Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Cumulative frequency & box plots · Core · Paper 1 style · no calculator · 5 marksThe times, in minutes, that 15 students took to walk to a sports ground are listed in order:12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 31, 33, 40
- (a) Find the median time. [1]
- (b) Find the lower quartile and the upper quartile. [2]
- (c) Find the interquartile range and the range. [2]
Show answer
- (a) 23 minutes
n = 15, median is the 8th value = 23. B1
- (b) LQ = 17, UQ = 29
LQ = 4th value = 17 B1; UQ = 12th value = 29 B1
- (c) IQR = 12 minutes, range = 28 minutes
29 − 17 = 12 B1; 40 − 12 = 28 B1
Box plots show the midday temperatures, in °C, in two cities during one month.City A: minimum 12, lower quartile 18, median 23, upper quartile 27, maximum 35.City B: minimum 20, lower quartile 24, median 26, upper quartile 28, maximum 31.
- (a) Find the range and the interquartile range for each city. [2]
- (b) Make two comparisons between the temperatures in the two cities. [2]
Show answer
- (a) City A: range 23, IQR 9. City B: range 11, IQR 4.
A: 35 − 12 = 23, 27 − 18 = 9; B: 31 − 20 = 11, 28 − 24 = 4. B2 (B1 for two correct)
- (b) City B had a higher median (26 °C against 23 °C), so it was generally warmer; City B had a smaller IQR (and range), so its temperatures were more consistent.
B1 compares medians in context; B1 compares IQR or range in context.
The masses, m kg, of 80 parcels are recorded.0 < m ≤ 2: 62 < m ≤ 4: 144 < m ≤ 6: 286 < m ≤ 8: 208 < m ≤ 10: 12
- (a) Complete a cumulative frequency table for m ≤ 2, m ≤ 4, m ≤ 6, m ≤ 8 and m ≤ 10. [2]
- (b) Assuming the parcels are spread evenly within each class, estimate the median mass. Give your answer correct to 3 significant figures. [2]
- (c) In the same way, estimate the interquartile range. [2]
- (d) Estimate the number of parcels with a mass greater than 7 kg. [1]
Show answer
- (a) 6, 20, 48, 68, 80
Running totals B2 (B1 for 3 correct)
- (b) 5.43 kg
40th parcel lies in 4 < m ≤ 6: 4 + (40 − 20)/28 × 2 M1 = 5.43 A1
- (c) 3.2 kg
LQ = 20th = 4; UQ = 60th = 6 + (60 − 48)/20 × 2 = 7.2 M1; IQR = 3.2 A1
- (d) 22
20/2 + 12 = 22 B1
A cricketer's scores in 12 innings are:4, 31, 17, 0, 52, 23, 9, 45, 12, 68, 27, 38
- (a) Find the median. [1]
- (b) The lower quartile is the median of the lowest six scores and the upper quartile is the median of the highest six. Find the interquartile range. [3]
- (c) In the next innings the cricketer scores 30. Does the median increase, decrease or stay the same? Give a reason. [1]
Show answer
- (a) 25
Order the data; (23 + 27) ÷ 2 B1
- (b) LQ = 10.5, UQ = 41.5, IQR = 31
LQ (9 + 12)/2 = 10.5 B1; UQ (38 + 45)/2 = 41.5 B1; 31 B1
- (c) It increases, from 25 to 27: there are now 13 scores and the middle (7th) one is 27.
B1 (increases, with reason)
The lengths, l cm, of 120 fish caught at a fish farm are recorded.10 < l ≤ 20: 820 < l ≤ 30: 2230 < l ≤ 40: 4040 < l ≤ 50: 3450 < l ≤ 60: 16
- (a) Write down the cumulative frequencies for l ≤ 20, l ≤ 30, l ≤ 40, l ≤ 50 and l ≤ 60. [2]
- (b) Assuming the lengths are spread evenly within each class, estimate the median length. [2]
- (c) Estimate the upper quartile and hence the interquartile range, correct to 1 decimal place. (The lower quartile is 30 cm.) [2]
- (d) Estimate the number of fish longer than 45 cm. [2]
Show answer
- (a) 8, 30, 70, 104, 120
B2 (B1 for three correct)
- (b) 37.5 cm
60th fish is in 30 < l ≤ 40: 30 + (30/40) × 10 M1 A1
- (c) UQ = 45.9 cm, IQR = 15.9 cm
90th fish: 40 + (20/34) × 10 M1; IQR A1
- (d) 33
Up to 45 cm: 70 + 17 = 87 M1; 120 − 87 A1
200 students took a test. Points on their cumulative frequency curve are (20, 18), (30, 50), (40, 100), (50, 150), (60, 182) and (70, 200), where the first number is the mark and the second is the cumulative frequency. Assume the curve is a straight line between these points.
