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.
Arvin is 5 years older than Michael. The sum of their ages is 37. What are their ages?
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.
16 + 21 = 37.
Mia is 3 years older than Pia. In 4 years, the sum of their ages is 35. What are their ages?
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.