How to Solve Number Word Problems Part 4

This is the fourth part and the conclusion to the Number Word Problem Series. In the introduction to this series, we have learned How to Solve Number Problems Mentally. In Part 1 and Part 2, we have discussed the basic number word problems, and in Part 3, we have learned how to solve word problems about consecutive numbers.

In this post, we discuss about more complicated problems especially problems that involve fractions. We have already discussed 9 problems in the previous parts of this series, so, we now solve the 10th problem.

Problem 10

There are consecutive numbers. The sum of the second and the fourth number is . What is the largest number?

Scratch Work

Since we have five consecutive numbers, if we let be the smallest, then the other numbers are , , , and .

Now, the second number is and the fourth is . Their sum is . And that’s where we get our equation.

Solution

Let , , , and be the five consecutive numbers.

Second Number:

Fourth Number:

Second Number + Fourth Number =

Subtracting 4 from both sides,

So, the five consecutive numbers are , , , and

The largest among them is .

Check: Yes, they are consecutive numbers and .

Problem 11

A number added to of itself is equal to . What is the number?

Scratch Work

If that number is equal to , then of that number is or . Therefore, if the number is , it’s one fourth is . Now, if we add and , the sum is 80. That is where we get our equation.

Solution

Let be the number and a fourth of it as

To get rid of the fraction, we multiply everything by . This gives us

Check: of is . Now, . This means that we are correct.

Problem 12

One number exceeds the other by . One third the larger subtracted by one one half the smaller is equal to . What are the numbers?

Scratch Work

One number exceeds the other by 22 means that the other number is 22 more than the smaller. So, if we let be the smaller number, the larger number is .

Now, we will subtract one-third by one fourth of . The difference will be . In equation form we have

.

Solution

Let be the smaller number and be the larger.

The least common multiple of the denominators of the fraction (3 and 2) is , so we multiply everything with to eliminate the fractional parts.

By the distributive property, we have

Multiplying both sides by gives us which is the smaller number. The larger number is .

Check: .

That’s it! I wanted to discuss more number problems, but we have other problems to discuss, so I’ll leave this topic for now. In the next post, we will be solving problems about age.