Cambridge O Level Mathematics: Histograms and Frequency Density with Unequal Class Widths (4024) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 9.7, Histograms, of the Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027. The content is the same for every candidate. Read the study guide first for the methods, and keep the revision notes open for the formulas. Other topics are on the course hub, progress can be tracked on the printable checklist, and the free 10-minute diagnostics show where to start. Show your working on every question.

Questions

1. (a) A class 8 ≤ x < 14 has frequency 27. Find its frequency density. [1] (b) A histogram bar has width 6 and frequency density 2.5. Find the frequency it represents. [1]

2. The times, in minutes, that 68 people took to finish a puzzle are grouped below. Calculate the frequency density of each class.

Time t (minutes)	Frequency
0 ≤ t < 5	10
5 ≤ t < 10	17
10 ≤ t < 20	26
20 ≤ t < 30	9
30 ≤ t < 60	6

[3]

3. Use your answers to question 2 to describe how you would draw the histogram: the labels on both axes, the vertical scale, and the position and height of every bar. [3]

4. A histogram of the masses, in kg, of some fruit crates has these bars.

Mass m (kg)	Frequency density
0 ≤ m < 4	1.5
4 ≤ m < 10	3.5
10 ≤ m < 16	2
16 ≤ m < 30	0.5

Copy and complete a frequency table, and find the total number of crates. [4]

5. A histogram has no vertical scale. A bar 2 cm wide and 3 cm tall represents 24 values. (a) How many values are represented by a bar 3.5 cm wide and 2 cm tall? [2] (b) A bar 2.5 cm wide represents 10 values. How tall is it? [2]

6. The distances travelled to work by 60 employees are summarised in a table. The bar for 10 ≤ d < 20 has frequency density 0.9.

Distance d (km)	Frequency
0 ≤ d < 2	14
2 ≤ d < 5	27
5 ≤ d < 10	x
10 ≤ d < 20	y

(a) Find y. [2] (b) Find x and the frequency density of the bar for 5 ≤ d < 10. [2]

7. On a histogram, the bar for 10 ≤ x < 15 has height 4 and the bar for 15 ≤ x < 25 has height 2.4. Rosa says that the first class contains more values because its bar is taller. Is she right? Give a reason. [3]

8. A histogram of waiting times, in minutes, has these bars: 0 ≤ t < 20 with height 0.9, 20 ≤ t < 30 with height 1.6, and 30 ≤ t < 60 with height 0.4. (a) Find the total number of waiting times shown. [3] (b) Estimate how many waiting times are between 24 and 30 minutes. State any assumption. [2]

9. The weekly reading times of 70 students are shown. Part of the table is missing.

Time h (hours)	Frequency
0 ≤ h < 2	8
2 ≤ h < 4	13
4 ≤ h < 8	22
8 ≤ h < 14	y
14 ≤ h < 26	z

The bar for 14 ≤ h < 26 has height 0.25. (a) Find z. [2] (b) Find y. [2] (c) Find the frequency density of every class. [2]

10. In a histogram, the bar for 10 ≤ x < 20 is three times as tall as the bar for 20 ≤ x < 40. Together the two classes contain 100 values. Find the height of each bar and the frequency of each class. [5]

11. The lengths of 120 phone messages are shown on a histogram with no vertical scale. The horizontal scale is 1 cm for every 5 seconds. The bars are:

Class (seconds)	Width on diagram	Height on diagram
0 to 5	1 cm	3 cm
5 to 15	2 cm	4 cm
15 to 30	3 cm	2 cm
30 to 40	2 cm	1.5 cm

The bar for 5 to 15 seconds represents 48 messages. (a) Find the number of messages represented by 1 cm² on the diagram. [2] (b) Find the frequency of each class. [3] (c) Find the percentage of the messages that are 15 seconds or longer. [2] (d) The vertical axis is to be labelled in frequency density. Find the frequency density that 1 cm of height represents. [2]

Answers

1. (a) Width = 14 − 8 = 6 and frequency density = 27 ÷ 6 = 4.5 [1]. (b) Frequency = 2.5 × 6 = 15 [1]. Examiner insight: Showing the division or multiplication line makes your method visible, and writing the width first helps you avoid using the wrong number.

2. Widths are 5, 5, 10, 10 and 30 [1]. Frequency density = frequency ÷ width, so 10 ÷ 5 = 2 and 17 ÷ 5 = 3.4 [1]. 26 ÷ 10 = 2.6, 9 ÷ 10 = 0.9 and 6 ÷ 30 = 0.2, so the densities are 2, 3.4, 2.6, 0.9, 0.2 [1]. Examiner insight: With five classes, one wrong width is easy to make. Writing all the widths first gives you something to check, and clear working shows your method.

3. Label the vertical axis Frequency density and the horizontal axis with the time in minutes, using a continuous scale [1]. Use an equal-step vertical scale up to at least 3.4 [1]. Draw touching bars: 0 to 5 at height 2, 5 to 10 at 3.4, 10 to 20 at 2.6, 20 to 30 at 0.9 and 30 to 60 at 0.2, with no gaps [1]. Examiner insight: When asked to describe or draw a histogram, the axis labels, the scale and the bar positions are separate things an examiner can look for. An unlabelled axis can lose credit even when every bar is correct.

