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There are many quant topics that you must know to succeed on the GRE. Some topics are more technically based, such as Linear and Quadratic Equations or Number Properties. Solving questions on these topics is tough enough, but the GRE adds a layer of difficulty by presenting word problems that require the use of these concepts. To tackle those problems, you must first create the equations that you’ll need to solve, based on the information provided in the question.
So, in this article, I’ll provide a number of strategies for efficiently solving GRE word problems, as well as examples, so we can see these strategies in action.
Here are the topics we’ll cover:
- The Substitution Method of Solving GRE Algebra Problems
- The Combination Method of Solving Algebra Problems
- When Dealing With Word Problems, Translation Is the Name of the Game
- You Must Balance Your Equations
- “Number of Items” Questions
- Age Problems
- Money Problems
- Word Problems With Variables in the Answer Choices
- Summary
- What’s Next?
We’ll start with the substitution method of solving GRE algebra problems.
Here are some sample GRE quantitative reasoning word problems that you might encounter:
Greta buys apples, which cost $1 each, and bananas, which cost $2 each, at the market. If she buys twice as many bananas as apples and spends $30 at the market, how many bananas does she buy?
John and Greg have a total of 40 marbles. If John has 3 times the number of marbles that George has, how many marbles does George have?
At their core, word problems present a scenario we must translate into variables and math equations and solve. This article will focus on those general word problems and how to solve them using various algebra techniques. But, before jumping into GREl word problems practice questions, let’s review the primary techniques we will use to solve these word problems. Since most GRE word problems have two variables and two equations, let’s practice the substitution method.
The Substitution Method of Solving GRE Algebra Problems
You will find that when dealing with algebra on the GRE, the majority of the time you solve equations, you are using substitution. (Alternatively, you may use the combination method, which we will discuss later in this article.) The good news is that substitution is pretty easy, and you can master it with just a bit of practice.
The general basis of substitution is that we isolate a variable in one equation and substitute what it equals into another equation. We do this so that we get one equation with just one variable, and then we solve for that single variable.
TTP PRO TIP:
We use substitution when solving most algebra problems that have two equations and two variables.
For example, let’s say we have the following two equations:
Equation 1: 2x = y
Equation 2: x + y = 21
Since equation 1 has y isolated, we can substitute y for 2x into equation 2. Let’s do so, and then solve for x:
x + 2x = 21
3x = 21
x = 7
We can easily solve for y by substituting 7 for x in equation 1:
2(7) = y
14 = y
Now, let’s practice with an example.
Substitution Method: Example 1
If 2m – 4 = 3m – 2n and 2m = 3n, then m equals which of the following?
- 2
- 4
- 7
- 8
- 12
Solution:
First, we can simplify the first equation by isolating m:
2m – 4 = 3m – 2n
2n – 4 = m
Next, we can substitute 2n – 4 from the first equation for m in the second equation, giving us:
2(2n – 4) = 3n
4n – 8 = 3n
n = 8
Since n = 8, we now can determine m:
2m = 3 * 8
2m = 24
m = 12
Answer: E
Remember, knowing the substitution method will be very helpful when dealing with word problems.
Now, let’s discuss another way of solving equations: the combination method.
The Combination Method of Solving Algebra Problems
As we just saw, substitution is a great way to solve algebra problems. However, substitution works well only when we can easily and cleanly isolate a variable in an equation. So, what happens when we cannot? Well, we have another fun technique we can use, the combination method.
The basis of the combination method is that we add two equations together to eliminate one of the variables, thereby leaving us with one equation that contains one variable. So, in essence, the destination of solving for a variable has not changed, but the vehicle getting us there has!
For example, let’s say we have the following two equations:
Equation 1: 2x + 3y = 28
Equation 2: 4x – 3y = 20
So, although we technically could go the route of substitution, doing so would produce fractions and get pretty messy. So, instead, the move is to add the equations together (using the combination method) and eliminate the y variables.
Let’s add the equations. When doing so, we combine all the like terms. So, we have:
2x + 4x = 6x
3y – 3y = 0
28 + 20 = 48
Thus, we are left with:
6x = 48
x = 8
TTP PRO TIP:
We use the combination method of solving equations when we cannot easily isolate a variable.
Let’s practice with another example.
Combination Method: Example 1
If 5x – 2y = 21 and 7x + 2y = 39, then x equals which of the following?
- 5
- 8
- 10
- 12
- 26
Solution:
Looking at the given equations, we see that there is no way to isolate a variable easily, so the substitution method isn’t optimal. Thus, we will use the combination method and ultimately add the two equations together.
When we add 5x – 2y = 21 and 7x + 2y = 39, we are left with:
12x = 60
x = 5
Answer: A
Now that we have a general overview of solving and writing equations, we can discuss the star of the show: word problems!
When Dealing With Word Problems, Translation Is the Name of the Game
Having read the previous two sections, you are now solid at solving equations. However, word problems are not just about solving equations; they are also about translating words into equations! So, let’s discuss some phrases and words to look out for and how to translate them.
