Averages, Range and Quartiles: Edexcel 4MA1 Past Paper Questions and Worked Solutions
This page collects every averages, range and quartiles question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 22 questions worth 82 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.
80 students entered a dancing competition. The table gives information about the length of time, in minutes, for which each student spent dancing.
| Time (m) | Frequency |
|---|---|
| 0 < m ≤ 12 | 11 |
| 12 < m ≤ 24 | 25 |
| 24 < m ≤ 36 | 23 |
| 36 < m ≤ 48 | 15 |
| 48 < m ≤ 60 | 6 |
Work out an estimate for the mean length of time the students spent dancing.
Show solution
- Midpoints: 6, 18, 30, 42, 54
- Midpoint × frequency: 66 + 450 + 690 + 630 + 324 = 2160
- Mean ≈ 2160 ÷ 80
Answer: 27 minutes
The table shows information about the frame size, in cm, of 60 bicycles sold in a shop.
| Frame size (S cm) | Frequency |
|---|---|
| 30 < S ≤ 36 | 4 |
| 36 < S ≤ 42 | 14 |
| 42 < S ≤ 48 | 18 |
| 48 < S ≤ 54 | 19 |
| 54 < S ≤ 60 | 5 |
(a) Write down the modal class. (1)
(b) Work out an estimate for the mean frame size. (4)
Show solution
Part (a)
- The modal class is the class with the highest frequency (19).
Answer: 48 < S ≤ 54
Part (b)
- Midpoints: 33, 39, 45, 51, 57
- Midpoint × frequency: 132 + 546 + 810 + 969 + 285 = 2742
- Mean ≈ 2742 ÷ 60
Answer: 45.7 cm
Here are six cards. Five of the cards have a number written on them: 16, 15, 3, 2, 9. The sixth card is blank. Work out the number that should be written on the last card so that the mean of the six numbers will be 11.
Show solution
- Mean = total ÷ number of values, so the total of the six numbers must be 6 × 11 = 66.
- Total of the five numbers we know: 16 + 15 + 3 + 2 + 9 = 45
- The last card must make up the difference: 66 − 45
Answer: 21
The table gives information about the distances 100 adults travel to work.
| Distance (d km) | Frequency |
|---|---|
| 0 < d ≤ 5 | 26 |
| 5 < d ≤ 10 | 40 |
| 10 < d ≤ 15 | 16 |
| 15 < d ≤ 20 | 10 |
| 20 < d ≤ 25 | 8 |
(a) Write down the modal class. (1)
(b) Work out an estimate for the mean distance. (4)
Show solution
Part (a)
- The modal class is the class with the highest frequency (40).
Answer: 5 < d ≤ 10
Part (b)
- We do not know the exact distances, so use the midpoint of each class as an estimate.
| Class | Midpoint | Frequency | Midpoint × frequency |
|---|---|---|---|
| 0 < d ≤ 5 | 2.5 | 26 | 65 |
| 5 < d ≤ 10 | 7.5 | 40 | 300 |
| 10 < d ≤ 15 | 12.5 | 16 | 200 |
| 15 < d ≤ 20 | 17.5 | 10 | 175 |
| 20 < d ≤ 25 | 22.5 | 8 | 180 |
- Total of midpoint × frequency = 65 + 300 + 200 + 175 + 180 = 920
- Estimated mean = 920 ÷ 100 (the total frequency)
Answer: 9.2 km
The table gives information about the speeds, in kilometres per hour, of 80 motorbikes as each pass under a bridge.
| Speed (s km/h) | Frequency |
|---|---|
| 40 < s ≤ 50 | 10 |
| 50 < s ≤ 60 | 16 |
| 60 < s ≤ 70 | 19 |
| 70 < s ≤ 80 | 23 |
| 80 < s ≤ 90 | 12 |
(a) Write down the modal class. (1)
(b) Work out an estimate for the mean speed of the motorbikes as they pass under the bridge. Give your answer correct to 3 significant figures. (4)
Show solution
Part (a)
- The class with the highest frequency (23).
