Age Word Problems

Age Word Problems in Math

Hi and welcome to this is video lesson on age word problems. Yes, the crazy, confusing, tricky age word problems where you’re given a bunch of names, and ages, and code, but don’t worry we’re going to find the straightforward way to solve for one of these problems.

Let’s take a look. Mike, Gavin, and Bella are siblings. Mike is 3 years older than twice Bella’s age. Gavin is 5 years younger than Mike. If Bella is 4, determine the age of her siblings. Ok, so we need to sort through these names and ages before we get confused. Let’s take three steps to do that.

Step 1: Create a little table of things we know. Since there are three people, let’s put down their names, Bella, Mike, Gavin.

Step 2: Let’s put their age equations into our table. First, let’s say Bella’s age is \(X\), because her brothers ages depend on hers, so Bella is \(X\) years old. Mike is twice Bella’s age, plus 3 years, so let’s do 2 times \(X\), plus 3 for Mike. Gavin is 5 years younger than Mike, so let’s go ahead and use Mike’s equation, minus 5, which is \(2X\) minus 2.

Now that we know that let’s go ahead and work on Step 3: To solve for the ages of Bella and her brothers. In the last part of the problem we were told to substitute 4 for Bella’s age, so let’s go ahead and say that \(X\) is equal to 4.

If that’s a given, then we know Mike’s and Gavin’s age because all we have to do is plug 4 in for \(X\). For Mike, 2 times 4, plus 3 is equal to 11, for Gavin, 2 times 4, minus 2 is equal to six, so we know that Bella is 4, Mike is 11, and Gavin is 6.

I hope that helps. Thanks so much for watching this video lesson, and until next time, happy studying.



by Mometrix Test Preparation | This Page Last Updated: July 21, 2023