Word Problems

“Word problems are what I do worst” is something I hear a lot. If you feel that way, I hope this page will help.

Word problems are trying to intimidate you. Knowing that can help you stand up to them. You don’t have to let them get you. You do need to follow some steps.

Read each problem as many times as you have to. All but the simplest word problems are impossible to grasp until you’ve read them a few times. Don’t try to write an equation right away. Unless the problem is unusually simple, you need your full concentration to read the problem and see what’s going on.

Taking notes may help you at this stage. Myself, I like to write things down as I read the problem. The notes may not be useful, but writing them helps me process. It’s a matter of personal preference. I suggest you try it and see if it helps.

Name things. You can’t talk about things until you’ve named them. You can’t create an equation to solve your word problem until you’ve given letter names to some of the problems elements. Once you’ve read the problem enough times to get a grip on it, give a letter name to the thing that is asked for. For example, if the problem asks how long it will take John and Mary to shovel the snow off the sidewalk if they work together, then give that a name, say (or whatever letter). 

Do write down variable names and definitions. Words slip around and change meaning as you use them. You need to know exactly what you mean by that variable name. If you use n to stand for, say, the number of sandwiches sold and you come up with $4.76 for an answer, it may be that you became confused about what you meant by n. The remedy is to write the name down in a place where you can refer to it easily.

Create an equation or a set of equations. This is the hard part. Break it into steps small enough to handle. Express the problem in words. Then work those words into an equation. For an example of developing an equation, see sample problem #2 below.

Solve. After what you’ve just been through, this is the easy part.

Check. Two steps:

First review the question. It may be so much work to find x that when you do find it you may think that’s the answer — but maybe the question asked you for 2x, not x. Take a quick glance back at the question to make sure the question you’re answering is the one that was asked and that your answer makes sense, at least approximately.

Then substitute your answer back into the equation and simplify. If that substitution gets you a true statement, you’re good. If not, you may need to hunt down your error.


Don’t let the problem scare you. It’s designed to scare you. You can stand up to it.

  • Read the problem as many times as necessary to understand what the story is and what you’re being asked.
  • Name variables and write down their definitions.
  • Write an equation. If it helps, work your way gradually from a sentence to an equation.
  • Look back to make sure you’re answering the question that was asked. Then make sure your answer works.

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Sample Problems

Let’s try that with some examples. Most of these are based on official SAT practice problems, revised for copyright reasons but with the same math. References after problems are to the problems on which these are based in The Official SAT Study Guide, 2016 edition.

Sample Problem #1

Question. Jana, who works from home, is trying to set up her work schedule. Jana’s employer gives her a lot of scheduling latitude, but she must work in 30-minute increments (meaning the length of each shift must be a multiple of 30 minutes), and all her working hours must be between 6 am and 6 pm. If Jana does not need to stop working to eat or for anything else, how many 30-minute time intervals can she schedule into the hours between 6 am and 6 pm on Thursday and Friday?

Solution.  Call the number of intervals p. Jana’s got 12 hours per day to work for two days. How many 30-minute intervals are in two 12-hour days? That’s p.

Thirty minutes is one half hour. There are two half-hour intervals in one hour, twenty-four in 12 hours, and forty-eight in two 12-hour days. So p equals 2 intervals per hour times 12 hours per day times 2 days. That is 48 intervals. Jana can work out forty-eight 30-minute intervals.

This problem is like problem #34 on p. 387 of The Official SAT Study Guide, 2016 edition.

Using two different kinds of units for one kind of thing can be confusing. This problem measures time in both minutes and and hours. When a mix like that happens, change into one kind of units as one of your first steps. 

Sample Problem #2: Discount and Tax

Question: Benny bought a pair of pants on sale for 10% off the original price. At the checkout he paid a final price of $37.80, which included 5% sales tax. What was the original price of the pants?


Define variables

What are you looking for? The original price. Call it p. Then make a note: “p: original price” or something like that. Then write an equation for p.

You know the final price and you’re looking for the original price, so write an equation for the final price in terms of the original price. Then use algebra to solve for the original price.

