Edexcel International GCSE Maths Histograms and Cumulative Frequency Past Paper Questions and Solutions (4MA1 Higher)

Part (a)

1. Running totals: 9, 28, 45, 63, 76, 80

Answer: 9, 28, 45, 63, 76, 80

Part (b)

1. Plot (20, 9), (30, 28), (40, 45), (50, 63), (60, 76), (70, 80), starting at (0, 0), and join with a curve.

Answer: Curve through (0, 0), (20, 9), (30, 28), (40, 45), (50, 63), (60, 76), (70, 80)

Part (c)

1. The median is the 80 ÷ 2 = 40th person. Read across from 40 to the curve, then down.

Answer: about 37 years (35 to 38 accepted)

Part (a)

1. The median is the 90 ÷ 2 = 45th bag. Read across from 45 to the curve and down.

Answer: about 146 grams (145 to 147 accepted)

Part (b)

1. Number of large bags = 93.75 ÷ 3.75 = 25
2. The 25 heaviest bags are large, so 90 − 25 = 65 bags weigh d grams or less.
3. Read across from 65 on the cumulative frequency axis to the curve, then down.

Answer: d = about 151 (150 to 152 accepted)

Part (a)

1. Running totals of the frequencies.

Answer: 8, 23, 40, 68, 101, 120

Part (b)

1. Plot (5, 8), (10, 23), (15, 40), (20, 68), (25, 101), (30, 120), start at (0, 0), and join with a curve.

Answer: Curve through (0, 0), (5, 8), (10, 23), (15, 40), (20, 68), (25, 101), (30, 120)

Part (c)

1. Median = the 120 ÷ 2 = 60th person. Read across from 60, then down.

Answer: about 18.5 minutes (17 to 20 accepted)

Part (d)

1. Read up from 23 minutes: about 88 people waited 23 minutes or less.
2. Longer than 23 minutes: 120 − 88 = 32 people
3. Percentage = ³²⁄₁₂₀ × 100 = 26.7%

Answer: about 27% (25% to 29% accepted)

Part (a)

1. Running totals: 12, 12 + 24 = 36, 36 + 28 = 64, 64 + 12 = 76, 76 + 10 = 86, 86 + 4 = 90

Answer: 12, 36, 64, 76, 86, 90

Part (b)

1. Plot at the upper class boundaries: (35, 12), (45, 36), (55, 64), (65, 76), (75, 86), (85, 90), starting at (25, 0).

Answer: Curve through (25, 0), (35, 12), (45, 36), (55, 64), (65, 76), (75, 86), (85, 90)

Part (c)

1. Read the cumulative frequency at 52 minutes (about 56) and at 42 minutes (about 29).
2. Silver medals ≈ 56 − 29

Answer: about 27 runners (25 to 29 accepted)

Part (a)

1. Running totals of the frequencies: 7, 7 + 10 = 17, 17 + 15 = 32, 32 + 32 = 64, 64 + 16 = 80

Answer: 7, 17, 32, 64, 80

Part (b)

1. Plot each cumulative frequency at the end of its class: (10, 7), (20, 17), (30, 32), (40, 64), (50, 80). Start at (0, 0) and join with a smooth curve.

Answer: Curve through (0, 0), (10, 7), (20, 17), (30, 32), (40, 64), (50, 80)

Part (c)

1. Median = the 80 ÷ 2 = 40th patient. Read across from 40, then down to the time axis.

Answer: about 33 minutes (32 to 34 accepted)

Part (d)

1. Lower quartile: read across from 80 ÷ 4 = 20, giving about 22 minutes.
2. Upper quartile: read across from 3 × 80 ÷ 4 = 60, giving about 38 minutes.
3. IQR = 38 − 22

Answer: about 16 minutes (15 to 17 accepted)

Part (a)

1. Lower quartile: the 80 ÷ 4 = 20th person. Read across from 20 to the curve and down: about 38 minutes.
2. Upper quartile: the 60th person: about 56 minutes.
3. IQR = 56 − 38

