This section covers:

**English to Math Translation****“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****Joint Variation Word Problem****Work Word Problems**(in Rational Functions and Equations)

Note that **Using Systems to Solve Algebra Word Problems** can be found here in the **Systems of Linear Equations** 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

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 two 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!**

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!

**“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:**

Ok, so 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!) 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 “18 – *n*“); the problem will just work out this way!

So let’s translate the English into math:

**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:**

Ok, so 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 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**. So 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:

# 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 Increase Word Problem:

The price of a pair of (cheap) shoes has increased by 15%. The original price of the shoes was $20. What is the new price?

### Solution:

# 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

*” to “one*

**y***“, or in our case, how many “*

**x***” to one “*

**m***“. We will see later that this is like a*

**p****Slope**

**that we’ll learn about in the**

**Coordinate System and Graphing Lines including Inequalities**section. Here’s the math:

**Ratio/Proportion Problem:**

The ratio of boys to girls in your new class is 5 : 2 (yeah!). 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 ) 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.

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.

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:** 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!

# 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:

Ok, we always have to define a variable, and we can look at what they are asking.

So let ** x** = what you need to make on the final. So 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:

**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). So when using decimals, your denominator should be 1:

# 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.

Note: If the problem asks for **even** or **odd** consecutive numbers, use “*n*“, “*n* + 2″, “*n* + 4″, and so on – for both odd and even 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.

So let ** M** = the age of sister Molly now. Then we know that your mom is 3

**(make it into an easier problem – if Molly is 10, your mom is 30).**

*M*Turn English into math (second sentence), and remember that we have to add 12 years to **both** **ages** (*M* + 12 for Molly) and (3*M* + 12 for your mom), since we’re talking about 12 years from now (unfortunately, moms have to age, too):

# 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:

So we always have to define a variable, and we can look at what they are asking.

Let ** Q** = the number of quarters that Briley has. Then we know that she has 10 –

**dimes (turn into easier problem – if she has 2 quarters, she has 10 minus 2, or 8 dimes).**

*Q*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).

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

# Mixture Word 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 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

**10 –**equals the number of pounds of the chocolate candy.

*J*# 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

**will be the number of liters from the 60% concentrate. (Put in real numbers to check this).**

*80 – T*Let’s put this in a chart again – it’s not too bad:

This one was a little more difficult since we had to multiply across for the **Total** row, too, since we wanted a 30% solution of the total. 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:

Remember always that** distance = rate x time**. Let’s 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.

It’s always good to draw pictures for these types of problems:

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 is 100 – 20, so the number sold would be

*x***– 20**.

# Converting Repeating Decimal to Fraction Word Problem:

Convert (.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**:

# Inequality Word Problem:

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).

Again, for an inequality, we’ll set up and solve inequalities like we do regular equations. So let “*x*” be the number, and translate the problem word-for-word: . Now solve:

# Another Inequality Word 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 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 work-study 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 “”.

So 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”. This number must be at least 12. So let’s set up the equation and solve:

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 **Coordinate Systems and Graphing Lines including Inequalities** – you are ready!

I work with independent study high school students- most of whom struggle in math. I have been looking for clear word problem explanations and your site is by far the best I’ve seen. I also really like the other instruction I’ve viewed so far. My students generally prefer this kind of delivery over video lessons, and you have organized everything clearly and thoughtfully.

I am excited to recommend your site to my students next fall!

Thank you so much for taking the time to leave a comment; it means so much to me. When I receive comments like yours, it gives me a reason to continue to write and blog! Please share the site with your students and give me any feedback you have to make it better. I truly love math and writing about it. Thanks again! Lisa

I will be sharing the site not only with my students, but also with the other Education Specialists in our program. You are a rare gem because 1) you good at math AND, more important, 2) you can clearly communicate the concepts for those of us who are not so good at it.

Thank you so much for explaining algebra clearly and concisely! I am a middle school student with troubles in algebra. I can do algebra to find X, but I just can’t “translate” the word problems into algebra. I’m used to simple word problems seen in elementary school. When the questions got harder, I couldn’t just think of the answer in my head; I had to actually run through the steps of algebra on paper.

This has given me new hope to try and strengthen my basics/foundation of algebra so that I can tackle the harder questions.

Again, thank you so much!!

