# 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 $4$ years. So, if Jim is $x$ years old, then Simon is $x + 4$ years old. The sum of their ages is $52$. This means that if add $x$ and $x + 4$, then the sum is $52$. That is the equation.

Solution

Let $x$ be Jim’s age and $x + 4$ be Simon’s age.

Now,

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

$x + (x + 4) = 52$.

Simplifying we have

$2x + 4 = 52$.

Subtracting $4$ from both sides, we have

$2x = 48$

$x = 24$

So, Jim is $24$ years old. Now, the question asks for the age of Simon. Simon is $24 + 4 = 28$ years old.

Check

Simon is $28$ and Jim is $24$, so he is indeed four years older. The sum of their ages is $28 + 24 = 52$ which agrees with the given in the problem. Therefore, we are correct.

Problem 5

Allan is $5$ times as old as Leah. Five years from now, he will be $3$ times as old. How old is Allan?

Scratch Work

Now, if Leah is, for example, 7 years old, then Allan is $5(7)$ years old. This means that if Leah is $x$ years old, then Allan is $5x$ years old. Five years from now, Leah will be $x + 5$ years old and Allan will be $5x + 5$ 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 $3$, then, their ages will be equal. That is, if we multiply $x + 5$ by $3$, it will be equal to $5x + 5$. In equation form,

$3(x + 5) = 5x + 5$

which is the final equation.

Solution

Let $x$ be Leah’s age and $5x$ be Allan’s age.

Five years from now, Leah will be $x + 5$ years old and Allan will be $5x + 5$ years old.

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

$3(x + 5) = 5x + 5$.

By Distributive Property, we have

$3x + 15 = 5x + 5$.

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

$15 - 5 = 5x - 3x$

$10 = 2x$.

Dividing both sides by $2$, we have

$5 = x$.

So, Leah is $5$ years old and Allan is $5(5) = 25$

Check

Allan is $25$ and Leah is $5$ so he is indeed $5$ times as old. In $5$ years, Allan will be $30$ and Leah will be $10$. Thirty is indeed three times $10$, so we are correct.

Problem 6

Philip is twice as old as Ben. If $5$ is subtracted from Philip’s age and $10$ is added to Ben’s age, then their ages will be equal. How old are both of them?

Scratch Work

Ben is $x$ years old and Philip is $2x$. If we subtract $5$ from Philip’s age, it will become $2x - 5$. If we add $10$ to Ben’s age, it will be $x + 10$. Now, after the results to these operations, their ages will be equal or

$2x - 5 = x + 10$

Solution

Let $x$ be Ben’s age and $2x$ be Philip’s age.

$2x - 5 = x + 10$

$2x - x = 10 + 5$

$x = 15$.

So, Ben is $15$ and Philip is $30$.

Check

Philip is $30$ and Ben is $15$ so, Philip is twice as old Ben. Subtracting $10$ from Philip’s age results to $20$. Adding $5$ to Ben’s age is $20$. Well, $20$ equals $20$, so we are correct.

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

### 2 Responses

1. Nobody says:

I noticed that there is a discrepancy between the given numbers of subtracted from and added to in Problem #6 and its checking part. I was left confused, thats all. Thanks

1. March 27, 2014

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