Answer: about 18 minutes (17 to 19 accepted)

Part (b)

1. Read the cumulative frequency at 50 minutes: about 42 people took 50 minutes or less.
2. No women took 50 minutes or less, so these 42 people are all men, and they are 60% of the men.
3. Number of men = 42 ÷ 0.6

Answer: 70

Part (a)

1. There are 60 people, so the median is the 30th value. Read across from 30 to the curve and down.

Answer: about 23 minutes (22 to 24 accepted)

Part (b)

1. Lower quartile (15th value) ≈ 16 minutes. Upper quartile (45th value) ≈ 28 minutes.
2. IQR ≈ 28 − 16

Answer: about 12 minutes (10 to 14 accepted)

Part (c)

1. Compare an average: Hospital A’s median (23) is lower than Hospital B’s (28).
2. Compare the spread: Hospital A’s IQR (12) is smaller than Hospital B’s (19).

Answer: On average waiting times are shorter at Hospital A (lower median), and they are more consistent at Hospital A (smaller IQR)

Part (a)

1. Running totals.

Answer: 7, 33, 57, 71, 78, 80

Part (b)

1. Plot (10, 7), (20, 33), (30, 57), (40, 71), (50, 78), (60, 80), start at (0, 0), and join with a curve.

Answer: Curve through (0, 0), (10, 7), (20, 33), (30, 57), (40, 71), (50, 78), (60, 80)

Part (c)

1. Median = the 40th customer. Read across from 40 and down.

Answer: about 23 minutes (21 to 24 accepted)

Part (d)

1. Read up from 42 minutes: about 72 customers took 42 minutes or less.
2. More than 42 minutes: 80 − 72 = 8 customers

Answer: about ⁸⁄₈₀ (= 0.1)

Part (a)

1. There are 50 plants, so the median is the 25th value. Go across from 25 on the cumulative frequency axis to the curve, then down.

Answer: about 28 cm

Part (b)

1. Read the cumulative frequency at 40 cm (about 44) and at 30 cm (about 30). The frequency for the class is the difference.
2. 44 − 30

Answer: 14

Part (c)

1. Go up from 35 cm to the curve and across: about 38 plants are 35 cm or shorter.
2. Plants taller than 35 cm = 50 − 38

Answer: about 12

Part (a)

1. Cumulative frequency is a running total: add each frequency to the total so far.

Answer: 12, 27, 47, 65, 74, 80

Part (b)

1. Plot each cumulative frequency at the upper end of its class: (20, 12), (30, 27), (40, 47), (50, 65), (60, 74), (70, 80). Start at (10, 0) and join the points with a smooth curve.

Answer: Curve through (10, 0), (20, 12), (30, 27), (40, 47), (50, 65), (60, 74), (70, 80)

Part (c)

1. Go up from 46 on the age axis to the curve, then across: the cumulative frequency is about 58. So about 58 people are 46 or younger.
2. People older than 46 ≈ 80 − 58 = 22
3. Percentage = ²²⁄₈₀ × 100 = 27.5%, which is about 28%

Answer: 28% (the mark scheme accepts 26% to 29%, depending on your curve)

Part (a)

1. Keep a running total of the frequencies.

Answer: 15, 31, 52, 66, 74, 80

Part (b)

1. Plot each running total at the upper end of its class: (60, 15), (70, 31), (80, 52), (90, 66), (100, 74), (110, 80), and start at (50, 0). Join with a smooth curve.

