Home / CIE iGCSE Maths E9.6 Cumulative frequency diagrams Exam Style Practice Questions- Paper 2

CIE iGCSE Maths E9.6 Cumulative frequency diagrams Exam Style Practice Questions- Paper 2

CIE iGCSE Maths E9.6 Cumulative frequency diagrams Exam Style Practice Questions- Paper 2

Question

Cumulative frequency diagram

The time taken for each of 200 students to complete a calculation is measured. The cumulative frequency diagram shows the results.
Use the diagram to find an estimate for:

(a) the interquartile range

(b) the number of students taking more than 40 seconds to complete the calculation.

▶️ Answer/Explanation
Solution

(a) Ans: 11 seconds

Graph with quartiles marked

Calculating Interquartile Range (IQR):

  1. Total students = 200
  2. Lower quartile (Q₁) position = 25% of 200 = 50th student
  3. Upper quartile (Q₃) position = 75% of 200 = 150th student
  4. From graph:
    • Q₁ ≈ 16 seconds (when CF = 50)
    • Q₃ ≈ 27 seconds (when CF = 150)
  5. IQR = Q₃ – Q₁ = 27 – 16 = 11 seconds

(b) Ans: 6 students

Finding students taking >40 seconds:

  1. At 40 seconds, cumulative frequency ≈ 194
  2. Students taking >40s = Total – CF at 40s = 200 – 194 = 6

Key points to remember:

  • IQR measures the middle 50% of data (Q₃ – Q₁)
  • Cumulative frequency graphs show running totals
  • For “more than” questions, subtract from total
Question

The speed-time graph shows information about a car journey.

(a) Find the deceleration of the car between 240 and 320 seconds.

(b) Calculate the total distance the car travels during the 320 seconds.

▶️ Answer/Explanation
Solution

(a) Ans: 0.2 m/s²

Deceleration = \(\frac{\text{Change in speed}}{\text{Time}}\)

From graph:
• Speed decreases from 16 m/s to 0 m/s
• Time interval = 320 – 240 = 80 s

\(\text{Deceleration} = \frac{16 – 0}{80} = 0.2 \, \text{m/s}^2\)

(b) Ans: 4240 m

Total distance = Area under graph (3 parts):

  1. 0-30s (Triangle): \(\frac{1}{2} × 30 × 16 = 240 \, \text{m}\)
  2. 30-240s (Rectangle): \(210 × 16 = 3360 \, \text{m}\)
  3. 240-320s (Triangle): \(\frac{1}{2} × 80 × 16 = 640 \, \text{m}\)

Total distance = \(240 + 3360 + 640 = 4240 \, \text{m}\)

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