=29\,\,\,\,\,\,n+1=30\,\,\,\,\,\,n+2=31$

The three consecutive numbers are 29, 30, and 31. Check our answer: $29+31=60\,\,\, \surd$

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:

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: Define a variable by looking at what the problem is asking. Let $M=$ the age of sister Molly now. From this, 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). 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. Get the variables to one side, and the constants to the other:

\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\\M=12\,\,\,&\,\,\,3M=36\end{align}

Molly is 12, and your mother is 36. Check our answer: 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:

Note that problems like this with two variables to solve for are more easily solved with a System of Equations, which we will cover in the next section.

Problem: Suppose Briley has 10 coins in quarters and dimes and has a total of $1.45. How many of each coin does she have? Solution: Let$ Q=$the number of quarters that Briley has. Then we know that she has$ 10-Q$dimes (turn into easier problem – if she has 2 quarters, she has 10 minus 2, or 8 dimes). Each quarter is worth$.25 and each dime is worth $.10; thus, 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 = $1.00 total):$ \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}$Briley has 3 quarters, and 7 dimes. Check our answer: 3 quarters is$.75 and 7 dimes is $.70. If we add$.75 and $.70, we get$1.45.   $\surd$

We could have also done this problem (and many problems like these) with a table:

## Mixture Word Problem:

Mixture Word Problems are where different quantities are mixed together, and you are asked to find certain quantities. They are also solved with multiple variables, like here in the Systems of Linear Equations and Word Problems section. Here is an example using one variable:

Problem: 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 12 pounds of candy (both kinds) that sells for $80? Solution: First define a variable, and it helps to use a table. Let$ J=$the number of pounds of jelly candy that is used in the mixture. Then$ 12-J$equals the number of pounds of the chocolate candy. From the table:$ \begin{align}5J+10(12-J)&=80\\5J+120-10J&=80\\-5J&=-40\\J&=8\end{align}$We would need 8 pounds of the jelly candy and$ 10-8=2$pounds of the chocolate candy. ## Percent Mixture Word Problems: Problem: A 20% concentrate is to be mixed with a mixture having a concentration of 60% to obtain 80 liters of a mixture with a concentration of 30%. How much of the 20% concentrate and the 60% concentrate will be needed? Solution: First define a variable, and it helps to use a table. Let$ T=$the number of liters we need from the 20% concentrate, and then$ 80-T$will be the number of liters from the 60% concentrate. (Put in real numbers to check this). Put the information in a table; this one is a little more difficult since we have to multiply across for the Total row, too, since we want a 30% solution of the total. From the table:$ \begin{align}.2T+.6\left( {80-T} \right)&=24\\.2T+48-.6T&=24\\-.4T&=-24\\T&=60\end{align}$We would need 60 liters of the 20% solutions and$ 80-60=20$liters of the 60% solution. Remember that if the problem calls for a pure solution or concentrate, use 100% (if the percentage is that solution) or 0% (if the percentage is another solution). Here’s an example: Problem: How many ounces of pure water must be added to 100 ounces of a 30% saline solution to make it a 10% saline solution? Solution: Don’t worry if you don’t totally get these; as you do more, they’ll get easier. Again, we’ll do more of these here in the Systems of Linear Equations and Word Problems section. ## Rate/Distance Word Problem: Distance problems have to do with how far, how fast and how long objects have travelled. Problem: A train and a car start at the same place. The train is going 40 miles per hour and a car is going in the opposite direction at 60 miles per hour. How long will it be until they are 100 miles apart? Solution: Always draw pictures, and remember that$ \text{Distance}=\text{Rate}\,\times \,\text{Time}$. Usually a rate is “something per something”. The rates of the train and car are 40 and 60, respectively. Let$ t$equal the how long (in hours) it will be until the train and the car are 100 miles apart. We must figure the distance of the train and car separately, and then we can add distances together to get 100.$ \text{Distance}\,\,=\,\,\text{Rate}\,\,\times \,\,\text{Time}$Total Distance:$ 100=60t+40t$Solve:$ 100=60t+40t;\,\,100=100t;\,\,t=1$. In one hour, the train and the car will be 100 miles apart. Note that there’s an example of a System Equations Distance Problem here, and a Parametric Distance Problem here in the Parametric Equations section. ## Profit Word Problem Profit word problems have to do with the amount of money made for products or services in business. Here is an example: 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: It’s easiest to make a table to store the information. Let $x=$ the number of programs that Hannah bought. 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 $80=100-20$, so the number sold would be $x-20$.