Answer: Curve through (50, 0), (60, 15), (70, 31), (80, 52), (90, 66), (100, 74), (110, 80)

Part (c)

1. The median is the 80 ÷ 2 = 40th value. Go across from 40 to the curve, then down.

Answer: about 74 minutes (73 to 75 accepted)

Part (d)

1. Lower quartile: the 80 ÷ 4 = 20th value, about 63 minutes.
2. Upper quartile: the 3 × 80 ÷ 4 = 60th value, about 86 minutes.
3. IQR = 86 − 63

Answer: about 23 minutes (21 to 24 accepted)

1. In a histogram the bar heights are frequency densities: frequency density = frequency ÷ class width.

1. Draw each bar with no gaps, from the start to the end of its class, up to its frequency density.

Answer: Bars at heights 2.5, 4, 3.6, 1.4 and 0.6

1. Frequency = frequency density × class width. Find the frequency of every bar.

1. Total number of people = 25 + 20 + 54 + 10 + 17 = 126
2. More than 40 minutes: half of the 30–50 bar (40 to 50 is 10 of its 20 minutes) gives 0.5 × 10 = 5 people, plus all 17 in the 50–60 bar. That is 5 + 17 = 22 people.
3. Proportion = ²²⁄₁₂₆ = ¹¹⁄₆₃

Answer: ¹¹⁄₆₃ (about 0.175)

Part (a)

1. Frequency density = frequency ÷ class width: 35 ÷ 10 = 3.5, 45 ÷ 15 = 3, 75 ÷ 15 = 5, 40 ÷ 20 = 2, 8 ÷ 10 = 0.8

Answer: Bars at frequency densities 3.5, 3, 5, 2 and 0.8

Part (b)

1. From 40 to 50 is ¹⁰⁄₁₅ of the 35–50 class: ¹⁰⁄₁₅ × 75 = 50 plants
2. Add the classes above 50: 40 + 8 = 48
3. 50 + 48

Answer: 98

Part (a)

1. Use the 0–10 bar to find the scale: frequency density = 40 ÷ 10 = 4. That bar is 20 small squares tall, so each small square = 0.2.
2. 10–15 is 35 small squares tall: fd = 7, frequency = 7 × 5 = 35
3. 15–20 is 39 small squares tall: fd = 7.8, frequency = 7.8 × 5 = 39
4. 20–30 is 28 small squares tall: fd = 5.6, frequency = 5.6 × 10 = 56

Answer: 35, 39, 56

Part (b)

1. 30–60: frequency density = 30 ÷ 30 = 1, so draw a bar 5 small squares tall from 30 to 60.

Answer: Bar from 30 to 60 at frequency density 1

Part (c)

1. Total number of people = 40 + 35 + 39 + 56 + 30 = 200
2. More than 25 km: half of the 20–30 class (25 to 30) is 0.5 × 56 = 28, plus all 30 in 30–60, giving 58 people
3. Probability = ⁵⁸⁄₂₀₀

Answer: ²⁹⁄₁₀₀ = 0.29

1. The two missing frequencies add to 140 − 23 − 18 − 14 = 85.
2. The bars have no scale, so compare them. Frequency = width × height, so compare both.
3. The 15–30 bar is 1.5 times as wide as the 5–15 bar and 1.6 times as tall (20 small squares against 12.5).
4. So if the 5–15 bar has frequency f, the 15–30 bar has 1.5 × 1.6 × f = 2.4f.
5. f + 2.4f = 85, so 3.4f = 85 and f = 25. The 15–30 class has 85 − 25 = 60.
6. Frequency densities: 5–15 is 25 ÷ 10 = 2.5 and 15–30 is 60 ÷ 15 = 4. So each big square up is 1 and each small square is 0.2.
7. Now the missing bars: 0–5 has 23 ÷ 5 = 4.6, 30–40 has 18 ÷ 10 = 1.8 and 40–60 has 14 ÷ 20 = 0.7.

