iGCSE Mathematics (0580) - C2.9 Graphs in practical situations- Exam Style Questions Paper 1- New Syllabus
Question
Ky cycles from his office to a meeting and back again.
The travel graph shows his time at the meeting and his journey back.

(a) How far is the meeting from his office?
(b) How long is Ky at the meeting for?
(c) Write down the time Ky arrives back at his office after the meeting.
(d) Ky cycles from his office to the meeting at a constant speed of $21\text{ km/h}$. Complete the travel graph.
Most-appropriate topic codes (Cambridge IGCSE Mathematics 0580):
▶️ Answer/Explanation
(b) The horizontal line starts at $10:20$ and ends at $11:00$. Calculating the duration between these two times gives $40\text{ minutes}$.
(c) Following the straight sloping line representing his return journey down to the horizontal axis (distance = $0$), it lands precisely on the grid mark for $11:50$.
(d) To find out how long the outward trip took, we use the formula $\text{time} = \frac{\text{distance}}{\text{speed}} = \frac{14\text{ km}}{21\text{ km/h}} = \frac{2}{3}\text{ of an hour} = 40\text{ minutes}$. Since he arrived at the meeting at $10:20$, he must have started cycling from the office $40$ minutes prior, which is at $09:40$. To complete the travel graph, draw a straight line from $(09:40, 0)$ to $(10:20, 14)$.

✅ Answer:
(a) $14\text{ km}$
(b) $40\text{ min}$
(c) $11:50$
(d) Graph completed with a line drawn from $(09:40, 0)$ to $(10:20, 14)$.

The travel graph shows a student’s journey.
(a) Explain what is happening between 14 20 and 14 40.
(b) Complete the statement: The student is travelling fastest between the times ______ and ______ because ______.
▶️ Answer/Explanation
Ans:
(a) The student is stationary (not moving) between 14:20 and 14:40 as the distance doesn’t change.
(b) The student is travelling fastest between 13:00 and 13:20 because the gradient is steepest (distance changes most rapidly).