From the table:

\begin{align}3\left( {x-20} \right)-1.5x&=15\\3x-60-1.5x&=15\\1.5x&=75;\,x=50\end{align}

Hannah bought 50 programs to make a total profit of $15. Since she sold 20 less than she bought, she sold 50 – 20 = 30 programs. ## Converting Repeating Decimal to Fraction Word Problem: Problem: Convert$ .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=$the repeating fraction, and then figure out ways to multiply$ x$by 10, 100, and so on (multiples of 10) so we can subtract two numbers and eliminate the repeating part. In other words, multiply the repeating decimal by a multiple of 10 to get the repeating digit(s) just to the left of the decimal point, and then multiply the repeating digit again by a multiple of 10 to get repeating digit(s) just to the right of the decimal. Then subtract the two equations and solve for$ x$:$ \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,x=4.2525252…\\\text{(original repeating decimal)}\\\\\,\,\,\,\,\,\,1000x=425.252525…\\\,\,\,\,\,\,\,\,\underline{{-\,10x=\,\,\,\,\,\,4.252525…}}\\\,\,\,\,\,\,\,\,\,990x=421.0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\displaystyle \frac{{421}}{{990}}\end{array}$Check our answer: Put$ \displaystyle \frac{{421}}{{990}}$in your graphing calculator, and then hit Enter; you should something like$ .4252525253$√ ## Inequality Word Problems Note that inequalities are very common in real-world situation, since we commonly hear expressions like “is less than” ($ <$), “is more than” ($ >$),“is no more than” ($ \le $), “is at least” ($ \ge $), and “is at most” ($ \le $). There are several inequality word problems the Solving Inequalities sections; here’s an example: Problem: Erica must tutor at least 12 hour per week in order to be eligible for her work-study program at her university. She must also study 10 more hours than the time she’s tutoring, so she can keep up her grades in the program. What is the minimum number of hours Erica must study in order to be eligible for her work-study program? Solution: First, define a variable$ h$, which is the number of hours that Erica must study (look at what the problem is asking). We know from above that “at least” can be translated to “$ \ge$”. If Erica works, let’s say, 30 hours in her work study program, she’d have to study 40 hours (it’s easier to put in real numbers). So, the amount of time she works in her work study program would be “$ h-10$”, and this number must be at least 12. Set up the equation and solve:$ \begin{array}{l}h-10\ge 12\\\underline{{\,\,\,\,\,\,+10\ge \,\,\,+10}}\\\,h\,\,\,\,\,\,\,\,\,\,\,\,\ge 22\end{array}$Erica would have to tutor at least 22 hours. Notice that 22 hours works, since the problems asked for “at least”. Check our answer: Try numbers right around the answer, like 21 hours (which wouldn’t be enough), and 22 hours (which would work!) This inequality word problem is a little more advanced: Problem:$ \displaystyle \frac{4}{5}$of a number is less than 2 less than the same number. Solve the inequality and graph the results. Solution: This is a little tricky since we have two different meanings of the words “less than”. The words “is less than” means we should use “$ <$” in the problem; it’s an inequality. The words “2 less than the same number” means “$ x-2$” (try it with “real” numbers). Set up and solve inequalities like we do regular equations. Let “$ x$” be the number, and translate the problem word-for-word:$ \displaystyle \frac{4}{5}x<x-2$. Get all the “$ x$”s to one side and all the “numbers” to the other sign. (We could have also multiplied both sides by 5 earlier to get rid of the fraction.) Remember to change the sign when we multiply both sides by –5, since we’re multiplying by a negative number:$ \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{4}{5}x\,\,\,\,\,\,<\,\,\,\,\,\,\,x-2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,\,\,\,\,-x<-x\,\,\,\,}}\\\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{4}{5}x-x<\,\,\,\,\,\,\,\,-2\\\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \,-\frac{1}{5}x\,\,\,<\,\,-2\\ \displaystyle \left( {-\frac{1}{5}x} \right)\left( {-5} \right)>\left( {-2} \right)\left( {-5} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x>10\,\,\,\,\,\text{(watch sign!)}\end{array}$Once we get the answer, we can also graph the solution. Try numbers close to 10, like 9 and 11, to make sure it works. ### Rational Inequality Word Problem Technically, this next problem contains a Rational Function, but it’s relatively easy to solve. Problem: 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?

Solution: Since there is a one-time cost in addition to a per-person 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 $5 to bowl. Thus, each student will need to pay$ \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$. In this example, we can multiply both sides by$ x$without worrying about reversing the inequality sign, since$ x$can only be positive:$ \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 so that each student pays no more than$15.

### Integer Function Problem (a little bit more advanced…)

Problem: 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!

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