The hardest thing about doing word problems is taking the words and translating them into math. Usually, once you get the equation, you’re fine and the actual math involved is often fairly simple. But figuring out the actual equation can seem nearly impossible. The following tips should help, however the best way of learning to solve word problems is practice.
The first step in translating and solving word problems is to . Don’t start trying to solve anything until you understand the problem. First, make sure you know just exactly what the problem is asking for, then see what information they’ve given you.
The second step is to work in an organized manner. Pick variables to stand for the unknowns, clearly labeling them so you’ll remember what they stand for. It may help to make notes as you go along. It can be quite frustrating to spend ten minutes solving a word problem on a test, only to realize at the end that you no longer have any idea what “x” stands for and you have to do the whole problem over again! Staying organized will help you think clearly and help in translating your final answer back into words.
The third step is to look for key words. Certain words indicate certain mathematical operations. Below are lists of those commonly used.
Other words and phrases:
“Increased/decreased by a factor of” can involve either addition or subtraction and multiplication.
“Per” can mean “divided by”, as in “Charley drove 90 miles on three gallons of gas. How many miles did he get per gallon?” Also, “a” sometimes means “divided by”, as in “When Marsha got gas, she paid $16.48 for four gallons. What was the cost for a gallon?”
Be careful with the expression “less than.” If you see “1.5 less than x”, the temptation may be to write “1.5 – x”. Do not do this! Consider the statement, “Jack makes $1.50 an hour less than Dave.” You don’t figure Dave’s wage by subtracting $1.50 wage from Jack’s. Instead, you subtract $1.50 from Dave’s wage. So remember not to confuse the “less than” construction in word problems.
Also note that order is important in the “quotient/ratio of” and “difference between/of” constructions. If a problems says “the ratio of x and y”, it means x divided by y. If the problem says “the difference of x and y”, it means x – y.
Here are some examples of how to extract keywords from word problems.
“the sum of 8 and y” translates to 8 + y
“4 less than x” translates to x – 4
“x multiplied by 13” translates to 13x
“the quotient of x and 3” translates to x/3
“the difference of 5 and y” translates to 5 – y
“the ratio of 9 more than x to x” translates to (x + 9)/x
“nine less than the total of a number and two” translates to (n + 2) – 9, which simplifies to n – 7
Here are some more lengthy examples:
The length of a football field is 30 yards more than its width. Express the length of the field in terms of its width w.
Whatever the width is, the length is 30 yards more than that. Remember “more than” means that we’re adding so you’ll be adding 30 to w and the expression they’re looking for is “Length = w + 30”.
Twenty gallons of crude oil were poured into two containers of different size. Express the amount of crude oil poured into the smaller container in terms of the amount, g, poured into the larger container.
The “how much is left” question comes up often and tends to cause some confusion. The solution is found by this reasoning: There are twenty gallons total, and we’ve already poured g gallons of it into the large container; how many gallons are left? The answer would be the 20 gallons minus the amount poured into the large container, or “The amount in the smaller container is 20 – g”.
Typically, with this type of question, you will be given some total amount then a smaller amount that’s a portion of the total. You need to pick a variable to stand for the unknown amount and the remaining portion is whatever is left after subtracting the unknown amount from the total.
For example, they may tell you that a trip took ten hours and had two legs. You could name the time for the first leg as t, and the remaining time for the second leg as 10 – t.
In a similar problem, they may tell you that a hundred-pound order of animal feed was filled by mixing products from bins A, B, and C, and that twice as much was taken from bin C as was from bin A. If you let a stand for the amount from bin A then the amount from bin C would be 2a, and the amount from bin B would be the remaining portion of the hundred pounds, or 100 – a – 2a; simplified to 100 – 3a.
Once you’ve learned to translate words into math and the math back into words, you’re ready to dive into the real world where there are almost an infinate number of possible word problems…physics is all word problems, business math is all word problems, etc., and “real life” can feel like an essay question.
Sometimes math tests ask you how a problem could be solved instead of asking for the answer. For example, let’s say that you want to find the expression that will calculate how much Pat and Mike earned last week. Pat worked 30 hours and was paid $7 per hour. Mike was paid $8 per hour and worked 25 hours.
Note that the information was presented in different orders. Be careful with problems presented in this manner and concentrate on the data you have to work with and what they are asking.
- First, be sure of what you’re trying to find. Here, it’s the total pay for BOTH people.
- So you’ll need Pat’s pay PLUS the Mike’s pay.