Answer: 70 < s ≤ 80
Part (b)
- Midpoints: 45, 55, 65, 75, 85
- Midpoint × frequency: 450 + 880 + 1235 + 1725 + 1020 = 5310
- Mean ≈ 5310 ÷ 80 = 66.375
Answer: 66.4 km/h
Given that a < b < c, the four whole numbers a, a, b and c have a mode of 7, a median of 8.5 and a mean of 9. Work out the value of a, the value of b and the value of c.
Show solution
- a appears twice, so the mode is a: a = 7
- In order the numbers are 7, 7, b, c. The median is halfway between the 2nd and 3rd: 7 + b2 = 8.5, so b = 10
- Mean 9 means the total is 4 × 9 = 36: 7 + 7 + 10 + c = 36, so c = 12
Answer: a = 7, b = 10, c = 12
The table gives information about the number of minutes that Abby spent walking each day in September.
| Number of minutes (M) | Frequency |
|---|---|
| 0 < M ≤ 30 | 5 |
| 30 < M ≤ 60 | 6 |
| 60 < M ≤ 90 | 8 |
| 90 < M ≤ 120 | 9 |
| 120 < M ≤ 150 | 2 |
Work out an estimate for the total number of minutes that Abby spent walking in September.
Show solution
- We do not know the exact times, so use the midpoint of each class: 15, 45, 75, 105, 135.
- Multiply each midpoint by its frequency: 15 × 5 = 75, 45 × 6 = 270, 75 × 8 = 600, 105 × 9 = 945, 135 × 2 = 270
- Add them to estimate the total: 75 + 270 + 600 + 945 + 270
Answer: 2160 minutes
Here are five cards with the numbers 15, 7, −2, 23 and x. The mean of the five numbers is 12. Work out the value of x.
Show solution
- The total of the five numbers = mean × 5 = 12 × 5 = 60
- The four known numbers add to 15 + 7 + (−2) + 23 = 43
- x = 60 − 43
Answer: x = 17
Here is a list of six numbers written in order of size: x, 5, y, z, 10, 12. The numbers have a range of 9, a median of 8 and a mode of 10. Find the value of x, the value of y and the value of z.
Show solution
- Range: 12 − x = 9, so x = 3
- Mode is 10, so 10 must appear more than once: z = 10
- Median is the mean of the 3rd and 4th numbers: y + 102 = 8, so y = 6
Answer: x = 3, y = 6, z = 10
The table shows information about the weights, in kilograms, of 40 babies.
| Weight (w kg) | Frequency |
|---|---|
| 2 < w ≤ 3 | 12 |
| 3 < w ≤ 4 | 16 |
| 4 < w ≤ 5 | 9 |
| 5 < w ≤ 6 | 2 |
| 6 < w ≤ 7 | 1 |
(a) Write down the modal class. (1)
(b) Work out an estimate for the mean weight of the 40 babies. (4)
One of the 40 babies is going to be chosen at random.
(c) Find the probability that this baby has a weight of more than 5 kg. (2)
Show solution
Part (a)
- The class with the highest frequency (16).
Answer: 3 < w ≤ 4
Part (b)
- Midpoints: 2.5, 3.5, 4.5, 5.5, 6.5
- Midpoint × frequency: 30 + 56 + 40.5 + 11 + 6.5 = 144
- Mean ≈ 144 ÷ 40
Answer: 3.6 kg
Part (c)
- More than 5 kg: 2 + 1 = 3 babies out of 40.
Answer: 340
Here is a list of six numbers written in order of size: 4, 7, x, 10, y, y. The numbers have a median of 9 and a mean of 11. Find the value of x and the value of y.
