Introduction to Age Problems in School Mathematics
Age problems are very similar to number word problems. They are easy to solve when you know how to set up the correct equations. Most of the time, this type of problem discusses the age of a certain individual in relation to another in the past, present, or future.
Below, are some of the common phrases used in age problems. In all the phrases, we let x be the age of Hannah now.
- Hannah’s age four years from now (x + 4)
- Hannah’s age three years ago (x – 3)
- Karen is twice as old as Anna (Karen’s age: 2x)
- Karen is thrice as old as Anna four years ago (Karen’s age: 2(x-4))
In solving age problems, creating a table is always helpful. This is one of the strategies that I am going to discuss in this series. As a start, we discuss one sample problem.
Problem: Karen is thrice as old as Hannah. The difference between their ages is 18 years. What are their ages?
First, we let x be Hannah’s age. Since Karen is thrice as old as Hannah, we multiply her age by 3. For instance, if Hannah is 10 years old now, then Karen is 30 or 3(10).
Now, since Hannah’s age is x, then Karen’s age is thrice x or 3x.
Now, to set up the equation, we look at the second statement. The difference between their ages is 18 years. This means that
Karen’s Age – Hannah’s age = 18.
In equation form, we have .
This gives us which means that .
So, Hannah is years old and Karen is .
Checking Your Work
You can always verify your work afters solving a problem. Let us check if the answers above is correct. Let’s look at the first sentence in the problem.
Is Karen thrice as old as Hannah? Yes, Karen is 27 and 27 is thrice 9.
Is the sum of their ages 18? Yes, 27 – 9 = 18.
As you can see, age problems is not that difficult once you know how to set up the correct equation. In the next posts, we are going to discuss more complicated age problems. Stay tuned.