# How to Solve Problems by Working Backwards Part 2

In the **previous post**, we have learned how to solve number problems by working backward. In this post, we discuss age problems. We already had 2 examples in the previous post in this series, so we start with the third example.

**Example 3**

Arvin is 5 years older than Michael. The sum of their ages is 37. What are their ages?

**Solution**

This is very similar to the problems in the previous post in this series. Arvin is 5 years older than Michael, so if we subtract 5 from Arvin’s age, their ages will be equal. But if we subtract 5 from Arvin’s age, we also have to subtract 5 from the sum of their ages. That is,

37 – 5 = 32.

Now, their ages are equal, so we can divide the sum by 2. That is 32 ÷ 2 = 16. This means that the younger person is 16 since we subtracted 5 from Arvin’s age. Therefore, Michael is 16 and Arvin is 16 + 5 = 21.

**Check**

16 + 21 = 37.

**Example 4**

Mia is 3 years older than Pia. In 4 years, the sum of their ages is 35. What are their ages?

**Solution**

There are two of them and we added 4 to both ages, so subtracting 8 from 35 (the sum) will determine the sum of their present age. That is 35 – 8 = 27 is the sum of their present age.

Next, Mia is 3 years older than Pia, so if we subtract 3 from the sum of their present age, their ages will be equal. So, 27 – 3 = 24.

We can now divide the sum of their ages by 2. That is 24/2 = 12.

This means that 12 is the age of the younger person because we subtracted 3 from the age of the older person.

So, Pia is 12 and Mia is 15.

**Check** 12 + 15 = 22.

In the **next post**, we will discuss more problems that can be solved by working backward.