Introduction to Age Problems in School Mathematics

Age Problems is the second part of the Math Word Problem Solving Series here in Mathematics and Multimedia. In this series, we are going to learn how to solve math word problems involving age.

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. » Read more

Math Word Problems: Solving Age Problems Part 2

This is the second part of the 3-part installment posts on Solving Age Problems in the Math Word Problem Solving Series.

In this post, we continue with three more worked examples on age problems.  The first part of this series can be read here.

PROBLEM 4

Janice is four times as old as his son. In $5$ years, she will be as three times as old as his son. What is Janice’s present age?

Solution

Let $x$ be the present age of Janice’s son and 4x be her age. According to the problem, in five years, she will be three times as old as her son.

In five years, Janice age will be $4x + 5$ and her son’s age will be $x+5$. If she is three times as old as her son, if we multiply her son’s age by $3$, their ages will be equal. That is,

$3(x + 5) = 4x + 5$.

Simplifying, we have $3x + 15 = 4x + 5$ giving us $x = 10$. Therefore, the son is $10$ years old and Janice is $40$ years old.  » Read more

Math Word Problems: Solving Age Problems Part 1

This is the second topic in the Word Problems Series. In this series of posts, we are going to discuss age problems.  In age problems, most times, ages of persons at different points in time — past, present, and future — are asked about.

One of the most famous age problems in the history of mathematics is Diohpantus’ Riddle. After this series, you should be able to provide a solution to Diophatus’ age problem. 🙂

As we have done in the previous topic, we will start with simple problems. As we go on, the problems become more complicated.  Let us start with the first problem.

PROBLEM 1

Adrian’s age is one less than twice Carlo’s age. The sum of their ages is 50. What are their ages?

Solution

As you have probably noticed that this problem is very similar to the problems we discussed earlier.  Age problems just like most algebra problems which are number problems in disguise.  » Read more