Work Toward an Equation

To find that equation, start with words and then work your way toward algebraic expressions.

p: original price

Final price is sale price plus sales tax

Final price = sale price plus 5% of sale price

Final price = 1.05 sale price.

You’re told the final price is $37.80, so substitute that in:

37.80 = 1.05 sale price

You’re not told what the sale price is, but you are given some information about it and you do need to talk about it. Sale price is original price with 10% off.

Sale price = p – 10% of

Sale price = 90% of p

Sale price = 0.9 p

Take this expression for sale price and insert it into the equation for final price:

37.80 = 1.05 sale price


37.80 = 1.05 times 0.9


Doing the multiplication, you get

37.80 = 0.945

You need to find p, so divide both sides by 0.945:




Is that right? Ballpark check: The sale reduces the cost by 10%, and the tax adds 5%. So the cost at checkout should be slightly less than the original price. $37.80 is slightly less than $40.00. Check.

Now, for a thorough check, stick the value for p back into the original problem and see if it works:

Assume the original price is $40.00. The pants are on sale for 10% off. 10% of $40.00 is $4.00. Then the sale price is  \$40.00 - \$4.00 = \$36.00.

And the tax? That’s 5%, and it will be calculated on the sale price, $36.00. 5% of $36.00 is $1.80. Then the amount paid at the checkout is the sale price plus the tax: $36.00 + $1.80 = $37.80. Check.

If you don’t know how to get to an equation, try writing down in words what you do know. To move toward an equation, replace whatever words you can with the variable names you have set up, use symbols to indicate arithmetic operations, and insert equal signs for things that you can see are equal. 

Interesting aside: What if the discount and the tax were at the same rate? Would they cancel? Let’s see what happens.

Question: Candace bought a skirt on sale for 10% off the original price. The skirt’s original price was $40. At the checkout Candace paid a final price that was $40, minus the 10% discount, plus 10% tax. What was the final price?

Solution: 10% of 40 is 4, so the sale price is $40 – $4 = $36. Tax is 10% of the sale price. 10% of $36 is $3.60, so the final price with tax is $36.00 + $3.60 = $39.60. Take 10% away and then add 10% back and you end up with a little less than what you started with.

If that doesn’t sound right to you, consider a 10 x 10 square grid.
It has 10 x 10 = 100 cells. Take away 10% of those cells — that’s 10 cells —
and you’re left with 90 cells, or 90% of what you started with. Now add another 10%. With 90 cells, the 10% you add is just nine cells. You end up with just 99 cells. You’ve lost 1% of the cells, just like Candace saved 1% on the price of her skirt with 10% discount and 10% tax. 

Sample problem #3: Rate – time – distance

Question. Earth is 150,000,000 km from the Sun. Light travels at 300,000 km per second. How long does it take light to get from the Sun to Earth?


distance equals rate times time


  • d: distance
  • r: rate
  • t: time


You need to find time, so solve for t:


In this case,

t = 150,000,000 km /300,000 km per second

That comes to 500 seconds.

That’s about 8 minutes. That ray of sunshine coming through your window just now left the Sun about 8 minutes ago.

This question is a little like #15 on page 495 of The Official SAT Study Guide, 2016 edition.

For rate-time-distance problems remember the equation: distance equals rate times time, or d=rt.

To remember d=rt, make sense of it. You can apply this equation anywhere. Just now how far are you from a door? That distance is equal to your walking speed times the time you need to walk there. d = rt.  How long will it take you to drive to New York? That’s the distance from wherever you are to New York divided by the speed at which you drive: t = d/r, which is another form of the equation d = rt.

Some people remember the equation by pronouncing it “dirt.”

Sample problem #4: a work problem

Question: F (that’s F for “fast”) can shovel all the snow from a sidewalk in 30 minutes, but S (S for “slow”) needs 60 minutes to do the same job. Working together, how much time do F and S need?

Solution: This is a work problem. Many people find work problems most vexing.

Start by asking what it is you have to find. That’s the time the two need together. Give it a name: t.