- (a) Find the median, the lower quartile and the interquartile range of the marks. [3]
- (b) The lowest 30% of students must retake the test. Estimate the highest mark of a student who must retake it. [2]
- (c) Estimate the percentage of students who scored more than 55 marks. [2]
- (d) For a second group of students, the median mark is 44 and the interquartile range is 14. Make two comparisons between the two groups. [1]
Show answer
- (a) Median 40, LQ 30, IQR 20
100th → 40 B1; 50th → 30 B1; 150th → 50, so IQR = 20 B1
- (b) 32
30% of 200 = 60 M1; 30 + (60 − 50)/50 × 10 = 32 A1
- (c) 17%
CF at 55 = 150 + 32/2 = 166 M1; 34/200 = 17% A1
- (d) Second group did better on average (higher median) and their marks were less spread out (smaller IQR).
Both comparisons in context B1
The finishing times, t minutes, of 160 runners in a 10 km race are summarised in a cumulative frequency table.t ≤ 40: 12t ≤ 45: 38t ≤ 50: 82t ≤ 55: 124t ≤ 60: 148t ≤ 70: 160Assume the times are spread evenly within each class.
- (a) Estimate the median time, correct to 1 decimal place. [2]
- (b) The fastest 10% of runners receive a medal. Estimate the greatest time that earns a medal, correct to 1 decimal place. [2]
- (c) Estimate the percentage of runners who took between 47 and 57 minutes. Give your answer correct to 1 decimal place. [3]
Show answer
- (a) 49.8 minutes
80th runner is in 45 < t ≤ 50: 45 + (42/44) × 5 M1 A1
- (b) 40.8 minutes
10% of 160 = 16 B1; 40 + (4/26) × 5 A1
- (c) 48.8%
cf(47) = 38 + 44 × 2/5 = 55.6 M1; cf(57) = 124 + 24 × 2/5 = 133.6 M1; 78 ÷ 160 × 100 = 48.75 A1
The numbers of books read in one term by 8 students are:7, 3, 9, 4, 7, 10, 2, 6
- (a) Find the mean. [2]
- (b) Find the median. [1]
- (c) Write down the mode and find the range. [2]
Show answer
- (a) 6
Sum = 48 M1; 48 ÷ 8 = 6 A1
- (b) 6.5
Ordered: 2, 3, 4, 6, 7, 7, 9, 10; (6 + 7)/2 B1
- (c) Mode 7, range 8
B1 B1
Class A has 20 students. Their mean test mark is 64. Class B has 30 students. Their mean mark on the same test is 72.
- (a) Find the total of all the marks in Class A. [1]
- (b) Find the mean mark of all 50 students. [2]
Show answer
- (a) 1280
20 × 64 B1
- (b) 68.8
(1280 + 2160) ÷ 50 M1 A1 (68.8)
- (a) The mean of five numbers is 12. Four of the numbers are 9, 15, 11 and 14. Find the fifth number. [2]
- (b) A sixth number is added. The mean of the six numbers is now 13. Find the sixth number. [2]
Show answer
- (a) 11
5 × 12 = 60 M1; 60 − 49 = 11 A1
- (b) 18
6 × 13 = 78 M1; 78 − 60 = 18 A1
A football team recorded the number of goals it scored in each match of a season.Goals: 0, 1, 2, 3, 4Number of matches: 4, 7, x, 5, 2
- (a) The mean number of goals per match is 1.8. Find x. [3]
- (b) Find the median number of goals. [2]
- (c) Write down the mode. [1]
Show answer
- (a) x = 12
Σfx = 30 + 2x and Σf = 18 + x M1; (30 + 2x)/(18 + x) = 1.8 M1; x = 12 A1
- (b) 2
30 matches, 15th and 16th values M1; both 2 A1
- (c) 2
B1
Four whole numbers have a mode of 7, a median of 8 and a mean of 9. The largest number appears only once.
- (a) Find the four numbers. [4]
- (b) A fifth whole number is added and the mean becomes 10. Find the fifth number. [2]
Show answer
- (a) 7, 7, 9, 13
Mode 7 means two 7s, which must be the two smallest B1; median (7 + c)/2 = 8 so c = 9 M1; total 4 × 9 = 36 M1; d = 36 − 23 = 13 A1
- (b) 14
5 × 10 − 36 M1 A1
The heights, h cm, of 73 plants are recorded.140 < h ≤ 150: 10150 < h ≤ 155: 15155 < h ≤ 160: 22160 < h ≤ 170: 18170 < h ≤ 190: 8The data is to be shown on a histogram.