- Pat worked 30 hours and was paid $7 per hour.
- Mike worked 25 hours and was paid $8 per hour. Notice that the data was rearranged to make the expressions comparable.
- Pat earned (30 × 7) dollars and Mike earned (25 × 8) dollars.
- And the expression that will give you their total pay is: (30 × 7) + (25 × 8)
You can check to see if one of the given answers matches your expression and if not, calculate the answer from your expression then see which answer choice yields the same amount.
Read these problems carefully then decide which arithmetic expression would yield the answer to the question.
You want to buy 2 items that cost $5 each and 3 items that cost $4 each. Which expression shows the total cost?
A. (2 × 3) + (5 × 4)
B. (2 + 5) × (3 + 4)
C. (2 × 5) + (3 × 4)
D. (2 + 3) × (5 + 4)
Problem #2: (Hint: Distance = RATE times TIME)
You drove 2 hours at 50 miles per hour and 3 hours at 40 miles per hour. Which expression shows the total distance traveled?
A. (2 × 40) + (3 × 50)
B. (2 × 50) + (3 × 40)
C. (2 + 3) × (50 + 40)
D. (2 + 50) × (3 + 40)
Olivia ran for 20 minutes at 12 miles per hour and for 10 minutes at 8 miles per hour. Which expression shows the total distance she ran? What was the distance?
A. (20 × 12)+(10 × 8)
B. (12 × 8) + (20 × 10)
C. (20/60 × 12) + (10/60 × 8)
D. (20 × 12) + (10 × 8)60
Jessi bought four pounds of cheese at $3 a pound and six pounds of ham at $2 a pound. Which expression shows the total cost? What was the total cost?
A. (3 × 2) + (4 × 6)
B. (4 × 3) + (6 × 2)
C. (4 + 3) × (6 + 2)
D. (6 × 2) × (4 × 3)
Rodrigo worked six days last week and earned $60 each day; this week he worked four days and earned $75 each day. Which expression shows how much more he earned last week than this week? How much more did he earn last week?
A. (6 × 60) – (4 × 75)
B. (6 – 4) × (75 – 60)
C. (60 – 6) × (75 – 4)
D. (6 × 60) + (4 × 75)
Mixture problems involve creating a mixture from two or more things, then determining some quality (percentage, price, etc) of the resulting mixture. For example: Your school is holding a family event this weekend. Students have been pre-selling tickets to the event; adult tickets are $5.00, and child tickets are $2.50. From past experience, you expect about 13,000 people to attend the event. But this is the first year in which tickets prices have been reduced for the younger children, so you really don’t know how many child tickets and how many adult tickets you can expect to sell. Your boss wants you to estimate the expected ticket revenue. You decide to use the information from the pre-sold tickets to estimate the ratio of adults to children and figure the expected revenue from this information.
You consult with your student ticket-sellers and discover that they have not been keeping track of how many child tickets they’ve sold. The tickets are identical until the ticket-seller punches a hole in the ticket indicating that it is a child ticket. But they don’t remember how many holes they’ve punched. They only know that they’ve sold 548 tickets for $2460. How much revenue from each of child and adult tickets can you expect? To solve this, we need to figure out the ratio of tickets that have already been sold.
Let A stand for the number of adult tickets pre-sold, and C stand for the child tickets pre-sold. Then A + C = 548. Also, since each adult ticket cost $5.00, then ($5.00)A stands for the revenue brought in from the adult tickets pre-sold; likewise, ($2.50)C stands for the revenue brought in from the child tickets. Then ($5.00)A + ($2.50)C = $2460. But we can only solve an equation with one variable, not two. So look again at that first equation. If A + C = 548, then A = 548 − C (or C = 548 − A; it doesn’t matter at this stage which variable you solve for). Organizing this information in a grid, we get:
tickets sold $/ticket total $ adult 548 − C $5 $5(548 − C) child C $2.50 $2.50C total 548 — $2460
From the last column, we get (total $ from the adult tickets) plus (total $ from the child tickets) is (the total $ so far), or, as an equation:
(5.00)(548 − C) + (2.50)C =
2740 − (5.00)C + (2.50)C =
2740 − (2.50)C =
−280/−2.50C = 112
Then 112 child tickets were pre-sold, and A = 548 − 112 = 436 adult tickets were sold. Using A and C for our variables, instead of x and y, was helpful, because the variables suggested what they stood for. We knew that C = 112 meant 112 child tickets. This is a useful technique.
