Math Word Problems: Solving Age Problems Part 3

October 11, 2012 GB High School Algebra, High School Mathematics

This is the third part of the Solving Age Problems of the Math Word Problem Solving Series. In this post, we discuss more complicated age word problems.

Problem 7

Anna who is $6$ years old and his father Ben who is $27$ years old have the same birthday. In how many years will Ben be twice as old as Anna?

Solution

As years go by, the number of years added to Ben’s and Anna’s ages is the same. If we let the number of years that have gone by be $x$ , then in $x$ years, their ages will be

Ben’s Age: $27 + x$

Anna’s Age: $6 + x$

Since in $x$ years, Ben will be as twice as old as Anna, if we multiply Anna’s age by $2$ , their ages will be equal. So, we can now set up the equation

$2(6 + x) = 27 + x$ .

Solving, we have $12 + 2x = 27 + x$ . which means that $x = 15$ .

That means that in $15$ years, Ben will be twice as old as Anna.

Checking the Answer

In $15$ years, Ben will be $27 + 15 = 42$ years old. In $15$ years, Anna will be $6 + 15 = 21$ years old. Clearly, Ben will be twice as old as Anna in $15$ years, so we are correct.

Problem 8

The sum of the ages of James and Clark is $50$ . Five years ago, Clark’s age was three times as James’ age. How old are James and Clark now?

Solution

If James is $x$ years old, then Clark is $50 - x$ years old (Why?). Five years ago, their ages were

James’ age: $x - 5$

Clark’s age: $(50 - x) - 5 = 45 - x$ .

Since $5$ years ago, Clark’s age was three times as that of James, if we multiply James’ age by $3$ , their ages will be equal. So, we set up the equation

$3(x-5) = 45-x$

Solving we have, $3x - 15 = 45 - x$ . This means that $4x = 60$ and $x = 15$ . Therefore, James is $15$ years old and Clark is $35$ years old.

Checking the Answer

The sum of James age which is $15$ and Clark’s age which is $35$ is equal to $50$ . Five years ago, James was $10$ years old and Clark was $30$ . As we can see, Clark was three times as old as James. Therefore, we are correct.

Problem 9*

Pol is $10$ years younger than Greg. In $7$ years, he will be $10$ years more than one half as old as Greg. Find their age at present.

Solution

Let $x$ be Greg’s age and let $x - 10$ be Pol’s age. In 7 years, their ages will be

Greg’s age: $x + 7$

Pol’s age: $x - 10 + 7 = x - 3$ .

In the problem, it is said that in $7$ years, Paul’s age will be ten years more than half of Greg’s age. Now, in $7$ years, half of Greg’s age will be

$\frac{1}{2}(x + 7)$ .

Now, since Paul is 10 more than one half Greg’s age, if we add $10$ years to half Greg’s age which is $\frac{1}{2}(x + 7)$ , their ages will be equal. Therefore,

$\frac{1}{2}(x + 7) + 10 = x - 3$ .

Multiplying both sides by $2$ we have $x + 7 + 20 = 2x - 6$ . Simplifying, we have $x + 27 = 2x - 6$ which gives us $x = 33$ . Therefore, Greg is $33$ years old and Pol is $23$ .

Checking the Answer

Greg is $33$ and Paul is $23$ , so Pol is $10$ years younger. In $7$ years, their ages will be $40$ and $30$ respectively. Half of Greg’s age by then is $20$ and if we add $10$ , the result is $30$ which equal’s Pol’s age.