- (a) Calculate the frequency density for each class. [3]
- (b) Which class has the tallest bar on the histogram? [1]
Show answer
- (a) 1, 3, 4.4, 1.8, 0.4
FD = frequency ÷ class width M1; 10/10, 15/5, 22/5, 18/10, 8/20 A2 (A1 for 3 correct)
- (b) 155 < h ≤ 160
B1 (highest frequency density)
The waiting times, t minutes, of 100 customers are summarised.0 < t ≤ 10: frequency 12, frequency density 1.210 < t ≤ 15: frequency ?, frequency density 3.615 < t ≤ 20: frequency 25, frequency density ?20 < t ≤ 30: frequency 30, frequency density ?30 < t ≤ 50: frequency ?, frequency density 0.75
- (a) Find the two missing frequencies. [2]
- (b) Find the two missing frequency densities. [2]
- (c) Estimate the number of customers who waited more than 25 minutes. [2]
Show answer
- (a) 18 and 15
3.6 × 5 = 18 B1; 0.75 × 20 = 15 B1
- (b) 5 and 3
25 ÷ 5 = 5 B1; 30 ÷ 10 = 3 B1
- (c) 30
Half of 20–30 class: 15 M1; 15 + 15 = 30 A1
The speeds, v km/h, of 150 cars on a motorway are recorded.60 < v ≤ 80: 1880 < v ≤ 90: 2790 < v ≤ 100: 42100 < v ≤ 110: 39110 < v ≤ 140: 24
- (a) Calculate the frequency density for each class. [3]
- (b) Estimate how many of these cars were travelling faster than 120 km/h. [2]
Show answer
- (a) 0.9, 2.7, 4.2, 3.9, 0.8
Frequency ÷ class width M1; all five A2 (A1 for three)
- (b) 16
(140 − 120)/30 × 24 M1 A1
A histogram shows the ages, a years, of people at a library. The bar for the class 10 < a ≤ 20 has height 6 cm and represents 42 people.
- (a) The bar for 20 < a ≤ 25 has height 8 cm. How many people does it represent? [2]
- (b) The class 25 < a ≤ 40 contains 63 people. Find the height of its bar. [2]
- (c) Using these three classes only, calculate an estimate of the mean age. Give your answer correct to 1 decimal place. [3]
Show answer
- (a) 28
6 cm ↔ FD 42/10 = 4.2, so 1 cm ↔ 0.7 M1; 8 × 0.7 × 5 = 28 A1
- (b) 6 cm
FD = 63/15 = 4.2 M1; 4.2 ÷ 0.7 = 6 cm A1
- (c) 24.9 years
Midpoints 15, 22.5, 32.5 M1; Σfx = 3307.5, Σf = 133 M1; 24.9 A1
A histogram shows the lifetimes, h hours, of some batteries.0 < h ≤ 20: frequency density 0.520 < h ≤ 30: frequency density 2.430 < h ≤ 35: frequency density 635 < h ≤ 40: frequency 2640 < h ≤ 60: frequency density 1.1
- (a) Find the frequency of each class and the missing frequency density. [3]
- (b) Calculate an estimate of the mean lifetime, correct to 3 significant figures. [3]
- (c) Write down the class that contains the median. [1]
Show answer
- (a) Frequencies 10, 24, 30, 26, 22; missing density 5.2
Frequency = density × width M1; 10, 24, 30, 22 A1; 26 ÷ 5 = 5.2 B1
- (b) 33.5 hours
Midpoints 10, 25, 32.5, 37.5, 50 M1; Σfx = 3750 M1; ÷ 112 A1
- (c) 30 < h ≤ 35
56.5th value; cumulative 10, 34, 64 B1
The distances, d km, that 40 people travel to work are recorded.0 < d ≤ 5: 85 < d ≤ 10: 1410 < d ≤ 15: 1015 < d ≤ 20: 620 < d ≤ 25: 2
- (a) Write down the modal class. [1]
- (b) Find the class that contains the median. [1]
- (c) Calculate an estimate of the mean distance. [3]
Show answer
- (a) 5 < d ≤ 10
B1
- (b) 5 < d ≤ 10
20th/21st values: cumulative 8, 22 B1
- (c) 10 km
Midpoints 2.5, 7.5, 12.5, 17.5, 22.5 M1; Σfx = 400 M1; 400 ÷ 40 = 10 A1
The times, t hours, that 50 teenagers spent on their phones on one day are recorded.0 < t ≤ 1: 61 < t ≤ 2: 132 < t ≤ 3: 173 < t ≤ 4: 94 < t ≤ 6: 5
- (a) Calculate an estimate of the mean time. [3]
- (b) Explain why your answer is only an estimate. [1]
Show answer
- (a) 2.43 hours
Midpoints 0.5, 1.5, 2.5, 3.5, 5 M1; Σfx = 121.5 M1; ÷ 50 A1
- (b) The exact times are not known; each time is replaced by the midpoint of its class.