Now we need to figure out how many adult and child tickets we can expect to sell overall. Since 436 out of 548 pre-sold tickets were adult tickets, we can expect 436/548, or about 79.6%, of the total tickets sold to be adult tickets. Since we expect about 13,000 people, this works out to about 10,343 adult tickets. (You can also find this value by using proportions.) The remaining 2657 tickets will be child tickets. Then the expected ticket revenue is $58,357.50, of which ($5.00)(10,343) = $51,715 will come from adult tickets and ($2.50)(2,657) = $6,642.50 from child tickets.
Let’s try another one. This time, suppose you work in a lab. You need a 15% acid solution for a certain test, but your supplier only ships a 10% solution and a 30% solution. Rather than pay the hefty surcharge to have the supplier make a 15% solution, you decide to mix 10% solution with 30% solution, to make your own 15% solution. You need 10 liters of the 15% acid solution. How many liters of 10% solution and 30% solution should you use?
Let x stand for the number of liters of 10% solution, and let y stand for the number of liters of 30% solution. (In this case, contrary to the ticket problem, the labeling I just did is very important, because x and y are not at all suggestive of what they stand for. If we don’t label, we won’t be able to interpret our answer in the end.) For mixture problems, it is often helpful to do a grid:
liters sol’n percent acid total liters acid 10% sol’n x 0.10 0.10x 30% sol’n y 0.30 0.30y mixture x + y = 10 0.15 (0.15)(10) = 1.5
Since x + y = 10, then x = 10 − y. Using this, we can substitute for x in our grid, and eliminate one of the variables:
liters sol’n percent acid liters acid 10% sol’n 10 − y 0.10 >0.10(10 − y) 30% sol’n y 0.30 0.30y mixture x + y = 10 0.15 (0.15)(10) = 1.5
When the problem is set up like this, you can usually use the last column to write your equation: The liters of acid from the 10% solution, plus the liters of acid in the 30% solution, add up to the liters of acid in the 15% solution. Then:
0.10(10 − y) + 0.30y =
1 − 0.10y + 0.30y =
1 + 0.20y =
0.5/0.20y = 2.5
Then we need 2.5 liters of the 30% solution, and x = 10 − y = 10 − 2.5 = 7.5 liters of the 10% solution. (If you think about it, this makes sense. Fifteen percent is closer to 10% than to 30%, so we ought to need more 10% solution in our mix.)
Usually, these problems are fairly easy to solve once you’ve found the equations. To help you see how to set up these problems, below are a few more problems with their grids (but not solutions).
How many liters of a 70% alcohol solution must be added to 50 liters of a 40% alcohol solution to produce a 50% alcohol solution?
liters sol’n % alcohol total liters alcohol 70% sol’n x 0.70 0.70x 40% sol’n 50 0.40 (0.40)(50) = 20 50% mix 50 + x 0.50 0.50(50 + x)
From the last column, you get the equation 0.7x + 20 = 0.5(50 + x). Solve for x.
How many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a solution that is 10% salt?
ounces liquid % salt total ounces salt water x 0 0 15% sol’n 50 0.15 (50)(0.15) = 7.5 10% mix 50 + x 0.10 0.10(50 + x)
From the last column, you get the equation 7.5 = 0.1(50 + x). Solve for x.
(Note the percentage for water. Pure watercontains no salt, so the percent is zero. If, on the other hand, you were trying to increase the salt content by adding pure salt, the percent would have been one hundred.)
Find the selling price per pound of a coffee mixture made from 8 pounds of coffee that sells for $9.20 per pound and 12 pounds of coffee that costs $5.50 per pound.
pounds coffee $/pound total $ for coffee pricey 8 $9.20 (8)($9.20) = $73.60 cheapo 12 $5.50 (12)($5.50) = $66 mix 8 + 12 = 20 ? $73.60 + 66 = $139.60
From the last row, you see that you have 20 pounds for $139.60, or $139.60/(20 pounds). Simplify.
How many pounds of lima beans that cost $0.90 per pound must be mixed with 16 pounds of corn that costs $0.50 per pound to make a mixture of vegetables that costs $0.65 per pound?
pounds $/pound total $ for veggies lima beans x $0.90 $0.90x corn 16 $0.50 (16)($0.50) = $8 mix 16 + x $0.65 (16 + x)($0.65)
From the last column, you get the equation $0.90x + $8 = (16 + x)($0.65). Solve for x.
