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.

age problem 2

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

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