Thank you so much for taking the time out to comment on my site. Comments like these makes me want to write more and more 😉 Lisa

thank you!

thank you!

i can make it as my project

^_^

again thank you so much……

😉

it was awesome..i learned a lot but i really had a hard time in analyzing word problems especially in age and investment. it sucks!

yuh i agree with angel, word problems are really time consuming.. but no worries at all, this website really helps a lot.. thank you

You know, this site is really helpful. I got a lot out of this. I will continue to explore this site.

Thank you so much for explaining algebra clearly and concisely! I am a middle school student with troubles in algebra. I can do algebra to find X, but I just can’t “translate” the word problems into algebra. I’m used to simple word problems seen in elementary school. When the questions got harder, I couldn’t just think of the answer in my head; I had to actually run through the steps of algebra on paper.

This has given me new hope to try and strengthen my basics/foundation of algebra so that I can tackle the harder questions.

Again, thank you so much!!!!

Thank you so much for your note – and I’m glad the site helped you! Please spread the word about SheLovesMath.com and keep coming back 😉

Lisa

thanks, i really appreciate your work man. i needed this to study for our test, seems you explained it well, it might be a piece of cake. Gotta go back to singing now, im IA.

Thanks so much for your comment! I hope the test went OK? Please spread the word about She Loves Math! I have a lot to write though 😉

Lisa

i wanna ask for some inequality word problems regarding distance, coin, age and bussiness.. thanks! hope to get some reply..

Sure – send me the problems! Lisa

waah! thanks for giving soo many examples!! now i understand the lesson!!! gomawo! 😀

Thank you for the help! I am taking the PRAXIS 1 and needed extra help with word problems with one unknown. Any more advice for age questions?

Haley,

Great! Just remember that if the problem talks about, say, 5 years ago Joan was twice the age of Sarah, you’d have to subtract 5 from each: J – 5 = 2(S – 5). The same goes for the future. Let me know if you get stuck on any practice problems

Lisa

i really enjoyed the way you analyse and simplify the word problem to a layman point of view.thanks alot and keep it up.

Hi ! just wanna ask some explanation of my assignment (word problems) .. if you are free based on the problems above, you have explained it well.. i hope you can help me .. ^__^

I will try to help you; send me the problems. I may want to use examples like them on my web page 😉 Lisa

Thank you! I learned a lot from you!

christy spent 3.8 pesos for 60 stamps. she bought only 2-cent, 10 cent and 12 cent stamps. if there were twice as many 10 cent stamps as 12 cent stamps, how many of each kind did he buy?(Hope you can help me solve this math problem. tnxs)

Thanks for writing! I can help with this problem, but I’m not sure how pesos relate to cents – are there 100 cents in 1 peso? I need that information to do this problem. Thanks, Lisa

Thank you so much for the help! I’m an incoming high school student and I’ve been doing some extra studying for the entrance exams and your website helped me big time! ^__^

Keep writing because you and your website is pure awesome!!! : )

Thank you so much, Addie! Comments like yours makes me want to keep writing and writing! Hopefully I’ll get a lot done this summer 😉

Lisa

how can I solve a distance problem with only 1 piece of the equation?

It took a crew 2h 40min to row a boat 6 miles upstream and back and back again. If the rate of flow of the stream was 3 km/h, what was the rowing speed of the crew in still water?

You are an excellent teacher!! Thank you for sharing this!!

Thanks so much for your note! Feedback like yours make me want to keep writing and finish the site. Please let me know if there are any ways to improve the site. Lisa

I think with mixture problems that you have to know which type you’re dealing with. Then, you have to have a plan of action for dealing with each type. Some use charts, some I’ve found that you can use cross-multiplication, etc. Anyway, here’s one method that I don’t remember seeing on the well-informed list:

Here’s another type of mixture problem: Unknown + Unknown + Unknown, etc. = Total.

Example: An excellent solution for cleaning grease stains from cloth or leather consists of the following: carbon tetrachloride 80%(by volume), ligroin 16%, and amyl alcohol 4%. How many pints of each should be taken to make up 75 pints of solution[taken from a College Algebra book].

Since the percentages each represent a fraction of the solution, you can relate the pints as versions of x.

