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A cyclist travels from Town X to Town Y. The travel graph shows his journey. Calculate the average speed for the return journey from Y to X and state the period he travelled to Town Y after his first

Mathematics

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4 parts
i

60 km/h60 \text{ km/h}

ii

1:30 pm to 2:30 pm1:30 \text{ pm to } 2:30 \text{ pm}

iii

30 minutes30 \text{ minutes}

iv

20 km20 \text{ km}

−4324-\frac{43}{24}

Step-by-step solution

15 steps

here are the solutions to all the questions on this paper, including the final part:

6. a) The travel graph below describes the journey of a cyclist from Town X to Town Y. Study the graph and use it to answer the questions that follow.

Part i· 4 steps

What is the average speed for the return journey from Town Y to Town X?

  1. Identify the start and end points of the return journey from the graph.

    The return journey starts at Town Y (60 km from Town X) at 3 pm and ends at Town X (0 km) at 4 pm.

  2. Calculate the distance covered during the return journey.

    Distance=60 km−0 km=60 km\text{Distance} = 60 \text{ km} - 0 \text{ km} = 60 \text{ km}

  3. Calculate the time taken for the return journey.

    Time=4 pm−3 pm=1 hour\text{Time} = 4 \text{ pm} - 3 \text{ pm} = 1 \text{ hour}

  4. Calculate the average speed using the formula Speed=DistanceTime\text{Speed} = \frac{\text{Distance}}{\text{Time}}.

    Average Speed=60 km1 hour=60 km/h\text{Average Speed} = \frac{60 \text{ km}}{1 \text{ hour}} = 60 \text{ km/h}

    The average speed for the return journey is 60 km/h\boxed{60 \text{ km/h}}.

Part ii

1:30 pm to 2:30 pm1:30 \text{ pm to } 2:30 \text{ pm}

State the period within which he travelled to Town Y after his first rest?

  1. Locate the first rest period on the graph.

    The first horizontal segment (rest) is from 1 pm to 1:30 pm at a distance of 35 km.

  2. Identify the travel segment that follows this rest and leads to Town Y (60 km).

    After 1:30 pm, the cyclist resumes travel, reaching Town Y (60 km) at 2:30 pm.
    The period of travel to Town Y after his first rest is from 1:30 pm to 2:30 pm.
    The period is 1:30 pm to 2:30 pm\boxed{1:30 \text{ pm to } 2:30 \text{ pm}}.

Part iii

30 minutes30 \text{ minutes}

How many minutes did the cyclist spend at Town Y?

  1. Determine the arrival time at Town Y from the graph.

    The cyclist arrives at Town Y (60 km) at 2:30 pm.

  2. Determine the departure time from Town Y from the graph.

    The cyclist starts the return journey from Town Y (60 km) at 3 pm.

  3. Calculate the duration of the stop at Town Y.

    Time spent=3 pm−2:30 pm=30 minutes\text{Time spent} = 3 \text{ pm} - 2:30 \text{ pm} = 30 \text{ minutes}

    The cyclist spent 30 minutes\boxed{30 \text{ minutes}} at Town Y.

Part iv

20 km20 \text{ km}

What distance did the cyclist cover between 1pm and 2pm?

  1. Find the distance from Town X at 1 pm from the graph.

    At 1 pm, the distance from Town X is 35 km.

  2. Find the distance from Town X at 2 pm from the graph.

    At 2 pm, the distance from Town X is 55 km.

  3. Calculate the distance covered during this hour.

    Distance covered=55 km−35 km=20 km\text{Distance covered} = 55 \text{ km} - 35 \text{ km} = 20 \text{ km}

    The cyclist covered 20 km\boxed{20 \text{ km}} between 1 pm and 2 pm.

    6. b) Copy and complete the table below

    ValueNumber of significant figures
    142.4316
    0.0005623
    5.32 x 10^53
    648000028
    1.0072 x 10^25

    Explanation of significant figures:

    • 142.431: All non-zero digits are significant.
    • 0.000562: Leading zeros (before the first non-zero digit) are not significant. The digits 5, 6, and 2 are significant.
    • 5.32 x 10^5: In scientific notation, all digits in the mantissa (the number before the power of 10) are significant.
    • 64800002: All non-zero digits are significant, and zeros between non-zero digits are also significant.
    • 1.0072 x 10^2: All digits in the mantissa are significant, including the zeros between non-zero digits.

    6. c) Evaluate: 1−23−(34+2−58)1 - \frac{2}{3} - \left(\frac{3}{4} + 2 - \frac{5}{8}\right)

  4. Evaluate the expression inside the parenthesis first.

    34+2−58\frac{3}{4} + 2 - \frac{5}{8}

    Find a common denominator for 4, 1 (from 2=212 = \frac{2}{1}), and 8. The least common multiple is 8.

    3×24×2+2×81×8−58\frac{3 \times 2}{4 \times 2} + \frac{2 \times 8}{1 \times 8} - \frac{5}{8} 68+168−58\frac{6}{8} + \frac{16}{8} - \frac{5}{8} 6+16−58=22−58=178\frac{6 + 16 - 5}{8} = \frac{22 - 5}{8} = \frac{17}{8}

  5. Substitute this result back into the main expression.

    1−23−1781 - \frac{2}{3} - \frac{17}{8}

    Find a common denominator for 1, 3, and 8. The least common multiple is 24.

    1×2424−2×83×8−17×38×3\frac{1 \times 24}{24} - \frac{2 \times 8}{3 \times 8} - \frac{17 \times 3}{8 \times 3} 2424−1624−5124\frac{24}{24} - \frac{16}{24} - \frac{51}{24}

  6. Combine the fractions.

    24−16−5124\frac{24 - 16 - 51}{24} 8−5124\frac{8 - 51}{24} −4324\frac{-43}{24}

    The evaluated expression is −4324\boxed{-\frac{43}{24}}.

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