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# Dealing with Word Problems Study Guide (page 4)

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Updated on Oct 4, 2011

### Solutions to Sample Question

#### Question 1

Here's how to mark up the problem:

Four years ago, the sum of the ages of four friends was 42 years. If their ages were consecutive numbers, what is the

What we know:

• Four friends are involved.
• Four years ago, the sum (which means add) of their ages was 42.
• Their ages are consecutive (that means numbers in sequence, like 4, 5, 6, etc.).

The question itself:

How old is the oldest friend now?

Plan of attack:

Use algebra or trial and error to find out how old the friends were four years ago. After finding their ages, add them up to make sure they total 42. Then add 4 to the oldest to find his current age.

### Solution:

Let the consecutive ages of the four friends four years ago be represented by: f, f + 1, f + 2, and f + 3. Since their sum was 42 years, write and solve an equation to add their ages:

f + f + 1+ f + 2 + f + 3 = 42

4f + 6 = 42

4f = 36

f = 9

Since f represents the age of the youngest friend four years ago, the youngest friend is currently 13 years old (9 + 4 = 13). Since she is 13, the ages of the four friends are currently 13, 14, 15, and 16. Thus, the oldest friend is currently 16.

Check:

Add up the friends' ages of four years ago to make sure the total is 42: 9 + 10 + 11+ 12 = 42. Check the rest of your arithmetic to make sure it's correct.

Find practice problems and solutions for these concepts at Dealing with Word Problems Practice Questions.

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