B1
The masses, m grams, of 85 apples are recorded.100 < m ≤ 120: 11120 < m ≤ 140: 23140 < m ≤ 160: 31160 < m ≤ 180: 15180 < m ≤ 200: 5
- (a) Calculate an estimate of the mean mass. Give your answer correct to 1 decimal place. [3]
- (b) Explain why your answer to part (a) is only an estimate. [1]
- (c) Find the percentage of apples with a mass greater than 160 g. Give your answer correct to 1 decimal place. [2]
Show answer
- (a) 145.3 g
Midpoints 110, 130, 150, 170, 190 M1; Σfx = 12 350 M1; 12 350 ÷ 85 = 145.3 A1
- (b) The exact masses are not known; midpoints are used.
B1
- (c) 23.5%
20/85 M1; 23.5% A1
The masses, m kg, of 60 babies born at a hospital in Peshawar are recorded.2.0 < m ≤ 2.5: 52.5 < m ≤ 3.0: 143.0 < m ≤ 3.5: 233.5 < m ≤ 4.0: 134.0 < m ≤ 4.5: 5
- (a) Calculate an estimate of the mean mass, correct to 3 significant figures. [3]
- (b) Find the percentage of babies with a mass of more than 3.5 kg. [1]
- (c) At a second hospital, 20 babies had an estimated mean mass of 3.05 kg. Estimate the mean mass of all 80 babies, correct to 3 significant figures. [2]
Show answer
- (a) 3.24 kg
Midpoints M1; Σfx = 194.5 M1; ÷ 60 A1
- (b) 30%
18 ÷ 60 × 100 B1
- (c) 3.19 kg
(194.5 + 61) ÷ 80 M1 A1
The ages, a years, of the members of a chess club are grouped.10 < a ≤ 20: 620 < a ≤ 30: k30 < a ≤ 40: 1240 < a ≤ 60: 7
- (a) An estimate of the mean age is 30 years. Find k. [3]
- (b) Three more people join the club. Each of them is in the class 60 < a ≤ 80. Calculate the new estimate of the mean age. [2]
Show answer
- (a) k = 22
Σfx = 860 + 25k, Σf = 25 + k M1; 860 + 25k = 30(25 + k) M1; k = 22 A1
- (b) 32.4 years
1410 + 3 × 70 = 1620 M1; 1620 ÷ 50 = 32.4 A1
72 students chose their favourite sport. 30 chose football, 20 chose cricket, 12 chose hockey and the rest chose tennis. The results are to be shown in a pie chart.
- (a) Calculate the sector angle for football. [1]
- (b) Calculate the sector angles for cricket and hockey. [2]
- (c) The sector for tennis has angle 50°. How many students chose tennis? [1]
Show answer
- (a) 150°
360 ÷ 72 = 5° per student; 30 × 5 = 150 B1
- (b) Cricket 100°, hockey 60°
20 × 5 B1; 12 × 5 B1
- (c) 10
50 ÷ 5 = 10 B1
The ages of 15 shoppers are shown in a stem-and-leaf diagram.1 | 8 92 | 0 3 3 73 | 1 4 4 4 84 | 2 5 65 | 0Key: 2 | 3 means 23 years
- (a) Find the median age. [1]
- (b) Write down the mode and find the range. [2]
- (c) What fraction of the shoppers were older than 40? [1]
Show answer
- (a) 34
8th of 15 values B1
- (b) Mode 34, range 32
B1 B1
- (c) 4/15
42, 45, 46, 50 B1
A two-way table shows how 120 students travel to school.Boys: walk ?, bus 27, car ?, total 64Girls: walk 20, bus ?, car 13, total ?Total: walk 45, bus 50, car ?, total 120
- (a) Complete the table. [3]
- (b) What fraction of the girls travel by bus? [1]
- (c) What fraction of all the students travel by car? Give your answer in its simplest form. [2]
Show answer
- (a) Boys: walk 25, car 12. Girls: bus 23, total 56. Car total 25.
Boys walk 45 − 20 = 25 B1; girls bus 50 − 27 = 23 B1; boys car 64 − 25 − 27 = 12, girls total 56, car total 25 B1
- (b) 23/56
B1 ft
- (c) 5/24
25/120 M1 (or B1 for 25/120 unsimplified); 5/24 A1.