Two hundred liters of a punch that contains 35% fruit juice is mixed with 300 liters of another punch. The resulting fruit punch is 20% fruit juice. Find the percent of fruit juice in the 300 liters of punch.
liters punch % juice total liters juice 35% juice 200 0.35 (200)(0.35) = 70 other punch 300 x 300x mix 200 + 300 = 500 0.20 (500)(0.20) = 100
From the last column, you get the equation 70 + 300x = 100. Solve for x, and then convert to a percentage.
Ten grams of sugar are added to a 40g serving of a breakfast cereal that is 30% sugar. What is the percent concentration of sugar in the resulting mixture?
grams in bowl % sugar total grams sugar sugar 10 1.00 10 cereal 40 0.30 (40)(0.30) = 12 mix 50 ? 10 + 12 = 22
From the last row, you see that there are 22 grams of sugar in the 50 grams in the bowl, or 22/50. Simplifying and converting to a percentage gives us 44%.
In the slope-intercept equation y = mx + b), the slope m is multiplied on the x value and b is the y-intercept, where the line crosses the y-axis. Graphing from this format can be rather easy, particularly if m and b are relatively simple numbers rather than fractions or complicated decimals. But, let’s a look at the meanings that slope and y-intercept can have in solving word problems.
You’ve learned that slope defines the ratio of the “rise” (in y) divided by the “run” (in x) between two points on a line, in other words, the ratio of the altitude change to the horizontal distance between any two points on the line. For instance, in y = ( 2x/3 ) + 2, the slope is m = ⅔. This means that, from any point on this line, you can get to another point on the line by going up 2 units then going to the right 3 units.
But we could also view this slope as a fraction over 1:
The ratio remains the same but, for every one unit that x moves to the right, y goes up by two-thirds of a unit. While this doesn’t graph as easily as m = two-thirds, it can be a more useful way of viewing things when doing word problems.
Often, word problems deal with changes over time. For instance, a problem might deal with how tall a plant grows if it increases by a fixed amount every three months.
When x = 0, the corresponding y-value is the y-intercept. However, in the context of word problems, the y-intercept also refers to the starting value. In the equation y = ( 2x/3 ) + 2, the y-intercept, 2, would be the size of the plant when we started keeping track of the data and the slope would indicate its rate of growth. Let’s do a few examples, to see how this works:
The average lifespan of American women has been tracked, and the model for the data is y = 0.2t + 73, where t = 0 corresponds to 1960. Explain the meaning of the slope and y-intercept.
The slope is m = 0.2 and means that every year, the average lifespan of American women increased by 0.2 years.
The y-intercept = 73 and means that, when they started keeping track in 1960, the average lifespan was 73 years.
The equation for the speed of a ball that is thrown straight up in the air is given by v = 128 − 32t, where v is the velocity in feet per second and t is the number of seconds after the ball is thrown. What is the meaning of the slope and with what initial velocity was the ball thrown?
The slope is m = −32 and means that every second the speed decreases by 32 feet per second.
The y-intercept = 128 and means that the ball was launched at 128 feet per second.
Fishermen in the Great Lakes have been recording the dead fish they encounter while fishing in the region. The Department of Environmental Services monitors the pollution index for the Great Lakes and their model for the number of fish deaths y for a given pollution index x is y = 9.607x + 111.958. What are the meanings of the slope and y-intercept?
The slope of m = 9.607 and means that, for every increase of one unit in the pollution index there are approximately nine or ten more fish deaths during the year.
The y-intercept of 111.958 means that, even if the pollution index is zero there would still be about 112 fish deaths a year.
Word problems with linear equations almost always work this way: the slope is the rate of change and the y-intercept is the starting value.
A system of equations is a collection of two or more equations with the same set of unknowns. In solving a system of equations, we try to find values for each of the unknowns that will satisfy every equation. The equations can be linear or non-linear and can be solved several different ways. The following problems will be linear equations with two unknowns and we’ll demonstrate the substitution and elimination methods.
A barge has speed over water of 7 knots per hour. A river flows downstream at the speed of 3 knots per hour. How long will it take the barge to go upstream from point A to point B, then back, if the distance from A to B is 20 miles?
We need to find the effective speed upstream and determine the time is takes to go from point A to point B, then determine the effective speed downstream and determine the time it takes to go from point B to point A. The effective speed upstream is 7 knots − 3 knots so it’s 4 knots per hour upstream. The effective speed downstream is 7 knots + 3 knots so it’s 10 knots per hour downstream. Let’s say t1 equals the time upstream and t2 equals the time downstream and we get:t1 =
t1 =distance ÷ speed
20 ÷ 4
t2 =distance ÷ speed
20 ÷ 10
Now we just add t1 and t2: 5 hours + 2 hours = 7 hours
But we’re not quite done. Reread the question: “How long will it take the barge to go upstream from point A to point B, then back…?” So you answer should be written:
It will take 7 hours for the barge to go upstream and back.