- “Is” translates to equals (=)
Daphne is the same age as Paul
Daphne’s age = Paul’s age
- “More” translates to addition (+)
Francesca has 6 more marbles than Pablo
Francesca = Pablo + 6
- “Less/fewer” translates to subtraction (-)
Samantha has 3 fewer coins than Cindy
Samantha = Cindy – 3
- “Times as many” translates to multiplication (✖)
Harold has 5 times as many newspapers as Carl
Harold = Carl ✖ 5
Keep in mind while we have listed some common translations above, there are others. These just happen to be the most common.
Now, before jumping into word problem practice questions, let’s discuss one point of confusion students have when translating words into equations: properly balancing the equations.
You Must Balance Your Equations
A basic principle of translating words into equations is that, ultimately, the equations must be balanced. If they are not, once you solve for a variable, you may come up with a negative value. Let’s look at a few correct and incorrect ways to balance equations, and we’ll explain why the equations are balanced correctly.
Balancing Equations: Example 1
Thomas has 30 more dollars than Maxine.
Solution:
When considering this equation, we must understand that Thomas has more money than Maxine.
So, to properly balance the equation, we must add 30 dollars to Maxine’s amount and set it equal to Thomas’ amount.
If we let T = Thomas’s money and M = Maxine’s money, we have:
- T = 30 + M
Balancing Equations: Example 2
Naif has 4 fewer toys than Paula.
Solution:
When considering this equation, we must understand that Naif has fewer toys than Paula.
Equivalently, this means that Paula has more toys than Naif.
So, to properly balance the equation, we must subtract 4 from Paula’s amount and set that amount equal to Naif’s.
If we let P = the number of Paula’s toys and N = the number of Naif’s toys, we have:
- P – 4 = N
Balancing Equations: Example 3
There are 5 times as many blue balls as red balls.
Solution:
When considering this equation, we must understand that there are more blue balls than red balls.
So, to properly balance the equation, we must multiply the number of red balls by 5 and set that amount equal to the number of blue balls.
If we let B = the number of blue balls and R = the number of red balls, we have:
- 5R = B
Now that we are familiar with how to translate, balance, and solve equations, let’s jump into the various types of general word problems you will encounter on the GRE, and how to solve them. We will start with basic word translations.
“Number of Items” Questions
We have already covered the concepts tested in “number of items” questions in the previous sections, but we can practice a few more GRE math word problems here as well.
Essentially, a “number of items” question is one in which we have two (or sometimes more) relationships between the number of people or things. Those relationships are described in words, and it’s our job to create and solve equations based on the relationships.
Let’s practice with a couple of these questions.
“Number of Items”: Example 1
Haily and Todd have a total of 20 cookies. If Haily has 4 times as many cookies as Todd, how many cookies does Todd have?
- 4
- 8
- 10
- 12
- 15
Solution:
First, we need to define our variables.
We can let H = the number of cookies Haily has and T = the number of cookies Todd has. Next, let’s create equations based on what the question stem tells us.
“Haily and Todd have a total of 20 cookies” can be translated to:
H + T = 20
“Haily has 4 times as many cookies as Todd” can be translated to:
H = 4T
Next, we can substitute 4T for H, and we have:
4T + T = 20
5T = 20
T = 4
Answer: A
Let’s practice one more. But, this time, we will add a twist.
“Number of Items”: Example 2
Pedro has 3 times as many stamps as Holland. If Pedro gives Holland 3 stamps, Pedro will have 2 more stamps than Holland. How many stamps did Pedo originally have?
- 6
- 8
- 12
- 16
- 18
Solution:
First, let’s define our variables.
We can let P = the number of stamps Pedro has and H = the number of stamps Holland has. Next, we can create some equations.
Because Pedro has 3 times as many stamps as Holland, we have:
P = 3H
If Pedro gives Holland 3 stamps, Pedro will then have 2 more stamps than Holland, so we have:
P – 3 = H + 3 + 2
P = H + 8
Next, we can substitute 3H for P, and we have:
3H = H + 8
2H = 8
H = 4
Thus, Pedro originally had 3 x 4 = 12 stamps.
Answer: C
Next, let’s discuss another common type of general word problem: age problems.
Age Problems
Age problems are a special type of GRE problem that tests us on word translations but with a twist. Instead of a somewhat basic translation, as we practiced above, an age problem will usually force us to translate an equation based on an age in the past or an age in the future, or even both.
For example, we may have to compare Victoria’s age in 10 years to someone else’s. To do so, we follow a very simple process. If we let Victoria’s age today = V, then her age in 10 years is V + 10.
In another scenario, if we compare her age 15 years ago to someone else’s, we would represent her age as V – 15.
In either case, we would then translate the given information into equations and solve as we have already practiced in previous sections.
TTP PRO TIP:
Age problems generally force us to compare ages in the past or future.
Let’s now practice with an example.
Age Problem: Example 1
William is 8 years younger than Mara. If, in 6 years, Mara will be twice as old as William, how old is William today?
- 2
- 4
- 6
- 8
- 10
Solution:
First, we can define our variables.