Pints:

Amyl alcohol = x

Ligroin(since 16% is 4 times the 4% of amyl alcohol, it will be 4x) = 4x

Carbon tetrachloride(80% is 20 times what amyl alcohol is and 5 times what ligroin is, so) = 20x

THEN, x + 4x + 20x = 75 | 25x = 75; x = 3 |

LAST, 3 x 4 = 12[which is 4x] | 20 x 3 = 60[which is 20x] |

CHECKING: 3 + 12 + 60 = 75. | Answer: You’ll add 3 pints of amyl alcohol, 12 pints of ligroin, and 60 pints of carbon tetrachloride to form the 75 pint cleaning solution.

NOTES: Can be done by percentage, but what if you need the algebraic work or have a problem that is presented differently? Concepts of problem: Determining x and other unknowns in relation to x, then adding them to the total. That philosophy is a common one in word problems–in general–whether percentages are being used or not.

Thank you so much for the new mixture problem! Good idea – I’ll add one like that.

Lisa

This page is a modern super lamp in dark and dim rooms. A glowing tribute to its author!

Hi, I have been looking online for a website that would help my son solve word problems in his advanced algebra 2 class. Do you think that your page would be helpful to him? Or would he need something more advanced?

Thanks!

Kim

Thanks for writing! I have algebra word problems in 2 different places: first with just one variable

here, and then using systemshere. I also have word problems in other sections, such as the Exponents, Logs, Rational Functions, Piecewise Functions, and others.So he look at these sections and also search for “Applications” on the site. Hope this helps! Lisa

Thank you so much I’m reviewing for the ACT and every problem that you have here was part of my review that I’ve been trying to make these are the types of questions that “rear their ugly heads” and you have to know how to solve them otherwise you’ll make silly mistakes even though they are rather easy THANK YOU!!!!!!!!!!!!! this will go in my ” tricky questions that always appear on the ACT portion of my review journal”

omg, this site’s a saviour, i have exams tmrw and this like saved me. thanks a lot lisa, ur explanation are super clear and really helpful

Thanks so much for writing, and I’m glad the site helped. Please spread the word about She Loves Math Lisa

I need help solving the following problem: A train 88 metres long overtakes a man walking at 4 Km/h and passes him in 10 seconds. Find the speed of the train.

Thanks for writing! This is how I’d do this problem: Distance = rate * time, and the man is taking away speed by walking 4 km/h, so 88 = ( r – 4) * (10/3600) * 1000, or r = 35.68 km/hr. We had to divide the 10 by 3600, since there are 3600 seconds in an hour, and multiply by 1000 to get km instead of m. Lisa

Best website I’ve seen. I’m in Algebra 2 and for review we had to do these types of problems and I completely forgot what I had learned from Algebra 1. The examples were very clear and showed you how to set up table! Thank you so much! Please do more.

Thanks so much! Comments like these makes me want to keep writing! Thanks again – you made my day 😉 Lisa

is it possible to have negative answer?

For which problem?

Can you help me to make a problem using quadratic equations about coin and mixture?

I really need a help

Thanks for writing! I don’t have one, but how about number 43 here? https://books.google.com/books?id=JnJaAAAAcAAJ&pg=PA376&lpg=PA376&dq=quadratic+coin+mixture+problem&source=bl&ots=Abgt-ElOb9&sig=FzeEva_nFh_udKJEsJ8Eh0L6C_w&hl=en&sa=X&ved=0CEMQ6AEwBmoVChMInLrb4-SmyAIVwpANCh2_UQcD#v=onepage&q=quadratic%20coin%20mixture%20problem&f=false

Lisa

A lot of these “algebraic word problems” can be simply solved without using algebra. Pupils in Asian primary schools are being taught to use arithmetic to solve word problems. Something is definitely missing in US elementary education.

Great observation! I totally agree with you, but your point also makes me think about adding sections on Word Problems without Algebra – I need to do that! Thanks! Lisa

(I appreciate the clarity with which you write.)

As a H.S. Senior in Algebra class / final exam day came with a bit of a surprise.

9 coin find in 2 weighings

the odd coin problem as an “option B”

I jumped at the chance to score a 100, if i get it right, or a big fat zero, if i didnt solve it by the end of the class bell.

My question is,, how difficult was that type of problem for a 17 year old to crack, under extreme pressure, desperately needing a passing grade, that allowed me to graduate on stage

with my class of ’82 ?