That was pretty easy…let’s try one that’s a bit more challenging:
A total of $12,000 is invested in two funds paying 9% and 11% simple interest. If the yearly interest is $1,180, how much of the $12,000 is invested at each rate?
For this problem, interest is calculated by multiplying the amount invested times the interest rate.
First, we have to translate this from English to algebra:
“A total of $12,000 is invested in two funds paying 9% and 11%.” If we let x equal the 9% fund and y equal the 11% fund, we can say:
x + y = 12000
When we consider that “the yearly interest is $1,180”, and “interest is calculated by multiplying the amount invested times the interest rate”, we can also say:
0.09x + 0.11y = 1180
We’ve converted the word problem into equations and we’re ready to solve.
The Substitution Method:
Step 1: Solve for y in the first equation:
x + y =
12000 – x
Step 2: Substitute this value for y in the second equation. This will change it to an equation with just one variable:
0.09x + 0.11y =
0.09x + 0.11(12000 – x) =1180
Step 3: Solve for x in the translated equation:
0.09x + 0.11(12000 − x) =
0.09x + 1320 − 0.11x =
− 0.02x =
Step 4: Substitute this value of x in the first equation:
x + y =
7000 + y =
Don’t forget the original question. You now have to translate your answer from algebra to English:
The amount invested at 9% is $7000 and the amount invested at 11% is $5000
You can check your answers by substituting the values of x and y in each of the original equations. If, after the substitution, the left side of the equation equals the right side of the equation, you know that your answers are correct.
The Elimination Method:
We determine our algebraic equations as we did above so we’ll just use them again:
x + y = 12000
0.09x + 0.11y = 1180
With this method, we have to create an equivalent equation for one of these so that when it’s added to the other, one of the variables is eliminated, then solve for the remaining variable. We could probably find a way to multiply through the second equation to get 1x or 1y but we’d wind up with some pretty messy numbers so it would be easier to convert the first one:
Step 1: Multiply x + y = 12000 by −0.09 to obtain an equivalent equation:
x + y =
−0.09x − 0.09y =12000
Step 2: Add this equivalent equation to 0.09x + 0.11y = 1180:
0.09x + 0.11y = 1180
−0.09x − 0.09y = 10800.02y =100y = 5000
Step 3: Substitute y = 5000 in our first equation to solve for x.
x + y =
x + 5000 =
Again, remember that the solution isn’t x = 7000 and y = 5000. Look at the question and phrase your answer accordingly:
“How much of the $12,000 is invested at each rate?”
The amount invested at 9% is $7000 and the amount invested at 11% is $5000
Most problems of this type can be solved using either method but some are better suited to one or the other. As usual, when given the option, use the method you’re most comfortable with.
The age of one brother combined with twice the age of a second brother is 50. The difference between twice the age of the first brother and the age of the second brother is 10 years.
Let’s call the first brother x and the second brother y then work through the translation from English to algebra:
English: The age of one brother combined with twice the age of a second brother is 50 Algebra: x + 2y = 50
For the next bit of information we have to do some rearranging. We know the “difference” means subtracting one from the other, so:
English: twice the age of the first brother difference the age of the second brother is 10 Algebra: 2x − y = 10
We now have algebraic equations we can work with: x + 2y = 50 and 2x − y = 10 and need to combine them in a way that will eliminate one of the variables. Since we have a positive 2y in the first equation and a negative y in the second, let’s work on them. If we multiply through the second term by two we’ll get 4x − 2y = 20 and we can add like so:
x + 2y = 50
+ 4x − 2y = 205x= 70x = 14
Now we can plug x into either equation and solve for y. Since there’s only one y in the second equation let’s use that to avoid dividing:
2x − y =
2(14) − y =
28 − 28 − y =
− y =
10 − 28
Now, let’s plug our answers into the equations to see if they work:
x + 2y =
14 + 2(18) =
14 + 36 =
502x − y =
2(14) − 18 =
28 − 18 =
Now you try one:
Three ducks and two ducklings weigh 32 kg. Four ducks and three ducklings weigh 44kg. All ducks weigh the same and all ducklings weigh the same. What is the weight of two ducks and one duckling?