Show solution
- With 6 numbers, the median is halfway between the 3rd and 4th: x + 102 = 9, so x + 10 = 18 and x = 8
- Mean 11 means the total is 6 × 11 = 66: 4 + 7 + 8 + 10 + y + y = 66
- 29 + 2y = 66, so 2y = 37
Answer: x = 8, y = 18.5
The table gives information about the length of time, in minutes, that each of 60 students took to travel to school on Monday.
| Length of time (t minutes) | Frequency |
|---|---|
| 0 < t ≤ 10 | 4 |
| 10 < t ≤ 20 | 10 |
| 20 < t ≤ 30 | 15 |
| 30 < t ≤ 40 | 25 |
| 40 < t ≤ 50 | 6 |
(a) Write down the modal class interval. (1)
(b) Work out an estimate for the mean length of time taken by these 60 students to travel to school on Monday. Give your answer correct to one decimal place. (4)
Show solution
Part (a)
- The class with the highest frequency (25).
Answer: 30 < t ≤ 40
Part (b)
- Use the midpoints 5, 15, 25, 35, 45.
- Midpoint × frequency: 5 × 4 = 20, 15 × 10 = 150, 25 × 15 = 375, 35 × 25 = 875, 45 × 6 = 270
- Total = 20 + 150 + 375 + 875 + 270 = 1690
- Mean ≈ 1690 ÷ 60 = 28.166…
Answer: 28.2 minutes
Jenny has six cards. Each card has a whole number written on it so that the smallest number is 5, the largest number is 24, the median of the six numbers is 14 and the mode of the six numbers is 8. Jenny arranges her cards so that the numbers are in order of size: 5, …, …, …, …, 24.
(a) For the remaining four cards, write on each dotted line a number that could be on the card. (3)
A basketball team plays 6 games. After playing 5 games, the team has a mean score of 21 points per game. After playing 6 games, the team has a mean score of 23 points per game.
(b) Work out the number of points the team scored in its 6th game. (3)
Show solution
Part (a)
- The mode is 8, so 8 must appear at least twice. Put 8 in the 2nd and 3rd places.
- The median is the mean of the 3rd and 4th cards: 8 + 4th2 = 14, so the 4th card is 20.
- The 5th card must be bigger than 20 (so 8 stays the only mode) and less than 24: 21, 22 or 23.
Answer: 5, 8, 8, 20, 21, 24 (the fifth card could also be 22 or 23)
Part (b)
- Total after 5 games = 5 × 21 = 105
- Total after 6 games = 6 × 23 = 138
- 6th game = 138 − 105
Answer: 33 points
Yusuf sat 8 examinations. Here are his marks for 5 of the examinations: 68, 72, 75, 77, 80. For his results in all 8 examinations, the mode of his marks is 80, the median of his marks is 74 and the range of his marks is 16. Find Yusuf’s marks for each of the other 3 examinations.
Show solution
- Mode 80: 80 must appear more often than any other mark, so one of the missing marks is 80.
- Take the highest mark as 80 (it is the mode and the highest mark we know). The range is 16, so the lowest mark is 80 − 16 = 64. That is the second missing mark.
- Now in order: 64, 68, 72, ?, 75, 77, 80, 80. With 8 marks, the median is the mean of the 4th and 5th: ? + 752 = 74, so ? = 73.
Answer: 64, 73, 80
5 children are playing on a trampoline. The mean weight of the 5 children is 28 kg. 2 of the children get off the trampoline. The mean weight of these 2 children is 26.5 kg. Work out the mean weight of the 3 children who remain on the trampoline.
Show solution
- Total weight of 5 children = 5 × 28 = 140 kg
- Total weight of the 2 who get off = 2 × 26.5 = 53 kg
- Total of the 3 who remain = 140 − 53 = 87 kg
- Mean = 87 ÷ 3
Answer: 29 kg
Jethro has sat 5 tests. Each test was marked out of 100 and Jethro’s mean mark for the 5 tests is 74. Jethro has to sit one more test that is also marked out of 100. Jethro wants his mean mark for all 6 tests to be at least 77. Work out the least mark that Jethro needs to get for the last test.