We can let W = Williams’s age today and M = Mara’s age today.
Next, let’s create some equations.
Because William is 8 years younger than Mara, we have:
W = M – 8
Next, we see that we must represent both their ages 6 years from now. Thus, William will be W + 6, and Mara will be M + 6.
If, in 6 years, Mara will be twice as old as William, we have:
M + 6 = 2(W + 6)
M + 6 = 2W + 12
M = 2W + 6
Next, we can substitute M – 8 for W in equation 2, and we have:
M = 2(M – 8) + 6
M = 2M – 16 + 6
10 = M
Thus, William is 10 – 8 = 2.
Answer: A
Next, let’s discuss money problems.
Money Problems
Another type of general word problem involves money. Money problems generally will have various quantities of items, so solving these problems using the substitution method can get quite messy. Thus, we mainly resort to using the combination method for these problems. So, let’s keep this process in mind when doing the following example.
TTP PRO TIP:
Be prepared to use the combination method when dealing with money problems.
Money Problem: Example 1
At a certain store, the price for each painting is the same, and the price for each photograph is the same. If 5 paintings and 8 photographs cost a total of $468, and the total cost of 5 paintings and 6 photographs is $406, what is the cost of a single painting?
- $28
- $31
- $33
- $35
- $41
Solution:
First, let’s create our variables.
We can let the price of each painting = A and the price of each photograph = P.
Next, let’s create our equations.
Because 5 paintings and 8 photographs cost a total of $468, we have:
5A + 8P = 468
Because the total cost of 5 paintings and 6 photographs is $406, we have:
5A + 6P = 406
So, it’s pretty difficult to isolate any of our variables. Thus, we will resort to the combination method. However, we need to do one extra step before adding the equations together to cancel out a variable. That step is multiplying either equation by -1. Let’s do that with the second equation.
-1(5A + 6p = 406) = -5A – 6P = -406
Now, if we add together 5A + 8P = 468 and -5A – 6P = -406, we obtain:
2P = 62
P = 31
Answer: B
Next, let’s discuss what happens when we have to solve word problems with variables in the answer choices.
Word Problems With Variables in the Answer Choices
So far in this article, we have discussed various word problems in which we have to solve for a numerical value. There are additional types of word problems that contain mostly variables and have variables in the answer choices. While these problems may seem intimidating, the good news is that you can solve them using the very same translation and algebra skills that you’ve been using thus far.
Let’s look at a previous example we’ve already solved, but replace the numbers in the problem with variables.
Variable Word Problem: Example 1
Haily and Todd have a total of c cookies. If Haily has 4 times as many cookies as Todd, in terms of c, how many cookies does Todd have?
- c/5
- c/10
- c/15
- 5c
- 10c
Solution:
NOTE: When following the solution, you should notice that not much will change from the previous solution that contained all numbers.
First, we need to define our variables.
We can let H = the number of cookies Haily has and T = the number of cookies Todd has.
Next, let’s create some equations based on the question stem.
“Haily and Todd have a total of c cookies” can be translated to:
H + T = c
“Haily has 4 times as many cookies as Todd” can be translated to:
H = 4T
Next, we can substitute 4T for H, and we have:
4T + T = c
5T = c
T = c/5
Answer: A
Let’s practice with one more!
Variable Word Problem: Example 2
Maria has 5 times as many airline miles as her friend Vito. Together they have a total of z airline miles. If Vito were to obtain more airline miles, how many airline miles, in terms of the variable(s) above, would Vito then have?
- (z – w) / 2
- v + m
- V + m + w/2
- (v – 5m) / 2
- (z + 3w) / 6
Solution:
Let m = the number of airline miles that Maria has and v = the number of airline miles that Vito has.
Since Maria has 5 times as many airline miles as Vito, we can write equation 1 as m = 5v.
Since together they have a total of z airline miles, equation 2 can be v + m = z.
Let’s substitute the 5v from the first equation for m in equation 2, and then solve for v:
v + m = z
v + 5v = z
6v = z
v = z / 6
Finally, we add (1/2)w to Vito’s total airline miles:
v = z / 6
v = z / 6 + w / 2
v = z / 6 + 3w / 6
v = (z + 3w) / 6
Answer: E
Summary
Of the 20+ major math topics on the GRE, one of the most challenging is word problems. This topic combines the requirement to translate words into equations with traditional algebraic techniques for solving equations. Familiarity with the strategies for solving GRE word problems is a must for scoring well on GRE quant.
The two main techniques for solving many types of word problems are the substitution method and the combination method. Most word problems can be translated from words into two equations. Then, the values of the variables involved can be determined by using either the substitution method or the combination method to answer the question.
Some of the common types of word problems that we have covered here are “number of items” questions, age questions, money questions, and word problems with variables in the answer choices.
What’s Next?
In this article, we have covered the topic of word problems on the GRE. There are dozens of other major topics to learn in order to earn a good GRE score. Read our article on strategies to improve your quant score on the GRE for more quantitative reasoning tips and tactics.
