# How to Solve Age Problems Part 2

This is the second part of the Solving Age Problem Series. We will continue solving age problems that are slightly more complicated that the first part. We have already discussed 3 problems in the first part of this series, so we continue with the fourth problem.

**Problem 4**

Simon is four years older than Jim. The sum of their ages is 52. How old is Simon?

**Scratch Work**

This problem is a sort of review of first part of this series. Simon is older than Jim by years. So, if Jim is years old, then Simon is years old. The sum of their ages is . This means that if add and , then the sum is . That is the equation.

**Solution**

Let be Jim’s age and be Simon’s age.

Now,

Jim’s age + Simon’s age = 52 which means that

.

Simplifying we have

.

Subtracting from both sides, we have

So, Jim is years old. Now, the question asks for the age of Simon. Simon is years old.

**Check**

Simon is and Jim is , so he is indeed four years older. The sum of their ages is which agrees with the given in the problem. Therefore, we are correct.

**Problem 5**

Allan is times as old as Leah. Five years from now, he will be times as old. How old is Allan?

**Scratch Work**

Now, if Leah is, for example, 7 years old, then Allan is years old. This means that if Leah is years old, then Allan is years old. Five years from now, Leah will be years old and Allan will be years old as shown on the table below.

Note that $latex $5 years from now, Allan will be three times as old as Leah. This means that if we multiply Leah’s age by , then, their ages will be equal. That is, if we multiply by , it will be equal to . In equation form,

which is the final equation.

**Solution**

Let be Leah’s age and be Allan’s age.

Five years from now, Leah will be years old and Allan will be years old.

Now, we multiply Leah’s age and equate it to that of Allan’s

.

By Distributive Property, we have

.

Putting all x’s to the right and all numbers to the left, we have

.

Dividing both sides by , we have

.

So, Leah is years old and Allan is

**Check**

Allan is and Leah is so he is indeed times as old. In years, Allan will be and Leah will be . Thirty is indeed three times , so we are correct.

**Problem 6**

Philip is twice as old as Ben. If is subtracted from Philip’s age and is added to Ben’s age, then their ages will be equal. How old are both of them?

*Scratch Work*

Ben is years old and Philip is . If we subtract from Philip’s age, it will become . If we add to Ben’s age, it will be . Now, after the results to these operations, their ages will be equal or

**Solution**

Let be Ben’s age and be Philip’s age.

.

So, Ben is and Philip is .

**Check**

Philip is and Ben is so, Philip is twice as old Ben. Subtracting from Philip’s age results to . Adding to Ben’s age is . Well, equals , so we are correct.

In the next post, we will be discussing more age problems. Please keep posted.

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[…] How to Solve Age Problems Part 2 discusses a slightly more difficult 2-person problems particularly present-past and present-future age relationships. […]