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0580/41 Oct/Nov 2024, Question 4: Speed, Distance and Time

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 17 marks

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The question, in short

Using a table of lengths swum in 15 minutes by Enzo, Rashid and Blessy, work out percentages, a simplified ratio, speeds, times and a decreasing-lengths calculation.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (ai) · 2 marks

Find Enzo's lengths as a percentage of the total lengths swum by all three swimmers.

  1. 1Total = 11.25 + 18.75 + 20 = 50
  2. 211.25/50 × 100 = 22.5%

Answer22.5

Part (aii) · 2 marks

Write the ratio Enzo : Rashid : Blessy of lengths swum in simplest form.

  1. 111.25 : 18.75 : 20 = 1125 : 1875 : 2000
  2. 2Divide by 125: 9 : 15 : 16

Answer9 : 15 : 16

Part (aiiia) · 3 marks

With each length 25 m, work out Blessy's average speed over 15 minutes in metres per second.

  1. 1Distance = 20 × 25 = 500 m
  2. 2Time = 15 × 60 = 900 s
  3. 3Speed = 500 ÷ 900 = 5/9 ≈ 0.556 m/s

Answer5/9 or 0.556 or 0.5555 to 0.5556

Part (aiiib) · 4 marks

Rashid keeps the same rate; find the time to swim 5 km, in hours and minutes.

  1. 1Rashid swims 18.75 × 25 = 468.75 m in 15 min
  2. 2In 1 hour: 468.75 × 4 = 1875 m
  3. 3Time = 5000 ÷ 1875 = 2.666… h = 160 min
  4. 4= 2 h 40 mins

Answer2 h 40 mins

Part (aiv) · 3 marks

Blessy swims one hour, lengths dropping 5% each 15 minutes; find lengths in the final 15 minutes.

  1. 1Each 15 minutes the lengths are multiplied by 0.95
  2. 2Final 15 minutes = 20 × 0.95³
  3. 3= 17.1475 ≈ 17.1

Answer17.1 or 17.14 to 17.15

Part (b) · 3 marks

Adam swims 450 m to the nearest 25 m in 10 minutes to the nearest minute; find the minimum distance in one hour.

  1. 1Minimum distance = 450 − 12.5 = 437.5 m
  2. 2Maximum time = 10 + 0.5 = 10.5 min
  3. 3Minimum distance in 1 hour = 437.5 ÷ 10.5 × 60
  4. 4= 2500 m

Answer2500

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Also searched as: 0580/41/O/N/24 · 0580_w24_qp_41 · 0580 w24 41 · question 4.
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