Answer: Table: 25 and 60. New bars: 0–5 at 4.6, 30–40 at 1.8, 40–60 at 0.7

1. No scale is given, so measure each bar’s height in small squares: 18, 40, 20, 15 and 6.
2. Use width (cm) × height (squares) as “area units”, because frequency is proportional to area.
3. 155 to 170 cm: 5 × 40 + 10 × 20 = 200 + 200 = 400 area units = 160 students, so 1 student = 400 ÷ 160 = 2.5 area units.
4. Total area: 15 × 18 + 5 × 40 + 10 × 20 + 15 × 15 + 5 × 6 = 270 + 200 + 200 + 225 + 30 = 925 area units
5. Total students = 925 ÷ 2.5

Answer: 370

1. There is no scale, so measure each bar’s height in small squares: 200–250 is 10, 250–300 is 8, 300–400 is 6, 400–450 is 15, 450–600 is 5 and 600–800 is 1.
2. Frequency is proportional to area, so use width (cm) × height (squares) as “area units”.
3. 400 < h ≤ 500 = all of the 400–450 bar and the first 50 cm of the 450–600 bar: 50 × 15 + 50 × 5 = 1000 area units
4. 300 < h ≤ 400: 100 × 6 = 600 area units
5. The difference, 1000 − 600 = 400 area units, represents 8 trees, so 50 area units = 1 tree.
6. Greater than 500 cm: the last 100 cm of the 450–600 bar plus the whole 600–800 bar: 100 × 5 + 200 × 1 = 700 area units
7. 700 ÷ 50

Answer: 14

1. Use the 8–8.5 bar for the scale: frequency density = 16 ÷ 0.5 = 32. That bar is 32 small squares tall, so each small square = 1 watermelon per kg.
2. Read the other bars: 7–8 is 15, 8.5–9.5 is 23, 9.5–10.5 is 30, 10.5–12 is 12.
3. Frequencies = density × width: 15 × 1 = 15, 23 × 1 = 23, 30 × 1 = 30, 12 × 1.5 = 18
4. Total = 15 + 16 + 23 + 30 + 18

Answer: 102

1. Use the bar we know to find the scale. 10 children in 0–20 (width 20) gives frequency density 10 ÷ 20 = 0.5. That bar is 2.5 small squares tall, so each small square up = 0.2.
2. Read the bars we need: 40–60 has frequency density 2.9, 60–75 has 3.2 and 75–100 has 2.0.
3. 50 < m ≤ 80 covers half of the 40–60 bar (50 to 60), all of the 60–75 bar, and 75 to 80 of the last bar.
4. 50–60: 2.9 × 10 = 29; 60–75: 3.2 × 15 = 48; 75–80: 2.0 × 5 = 10
5. Total = 29 + 48 + 10

Answer: 87

1. There is no scale, but we only need a percentage, so we can measure each bar in small squares. Area is proportional to frequency.
2. Using width (in years) × height (in small squares): 20–25: 5 × 16 = 80, 25–30: 5 × 50 = 250, 30–40: 10 × 33 = 330, 40–50: 10 × 19 = 190, 50–75: 25 × 9 = 225
3. Total = 80 + 250 + 330 + 190 + 225 = 1075
4. Aged more than 55: the part of the last bar from 55 to 75, which is 20 × 9 = 180
5. Percentage = ¹⁸⁰⁄₁₀₇₅ × 100 = 16.744…

Answer: 16.7%

1. The vertical axis has no scale, so use the bar we know. The 15–20 bar has frequency 14 and width 5, so its frequency density = 14 ÷ 5 = 2.8. Its height is 28 small squares, so each small square up is 0.1.
2. Read the other bars: 20–35 has height 3.4 and 35–55 has height 0.4.
3. Frequency = frequency density × class width: 20–35 gives 3.4 × 15 = 51 adults; 35–55 gives 0.4 × 20 = 8 adults.
4. Adults shown so far = 14 + 51 + 8 = 73, so the missing class 5–15 has 100 − 73 = 27 adults.
5. Frequency density for 5–15 = 27 ÷ 10 = 2.7, so draw a bar from 5 to 15 with height 2.7.

Answer: Bar from 5 km to 15 km at frequency density 2.7