# 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. Continue reading

# 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.  Continue reading