Worded questions and working with three or more fractions

Sometimes a question written in words can appear more difficult. Written questions require the student to extract the important information, create a mathematical model and identify the type of calculation they are needing to perform.


Steve and Sheila buy a digital TV recorder. Steve fills of the recorder with his programmes and Sheila fills of the recorder with her programmes.

a) What fraction of the recorder is full?
b) What fraction of the recorder is empty?

Steve deletes one of his recordings which takes up of the recorder.

c) What fraction of the recorder do Steve’s recordings fill now?

Worked answer

From the question we have identified that there are two fractions representing two parts of the same whole, the digital recorder: and .

To answer part a) we need to make the denominators the same:

Visual representation for part a). On the left is a bar divided into five segments, with two segments shaded to represent Steve's share. On the right is a bar divided into three segments, with one segment shaded to represent Sheila's share. In the middle is a bar divided into fiftheeths, with two portions shaded to represent the two shares of the recordings: six fifteeths and five fifteeths respectively.

The denominators of our two fractions are 3 and 5, both prime numbers, so we use the lowest common multiple, 15, as our denominator.

To get our equivalent fractions of fifteenths, we multiply the numerator and denominator of each fraction as shown above. , multiplying top and bottom by 3, becomes and , multiplying top and bottom by 5, becomes .

We can now add the two fractions together as they have the same denominator:

Visual representation of part b). Three bars, each divided into fifteen segments. Six fifteenths plus five fifteenths equals eleven fifteenths

This means, of the recorder is full.

b) To find how much of the recorder is empty, we know that the ‘whole’ recorder can be represented as ‘1’, and this can be represented as (using the denominator of our previous answer). It’s then straightforward to take away from , to get . Alternatively we could say that the difference between and is , so of the disc is empty.

Visual representation of part c). Three bars, each divided into fifteen segments. Fifteen fifteenths (one whole), minus eleven fifteenths equals four fifteenths.

c) The final part of the question asks us to take of away from .

First we find the equivalent fraction of with 10 as the denominator. This gives .

Now we take away from to get .

Hence, Steve’s recordings now take up of the recorder.


Work through this question again, where are the points in this question where students might make a mistake? Can you come up with any general advice for tackling worded questions? Share in the comments below.

Problem worksheet

Now complete questions 9, 10 and 11 from this week’s worksheet.
As a reminder, the worksheet can be found in the first step of this week.

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This article is from the free online course:

Maths Subject Knowledge: Fractions, Decimals, and Percentages

National STEM Learning Centre