Word Problems for the DAT
/Learn key concepts related to word problems, plus practice questions and answers
Everything you need to know about word problems for the dat
Table of Contents
Part 1: Introduction to word problems
Part 2: Basic word problems
a) Overview
b) Relative age
c) Coin problems
d) Combined work
e) Distance/rate
Part 3: Equation-based problems
a) Percent change
b) Dilutions
c) Compound interest
Part 4: Questions and answers
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Part 1: Introduction to word problems
Word problems on the DAT can be somewhat tricky. The key is to translate the words into equations! This guide will walk you through different word problems you’ll see on the DAT. At the end of this guide, test yourself with practice questions and answers.
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Part 2: Basic word problems
a) Overview
We’re going to go over the most common types of word problems with examples of how to solve each. The types of questions we want to master for test day are:
Relative age
Coins
Combined work
Distance/rate
Percent changes
Dilutions
Compound interest
b) Relative age
The key to these and most other word problems is to translate the words of the question into equations. Specifically, the equations for relative age problems will relate the ages of the people in the question. For example, Abby is 10 years older than Brooks. If A is Abby’s age and B is Brooks’ age, then A = B + 10.
This is the basic concept for these types of questions. Let’s do a simple example.
Abby is 5 years older than Brooks, who is 10 years younger than Chris. What is the age difference between Abby and Chris?
Let’s look at the first part of the first statement, which states that Abby is 5 years older than Brooks. This means that Brooks needs 5 more years to match Abby. Alternatively, Abby must lose 5 years to match Brooks.
A = B + 5
Now, let’s look at the second part: Brooks is 10 years younger than Chris. This means that Brooks needs 10 years to match Chris.
C = B + 10
Since the question asks us to relate Chris to Abby, we want a final equation to include both C and A. To do so, set both equations equal to B.
A = B + 5
A − 5 = B
C = B + 10
C − 10 = B
Now, we are going to substitute B.
C − 10 = B
C − 10 = A − 5
C = A + 5
Translating this back into words, Abby needs 5 years to match Chris. Thus, Abby is 5 years younger than Chris.
Let’s look at a more difficult example of these types of questions.
Dallin is 4 years younger than Hannah. 3 years ago, Hannah was twice as old as Dallin. How old is Hannah now?
Again, let’s turn this problem into an equation. The first statement says that Dallin is 4 years younger than Hannah, so H = D + 4. The next statement says that Hannah’s age was double Dallin’s age three years ago. To express this mathematically, subtract 3 from Hannah’s age and set that equal to 2 times D - 3.
H - 3 = 2(D - 3)
At this point, you can rearrange the original equation to D = H - 4. Substitute H - 4 into the second equation for the value of D.
H - 3 = 2(H - 4 - 3)
H - 3 = 2H - 14
11 = H
Therefore, Hannah is 11 years old, making Dallin 11 - 4 = 7 years old. To check our answer, subtract 3 from each age, and we see that 3 years ago, Hannah was 8 and Dallin was 4.
c) Coin problems
Coin problems are similar to relative age problems in that you should translate the text into an equation. After doing so, simply plug in the values of the coins and solve.
Again, let’s walk through an example of these question types.
You have 15 quarters, and three times as many dimes as quarters. Additionally, you have 10 USD in bills. What is the total value of coins and dollar bills?
Since there are triple the number of dimes, it takes 3 times the number of quarters to match the number of dimes. Thus,
3 · Q = D
Since there are 15 quarters, there are 45 dimes. Now we have to find the total value:
(15 · 0.25) + (45 · 10) + 10
3.75 + 4.50 + 10 = 18.25
Coin problems are among the easier word problems on the DAT.
d) Combined work
The general idea for combined work problems is to:
Define variables
Make equations
Solve
You’re probably picking up by now that the approach to solving word problems is generally the same. Combined work problems can look something like this:
Abigail and Brooks are washing cars. Abigail can wash 10 cars in 5 hours, and Brooks can wash 5 cars in 5 hours. How many cars can they wash in 8 hours?
Remember the plan - define variables, make equations, and solve.
Our variables are the rates of work by Abigail and Brooks. Abigail washed 10 cars in 5 hours, so her variable A is equal to \(\frac{10\space cars}{5\space hours}=\frac{2\space cars}{1\space hour}\)
Likewise, Brooks washed 5 cars in 5 hours, so his variable B is equal to \(\frac{5\space cars}{5\space hours}=\frac{1\space car}{1\space hour}\)
Equations for combined work problems often take this form:
work = rate of work × time
To answer the question, we want to solve for work. We know the time is 8 hours. The rate of work is simply A + B, so
work = (A + B) × 8 hours
= (2 cars/hr + 1 car/hr) × 8 hours
= 24 cars
The answer to this question is 24 cars in 8 hours.
Let’s change up the problem a little bit.
Abigail can wash 2 cars in 1 hour, and Brooks can wash 1 car in 40 minutes. If Abigail and Brooks work together, how long will it take for them to wash one car?
This question differs from the previous one in that Abigail and Brooks are working together on the same job rather than separately. Abigail’s rate of washing cars is 2 cars per hour, or 1 car every 30 minutes. Brooks’ rate is 1 car every 40 minutes. To solve problems like this, we can use the following equation:
Plug the time that it takes for Abigail and Brooks to wash the car into the equation to solve for one over the total time.
\[\frac{4}{120}+\frac{3}{120}=\frac{7}{120}\] One over the total time is equal to \(\frac{7}{120}\). To determine the value for total time, simply flip both fractions.
When Abigail and Brooks work together to wash the same car, it will take them 17.1 minutes.
e) Distance/rate
These problems are very similar to the first type of combined work problems we just covered. Oftentimes, we make connections like
Distances ∼ Work
Rate (these are usually velocities) ∼ Rate of Work
Time is similar to well... Time!
The only real issue that makes them different is the direction of travel.
Let’s keep up with Abigail and Brooks. If they are 60 miles apart and driving towards each other with Abigail driving 20 MPH and Brooks 10 MPH, how long does it take for them to reach each other?
We’re going to keep with the same setup as the previous problems, where we define variables, make equations, and then solve them. The variables are the speeds that each person is traveling, where A = 20 MPH and B = 10 MPH. For the quantitative reasoning section, assume these speeds are constant unless you are told otherwise.
Rate is a little bit tricky here, as it is not equal to A or B alone. In our case, Abigail and Brooks are driving towards each other, so they are both working to shorten the distance between themselves. Therefore, the rate is equal to A + B.
Distance is given as 60 mi. Plug these variables in, and you can solve for time.
The final answer is 2 hours when Abigail and Brooks are driving towards each other.
The main takeaway from this problem is that, when two people or objects are traveling towards each other, the total rate is equal to the sum of the individual rates.
What if they are driving in the same direction? Let’s say that Abigail and Brooks drive at 20 and 10 MPH, respectively. But this time, they start driving at the same time from the same location. How far away will they be in 10 hours?
The idea is that we want to find the distances that Brooks travels and Abigail travels and take the difference. Once again, the variables are A = 20 MPH and B = 10 MPH.
DA = 20 MPH × 10 hours = 200 mi
Likewise,
DB = 10 MPH × 10 hours = 100 mi
The difference between DA and DB is how far apart they are, so
The final answer is that Brooks and Abigail will be 100 mi apart after 10 hours of driving. Another way to solve this problem, though, is to recognize that both people are traveling in the same direction. Therefore, their combined rate will be equal to the difference in their rates. In this case, the combined rate is 20 MPH - 10 MPH = 10 MPH, which would give us the same answer.
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