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Free Printable Histograms PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 7)

Download free histograms worksheets (PDF), GCSE maths past paper questions and worked solutions for Histograms (7R). This is a Grade 7 statistics topic from the GCSE questions by topic collection for Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Maths4Everyone, Corbett Maths, Maths Made Easy, Dr Frost, Naiker Maths, 1st Class Maths, Metatutor, Maths Takeaway, Maths DIY, Dr Austin, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our histograms revision notes and worked examples with diagrams, try real GCSE past paper histograms questions with step-by-step solutions, or download a PDF of all histograms worksheet links.

Free Histograms PDF Worksheets & Answers

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GCSE Maths Grade 7 Revision Notes & Method

Guide to Histograms: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

In a histogram for grouped continuous data with unequal class intervals, the vertical axis plots Frequency Density rather than frequency, and the area of each bar represents the frequency. Always add a column for class width and use the formula triangle:

Frequency Density = FrequencyClass Width  and  Frequency = Frequency Density × Class Width. If an interval is cut part-way through (e.g. estimating how many values exceed 25 in a 20 < t ≤ 40 bar), use the fraction of the bar's width.

Example 1: Calculate the frequency densities for the grouped table below.

Mass (m g)Freq (f)Width (w)FD = f ÷ w
0 < m ≤ 1015101.5
10 < m ≤ 152054.0
15 < m ≤ 3536201.8
Plot bars with heights 1.5, 4.0, and 1.8.

Example 2: Using the table in Example 1, estimate how many items have mass between 12 g and 25 g.

12 18 0 1 2 3 4 FD 0 10 12 15 25 35 Mass (g)

Solution:

Split 12 g to 25 g across both bars:
• 12 to 15 g: 3 × 4.0 = 12
• 15 to 25 g: 10 × 1.8 = 18
Answer: 12 + 18 = 30 items
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Histograms Questions & Answers

Past Paper Example 1 GCSE Nov 2017 2H Q17 5 marks
The table gives information about the heights of 150 students.
Height (h cm)Frequency140 < h ⩽ 15015150 < h ⩽ 15530155 < h ⩽ 16051160 < h ⩽ 16536165 < h ⩽ 18018
(a) On the grid, draw a histogram for this information. [3 marks]
(b) Work out an estimate for the fraction of the students who have a height between 150 cm and 170 cm. [2 marks]
Worked Solution & Mark Scheme (5 marks)
Part (a) Solution:
FD = 15÷10 = 1.5, 30÷5 = 6, 51÷5 = 10.2, 36÷5 = 7.2, 18÷15 = 1.2 M2 A1
140150160170180024681012Height (h cm)Frequency density
Answer (a): Histogram with frequency densities 1.5, 6, 10.2, 7.2, 1.2
Part (b) Solution:
30 + 51 + 36 + (5/15 × 18) = 117 + 6 = 123 M1
⇒ 123/150 = 41/50 A1
Answer (b): 123150 (or 4150 or 0.82)
Past Paper Example 2 AQA Nov 2018 3H Q18 3 marks
Here is some information about the ages of people at a concert.
@@TABULAR0@@
Ages of people at a concert0.00.51.01.52.02.50510152025303540455055606570Frequency densityAge, x (years)
Draw a histogram to represent the information.
Worked Solution & Mark Scheme (3 marks)
Calculate frequency density = FrequencyClass width: M1
• 10 ≤ x < 15: width = 5, FD = 85 = 1.6
• 15 ≤ x < 25: width = 10, FD = 2410 = 2.4 M1
• 25 ≤ x < 40: width = 15, FD = 3015 = 2.0
• 40 ≤ x < 70: width = 30, FD = 3930 = 1.3 A1
Ages of people at a concert0.00.51.01.52.02.50510152025303540455055606570Frequency densityAge, x (years)
Answer: Bars drawn with frequency densities: 1.6, 2.4, 2.0, 1.3
Past Paper Example 3 IGCSE Specimen 1H Q19 4 marks
150 people took part in a survey.
The histogram shows information about the ages of these people.
203040506070800Age (years)Frequencydensity
Work out how many of these 150 people are aged between 50 years and 55 years.
Worked Solution & Mark Scheme (4 marks)
In minor grid units (width × height in small squares): 20–30: 10 × 17 = 170 M1
30–50: 20 × 36 = 720
50–55: 5 × 44 = 220 M1
55–70: 15 × 20 = 300
70–80: 10 × 9 = 90 M1
Total area = 170 + 720 + 220 + 300 + 90 = 1500 small squares
Number of people aged 50–55 = (220 / 1500) × 150 = 22. A1
Answer: 22

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