Show solution
- Total of the first 5 tests = 5 × 74 = 370
- For a mean of 77 over 6 tests, the total must be at least 6 × 77 = 462
- Last test ≥ 462 − 370
Answer: 92
Here are the test marks of 15 students: 7, 10, 14, 15, 16, 17, 18, 19, 20, 20, 23, 25, 30, 39, 40. Find the interquartile range of these marks.
Show solution
- n = 15. Lower quartile = 14(15 + 1) = 4th value = 15
- Upper quartile = 34(15 + 1) = 12th value = 25
- IQR = 25 − 15
Answer: 10
Here are the numbers of goals scored by a hockey team in its 11 games this season: 0, 1, 2, 2, 3, 4, 4, 6, 7, 9, 11. Work out the interquartile range of the numbers of goals.
Show solution
- The data are already in order and n = 11.
- Lower quartile = the 14(n + 1)th = 3rd value = 2
- Upper quartile = the 34(n + 1)th = 9th value = 7
- Interquartile range = upper quartile − lower quartile = 7 − 2
Answer: 5
Here is the number of runs scored by a baseball team in each of its 15 games this season, in order of size: 0, 1, 1, 3, 5, 6, 7, 7, 8, 9, 9, 12, 12, 15, 16. Work out the interquartile range of the number of runs.
Show solution
- Lower quartile = 14(15 + 1) = 4th value = 3
- Upper quartile = 34(15 + 1) = 12th value = 12
- IQR = 12 − 3
Answer: 9
Here are the numbers of eggs laid by some hens on each of 11 days in June: 9, 10, 12, 15, 17, 18, 19, 19, 20, 25, 26. Work out the interquartile range of the numbers of eggs.
Show solution
- The 11 values are already in order.
- Lower quartile = 14(11 + 1) = 3rd value = 12
- Upper quartile = 34(11 + 1) = 9th value = 20
- IQR = 20 − 12
Answer: 8
Here is the number of goals that Henri’s team scored one summer in each water polo match: 5, 8, 9, 11, 13, 13, 14, 15, 16, 17, 20. Find the interquartile range of the numbers of goals. Show your working clearly.
Show solution
- n = 11. Lower quartile = 14(11 + 1) = 3rd value = 9
- Upper quartile = 34(11 + 1) = 9th value = 16
- IQR = 16 − 9
Answer: 7
The histogram and the table give some information about the amounts of time, in hours, that Year 11 students at Bergdesh Academy spent, in total, on their homework last week. No student in Year 11 spent longer than 9 hours on their homework.
| Time spent on homework (t hours) | Frequency |
|---|---|
| 0 < t ≤ 2 | 28 |
| 2 < t ≤ 4 | |
| 4 < t ≤ 5 | |
| 5 < t ≤ 6 | |
| 6 < t ≤ 9 |
Using the information in the histogram and in the table, work out an estimate for the mean amount of time the Year 11 students spent on their homework last week. Give your answer in hours correct to 3 significant figures.
Show solution
- Use the 0–2 bar to find the scale: frequency density = 28 ÷ 2 = 14. That bar is 14 small squares tall, so each small square = 1.
- Read the other bars: 2–4 is 20, 4–5 is 30, 5–6 is 24 and 6–9 is 12.
- Frequencies = frequency density × width: 2–4: 20 × 2 = 40; 4–5: 30 × 1 = 30; 5–6: 24 × 1 = 24; 6–9: 12 × 3 = 36
- Total students = 28 + 40 + 30 + 24 + 36 = 158
- Midpoint × frequency: 1 × 28 + 3 × 40 + 4.5 × 30 + 5.5 × 24 + 7.5 × 36 = 28 + 120 + 135 + 132 + 270 = 685
- Mean ≈ 685 ÷ 158 = 4.335…
Answer: 4.34 hours
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