A pie chart shows how people in a town travel to work. The four sector angles are x°, 2x°, (x + 30)° and 90°.
- (a) Find x. [2]
- (b) The pie chart represents 480 people. How many people are in the sector with angle 2x°? [2]
- (c) In a second pie chart for a different town, 72 people are represented by a sector of 54°. How many people does the whole of this second pie chart represent? [2]
Show answer
- (a) x = 60
x + 2x + (x + 30) + 90 = 360 M1; 4x = 240, x = 60 A1
- (b) 160
120/360 × 480 M1; 160 A1
- (c) 480
72 ÷ 54 × 360 M1; 480 A1
The table shows the hours of revision and the test score of 8 students.Hours: 1, 2, 3, 4, 5, 6, 7, 8Score: 35, 42, 46, 55, 58, 66, 70, 78
- (a) What type of correlation does the data show? [1]
- (b) Calculate the mean number of hours and the mean score. [2]
- (c) A line of best fit has equation y = 6x + 29.25, where x is hours and y is score. Use it to estimate the score of a student who revised for 5.5 hours. [1]
- (d) Explain why the line should not be used to estimate the score for 15 hours of revision. [1]
Show answer
- (a) Positive
B1
- (b) 4.5 hours, 56.25
36 ÷ 8 B1; 450 ÷ 8 B1
- (c) 62.25
6 × 5.5 + 29.25 B1
- (d) 15 is outside the range of the data (extrapolation); the score would be more than 100.
B1
A café records the midday temperature (°C) and the number of cups of hot chocolate sold on 8 days.Temperature: 2, 5, 8, 10, 13, 15, 18, 20Cups sold: 88, 80, 71, 30, 58, 52, 44, 38
- (a) One day does not fit the pattern. Write down its temperature. [1]
- (b) Describe the correlation and what it means in context. [2]
- (c) A line of best fit passes through (0, 94) and (20, 38). Find its equation in the form y = mx + c. [2]
- (d) Use your line to estimate the number of cups sold when the temperature is 12 °C, and the temperature when 66 cups are sold. [2]
Show answer
- (a) 10 °C
(10, 30) is the outlier B1
- (b) Negative correlation: as temperature rises, fewer cups are sold.
Negative B1; context B1
- (c) y = −2.8x + 94
Gradient (38 − 94)/20 = −2.8 M1; A1
- (d) 60 cups; 10 °C
94 − 33.6 = 60.4 ≈ 60 B1; (94 − 66)/2.8 = 10 B1
The table shows the practice time (x hours per week) and typing speed (y words per minute) of 6 people.x: 2, 4, 5, 7, 8, 10y: 14, 19, 22, 25, 29, 35
- (a) Calculate the mean practice time and the mean typing speed. [2]
- (b) The line of best fit passes through the mean point and the point (0, 9). Find its equation. [2]
- (c) Use the line to estimate the typing speed of a person who practises for 9 hours a week. [1]
- (d) Describe the strength and type of correlation. [1]
Show answer
- (a) (6, 24)
36 ÷ 6 = 6 B1; 144 ÷ 6 = 24 B1
- (b) y = 2.5x + 9
Gradient (24 − 9)/6 = 2.5 M1; A1
- (c) 31.5 words per minute
2.5 × 9 + 9 B1
- (d) Strong positive
B1
The table shows the engine size (litres) and fuel economy (km per litre) of 7 cars.Engine size: 1.0, 1.3, 1.5, 1.8, 2.0, 2.4, 3.0Fuel economy: 18.5, 16.8, 15.4, 13.9, 12.6, 11.0, 8.3
- (a) Describe the correlation and what it means in context. [1]
- (b) Calculate the mean engine size and the mean fuel economy. Give answers correct to 3 significant figures. [2]
- (c) The line of best fit passes through the mean point and the point (1.0, 18.4). Find the gradient of the line, correct to 3 significant figures. [2]
- (d) Use the line to estimate the fuel economy of a car with a 2.2 litre engine, correct to 3 significant figures. [2]
- (e) Explain why the line should not be used for a car with a 5.0 litre engine. [1]
Show answer
- (a) Strong negative correlation: cars with larger engines travel fewer kilometres per litre.
B1
- (b) 1.86 litres and 13.8 km per litre
B1 B1
- (c) −5.38
(13.79 − 18.4)/(1.857 − 1.0) M1 A1
- (d) 11.9 km per litre
18.4 + 1.2 × gradient M1 A1
- (e) 5.0 litres is outside the range of the data (extrapolation), so the trend may not continue.
B1