Time the two need together: t

Next ask what fraction of the job each can do in one unit of time. That’s one minute in this problem, because the information is given in minutes. F can do the job in 30 minutes, so the fraction he can do in one minute is 1/30.

Fraction of the job F can do in one minute: 1/30

And what fraction of the job can S do in one minute? She needs 60 minutes to do the whole thing, so in one minute she can do 1/60 of it.

Fraction of the job S can do in one minute: 1/60

Each person’s part of the job is the fraction that person can do in one minute — 1/30 for F and 1/60 for S — times the time F and S both work, t. So F‘s part is t/30 and S‘s is t/60. Together they do the whole job, so the sum of their parts adds up to one whole, 1. Then:

t/30 + t/60=1

To solve, clear the fractions by multiplying by the least common multiple of their denominators, 60 in this case. Then

60 X t/30 + 60 X t/60=60 X 1

2t + t = 60

= 20

Working together, F and S can clear the sidewalk in 20 minutes.

Check: First see if the answer is reasonable. Twenty minutes is less time than either person needs alone. And it’s close to half the average of 30 and 60, which is 22.5 minutes.

To be really thorough, put the answer back into the equation:

t/30 + t/60=1

If t=20, then the left-hand side of the equation is 20/30 + 20/60 and that equals 60/60, which equals 1. Check.

For a work problem, find the fraction of the job that each person can do in one unit of time. Each person’s fraction of the job is the part that person can do in one minute (or hour or day) times t, the number of minutes (or hours or days) the job takes. The sum of these fractions is 1, the whole job.

If one person needs a minutes to complete a job and the other needs minutes, then the equation for the time they need together is  {\color{Blue} \frac{t}{a}+\frac{t}{b}=1} .

For a quick mini-check, note that two people working together can always do the job in less time than either can alone; at the same time, they need at least half the time the faster one needs alone. (That’s true in problem-land, even if not in real life.) If the two both need the same time for a job, then together they need half that time. If their rates are close, the time they need together will be close to half the average of their times.

Sample problem #5: systems of equations and prob – stats

Question. This table summarizes the results of a survey about party membership and gender. Sort of. Some numbers are missing. As the table shows, 20 of the 90 people surveyed belong to Party A and the other 70 belong to Party B. Also, note these facts not stated in the table: There are 5 times as many females in Party B as in Party A, and there are two times as many males in Party B as in Party A.

What is the probability that someone in Party A selected at random is female?


What’s “random”?

“Random” means every member of a group is equally likely to be selected. That means the probability that someone in Party A selected at random is female is equal to the number of females in Party A divided by the whole Party A membership. That membership is 20, as you can see by the bottom line of Party A. The number of females in Party A is represented by the letter in the table, so the probability that someone in Party A selected at random is female is a divided by 20.

But the correct answer is not a letter, it’s a number. That means you need to find a. How can you do that?

In most problems you would define variable names. In this problem the names are given already.

You have four pieces of information:

  • a+c=20
  • b+d=70
  • b=5a
  • d=2c

That’s a system of four equations with four unknowns. To find a you need to solve at least part that system. How can you do that?

The two usual methods of solving systems of equations are substitution and elimination. You could use either of those methods. I’m going to use a combination. (I tried three ways — elimination, substitution, and a combination of the two — and found there’s not a lot of difference among the three in difficulty or in the amount of work involved.)

In the second equation, b+d=70, if you substitute 5a for b and 2c for d, you get  5a+2c=70, an equation with a and c as variables. The first equation also has variables a and c. You can solve those two equations together.

Now the system looks like this:



Solve it by elimination: Multiply the first equation by -2:


Now there is -2c in the first equation and there is 2in the second equation:



Add those together and you get




Now you’ve got what you need, a, from the system of equations. You may want to continue to solve for b, c, and d so that you can check.

As stated above, the probability you seek is a divided by 20. That’s 10/20, which equals 1/2. The probability that someone in Party A selected at random is female is 1/2.

(p. 613 #29)

I chose this problem because it illustrates two different methods, elimination and substitution. And because it’s about the hardest SAT word problem I could find, and it wouldn’t be fair for me to include only easy ones.

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