4. Widths are 4, 6, 6 and 14 [1]. Frequencies are 4 × 1.5 = 6, 6 × 3.5 = 21, 6 × 2 = 12 and 14 × 0.5 = 7 [1][1]. Total = 6 + 21 + 12 + 7 = 46 crates [1]. Examiner insight: Add the frequencies to get a total. Adding the bar heights gives a number with no meaning, and that is a common way to lose the final mark.

5. (a) The known bar has area 2 × 3 = 6 cm², so 24 ÷ 6 = 4 values per cm² [1]. The new bar has area 3.5 × 2 = 7 cm², so it represents 7 × 4 = 28 values [1]. (b) Area needed = 10 ÷ 4 = 2.5 cm² [1]. Height = 2.5 ÷ 2.5 = 1 cm [1]. Examiner insight: Finding the value of one square centimetre first keeps both parts short. Stating the units, such as values per cm², shows that you know what the number means.

6. (a) Width of the last class = 20 − 10 = 10, so y = 0.9 × 10 = 9 [1]. This uses frequency = frequency density × width [1]. (b) x = 60 − 14 − 27 − 9 = 10 [1]. Width = 5, so frequency density = 10 ÷ 5 = 2 [1]. Examiner insight: Start with the class where you are given a height, because it is the only one you can find directly. Then use the total to find the remaining class.

7. The first class has width 5, so its frequency is 5 × 4 = 20 [1]. The second has width 10, so its frequency is 10 × 2.4 = 24 [1]. Rosa is wrong: the second class has more values, because its bar is wider even though it is shorter [1]. Examiner insight: A good answer to an interpretation question gives the evidence as well as the verdict. The two frequencies, or a statement about the areas, are what support the conclusion.

8. (a) Widths are 20, 10 and 30 [1]. Frequencies are 20 × 0.9 = 18, 10 × 1.6 = 16 and 30 × 0.4 = 12 [1]. Total = 46 [1]. (b) The part from 24 to 30 is 6 wide, so 6 × 1.6 = 9.6, which is about 10 [1]. This assumes the values are spread evenly across the class 20 ≤ t < 30 [1]. Examiner insight: An estimate from part of a class depends on an assumption, and saying what it is shows that you understand why the answer is only approximate.

9. (a) Width = 12, so z = 0.25 × 12 = 3 [1][1]. (b) y = 70 − 8 − 13 − 22 − 3 = 24 [1][1]. (c) Widths are 2, 2, 4, 6 and 12. The densities are 8 ÷ 2 = 4, 13 ÷ 2 = 6.5, 22 ÷ 4 = 5.5, 24 ÷ 6 = 4 and 3 ÷ 12 = 0.25, so 4, 6.5, 5.5, 4, 0.25 [1][1]. Examiner insight: Write each step rather than only the final numbers. Clear working also lets a reader see exactly where a slip happened.

10. Let the height of the 20 ≤ x < 40 bar be k, so the other bar has height 3k [1]. The widths are 10 and 20, so the frequencies are 3k × 10 = 30k and k × 20 = 20k [1]. Then 30k + 20k = 100, so k = 2 [1]. The heights are 6 (for 10 ≤ x < 20) and 2 (for 20 ≤ x < 40) [1]. The frequencies are 60 and 40 [1]. Examiner insight: Introducing a letter for the unknown height and writing both frequencies in terms of it turns the problem into one equation. Treating the two heights as if they were the frequencies is an error to avoid.

11. (a) The 5 to 15 bar has area 2 × 4 = 8 cm² [1], so 1 cm² represents 48 ÷ 8 = 6 messages [1]. (b) 0 to 5: area 1 × 3 = 3 cm², so 3 × 6 = 18 [1]. 15 to 30: area 3 × 2 = 6 cm², so 36 [1]. 30 to 40: area 2 × 1.5 = 3 cm², so 18, and the 5 to 15 class has 48 [1]. (c) Total = 18 + 48 + 36 + 18 = 120, which matches the question. Messages of 15 seconds or longer = 36 + 18 = 54 [1], and 54 ÷ 120 = 0.45, so 45% [1]. (d) The 5 to 15 class has frequency density 48 ÷ 10 = 4.8 and its bar is 4 cm tall [1], so 1 cm of height represents 4.8 ÷ 4 = 1.2 [1]. Examiner insight: Check that your frequencies add to the stated total before using them in later parts. It is a quick way to catch a slip in the scale.

Where marks are usually lost

• Plotting frequency, not frequency density, when the class widths are different.
• Using the width of the wrong class, especially when widths change every row.
• Multiplying when you should divide, or the reverse, between a bar and a frequency.
• Forgetting to label the vertical axis “Frequency density”, or leaving out the units on the horizontal axis.
• Drawing gaps between bars, or bars that do not start and end at the class boundaries.
• Adding the heights of bars to get a total instead of adding the frequencies.
• Judging which class is largest by the height of its bar rather than its area.
• Giving an estimate from part of a class without stating the even-spread assumption.
• Skipping the final check that the frequencies add up to the stated total.

Next steps

• Review the formulas in the revision notes.
• Re-read the methods in the study guide.
• Go to the Cambridge O Level Mathematics hub and the printable checklist.
• Try the free 10-minute diagnostics to find your next topic.
• Book a free trial class if you would like help from a teacher.

Official syllabus

Cambridge O Level Mathematics (Syllabus D) 4024 syllabus for 2025, 2026 and 2027, version 2, published by Cambridge University Press & Assessment (Cambridge Assessment International Education). The relevant section is 9.7 Histograms.