CIE iGCSE Maths C9.3 Averages and range Exam Style Practice Questions- Paper 1
▶️ Answer/Explanation
(a) The mode is simply the value that appears most frequently. Here, $2$ occurs twice, while everything else appears once, so the mode is $2$.
(b) The range is the difference between the largest and smallest numbers: $7 – 0 = 7$.
(c) The data is already in ascending order. Since there is an even number of values ($6$), the median is the midpoint of the two middle terms ($2$ and $3$): $\frac{2 + 3}{2} = 2.5$.
(d) To calculate the mean, sum all values together and divide by the total count: $\frac{0 + 2 + 2 + 3 + 4 + 7}{6} = \frac{18}{6} = 3$.
✅ Answer: (a) 2 (b) 7 (c) 2.5 (d) 3
(b) The range is the difference between the largest and smallest numbers: $7 – 0 = 7$.
(c) The data is already in ascending order. Since there is an even number of values ($6$), the median is the midpoint of the two middle terms ($2$ and $3$): $\frac{2 + 3}{2} = 2.5$.
(d) To calculate the mean, sum all values together and divide by the total count: $\frac{0 + 2 + 2 + 3 + 4 + 7}{6} = \frac{18}{6} = 3$.
✅ Answer: (a) 2 (b) 7 (c) 2.5 (d) 3
▶️ Answer/Explanation
(a) (i) To find the median, we first arrange the numbers in ascending order: $5, 7, 8, 10, 22, 35, 75$. Since there are 7 numbers, the median is the middle value (the 4th number), which is 10.
(ii) The range is calculated by subtracting the smallest value from the largest value: $\text{Maximum} – \text{Minimum} = 75 – 5 = 70$.
(b) Since 1 hour equals 60 minutes, we can convert 75 minutes to hours by dividing by 60: $\frac{75}{60} = \frac{5}{4} = 1.25$ hours (or $1\frac{1}{4}$ hours).
Answers:
(a) (i) 10 min
(ii) 70 min
(b) 1.25 h or $1\frac{1}{4}$ h
