Histograms and cumulative frequency - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 36 exam-style questions on Histograms and cumulative frequency for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

36 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The histogram shows information about the masses of some apples.

Work out the number of apples with a mass between 140 g and 160 g.

Histogram drawn on a grid of small squares. The horizontal axis is "Mass (g)", from 100 to 200, numbered every 20, with 4 small squares between numbered values (each small square is 5). The vertical axis is "Frequency density", from 0 to 3.5, numbered every 0.5, with 5 small squares between numbered values (each small square is 0.1). The bars touch, with no gaps between them. Bars: from 100 to 120, height 0.8; from 120 to 130, height 2.2; from 130 to 140, height 3; from 140 to 160, height 1.8; from 160 to 200, height 0.4.

Model answer

The bar from 140 to 160 has a frequency density of 1.8 and a width of 20 g.
Frequency \(= 1.8 \times 20 = 36\) apples

Explanation

In a histogram the area of a bar gives the frequency: frequency = frequency density × class width. A common mistake is to give the height of the bar (1.8) as the frequency.

Mark scheme

1 mark
M1 for 1.8 × 20
frequency density × class width; correct answer with no working scores full marks
1 mark
A1 for 36
cao

Question 2

2 marks

The cumulative frequency graph shows information about the lengths of 50 phone calls. The classes are 0 to 5 minutes, 5 to 10 minutes, and so on.

Use the graph to find the modal class.

Cumulative frequency graph on a grid of small squares. The horizontal axis is "Length of call (minutes)", from 0 to 30, numbered every 5, with 5 small squares between numbered values (each small square is 1). The vertical axis is "Cumulative frequency", from 0 to 50, numbered every 10, with 10 small squares between numbered values (each small square is 1). A smooth curve starts at (0, 0) and passes through (5, 6), (10, 18), (15, 34), (20, 43), (25, 48) and (30, 50) (each point is marked with a cross).

Model answer

Cumulative frequencies at 5, 10, 15, 20, 25, 30: 6, 18, 34, 43, 48, 50
Frequencies: \(6\), \(18 - 6 = 12\), \(34 - 18 = 16\), \(43 - 34 = 9\), \(48 - 43 = 5\), \(50 - 48 = 2\)
The greatest frequency is 16, so the modal class is 10 to 15 minutes.

Explanation

The frequency of each class is the increase in cumulative frequency across it. The modal class has the biggest increase, which is where the graph is steepest.

Mark scheme

1 mark
M1 for finding at least three class frequencies from differences in cumulative frequency, e.g. 6, 12, 16, 9, 5, 2
correct answer with no working scores full marks
1 mark
A1 for 10 to 15 minutes
accept 10 < t ≤ 15

Question 3

3 marks

The cumulative frequency graph shows information about the masses of 80 parcels.

Use the graph to find an estimate for the number of parcels with a mass greater than 2.5 kg but not more than 4 kg.

Cumulative frequency graph on a grid of small squares. The horizontal axis is "Mass (kg)", from 0 to 6, numbered every 1, with 5 small squares between numbered values (each small square is 0.2). The vertical axis is "Cumulative frequency", from 0 to 100, numbered every 10, with 5 small squares between numbered values (each small square is 2). A smooth curve starts at (0, 0) and passes through (1, 6), (2, 20), (3, 44), (4, 66), (5, 76) and (6, 80) (each point is marked with a cross).

Model answer

Cumulative frequency at 4 kg \(= 66\)
Cumulative frequency at 2.5 kg \(\approx 31\)
Number of parcels \(\approx 66 - 31 = 35\)

Explanation

Each reading tells you how many parcels have at most that mass. The number between two masses is the difference between the two cumulative frequencies.

Mark scheme

1 mark
M1 for reading the cumulative frequency at 4 kg as 66
correct answer with no working scores full marks
1 mark
M1 for reading the cumulative frequency at 2.5 kg, in the range 29 to 34
1 mark
A1 for an answer in the range 32 to 37
ft 66 − their reading at 2.5

Question 4

4 marks

The histogram shows information about the lengths of 50 leaves. The frequency density axis has no numbers.

Work out the number of leaves with a length between 5 cm and 6 cm.

Histogram drawn on a grid of small squares. The horizontal axis is "Length (cm)", from 2 to 12, numbered every 1, with 2 small squares between numbered values (each small square is 0.5). The vertical axis is "Frequency density"; it has tick marks every 4 small squares but no numbers. The bars touch, with no gaps between them. Bars: from 2 to 4 (4 small squares wide), 8 small squares tall; from 4 to 5 (2 small squares wide), 18 small squares tall; from 5 to 6 (2 small squares wide), 24 small squares tall; from 6 to 8 (4 small squares wide), 13 small squares tall; from 8 to 12 (8 small squares wide), 4 small squares tall.

Model answer

Areas in small squares (width × height):
2 to 4: \(4 \times 8 = 32\); 4 to 5: \(2 \times 18 = 36\); 5 to 6: \(2 \times 24 = 48\); 6 to 8: \(4 \times 13 = 52\); 8 to 12: \(8 \times 4 = 32\)
Total area \(= 200\) small squares, which represents 50 leaves.
So 1 leaf is represented by \(200 \div 50 = 4\) small squares.
Leaves from 5 cm to 6 cm \(= 48 \div 4 = 12\)

Explanation

Frequency is proportional to area in a histogram, so the total area of all the bars represents all 50 leaves. Working out what one small square represents lets you turn the area of any bar into a frequency.

Mark scheme

1 mark
P1 for finding the area of the bars in small squares: 32, 36, 48, 52, 32
at least three correct; any consistent unit of area is fine; correct answer with no working scores full marks
1 mark
P1 for the total area, 200 small squares
1 mark
P1 for a correct scale, e.g. 200 squares represent 50 leaves so 4 squares represent 1 leaf, or \(\dfrac{48}{200} \times 50\)
1 mark
A1 for 12
cao

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