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A journey consists of two parts. The first part is 180 km at 60 km/h and the second part is 150 km at 50 km/h. Calculate (i) the total time for the journey and (ii) the average speed for the entire jo

Mathematics

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c)

i) The total time for the journey

  1. Calculate the time for the first part of the journey.

    Distance =180= 180 km
    Speed =60= 60 km/h
    Time =DistanceSpeed= \frac{\text{Distance}}{\text{Speed}}

    Time1=180 km60 km/h=3 hours\text{Time}_1 = \frac{180 \text{ km}}{60 \text{ km/h}} = 3 \text{ hours}

  2. Calculate the time for the second part of the journey.

    Distance =150= 150 km
    Speed =50= 50 km/h

    Time2=150 km50 km/h=3 hours\text{Time}_2 = \frac{150 \text{ km}}{50 \text{ km/h}} = 3 \text{ hours}

  3. Calculate the total time for the journey.

    Total Time=Time1+Time2=3 hours+3 hours=6 hours\text{Total Time} = \text{Time}_1 + \text{Time}_2 = 3 \text{ hours} + 3 \text{ hours} = 6 \text{ hours}

    The total time for the journey is 6 hours\boxed{6 \text{ hours}}.

    ii) The average speed for the entire journey

  4. Calculate the total distance for the journey.

    Total Distance=180 km+150 km=330 km\text{Total Distance} = 180 \text{ km} + 150 \text{ km} = 330 \text{ km}

  5. Calculate the average speed.

    Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} Average Speed=330 km6 hours=55 km/h\text{Average Speed} = \frac{330 \text{ km}}{6 \text{ hours}} = 55 \text{ km/h}

    The average speed for the entire journey is 55 km/h\boxed{55 \text{ km/h}}.

    a)

  6. Find the least common multiple (LCM) of the denominators 6, 12, and 4.

    The LCM of 6, 12, and 4 is 12.

  7. Convert each fraction to an equivalent fraction with a denominator of 12.

    56=5×26×2=1012\frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12} 712\frac{7}{12} 14=1×34×3=312\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}

  8. Perform the addition and subtraction.

    1012+712−312=10+7−312=17−312=1412\frac{10}{12} + \frac{7}{12} - \frac{3}{12} = \frac{10 + 7 - 3}{12} = \frac{17 - 3}{12} = \frac{14}{12}

  9. Simplify the fraction.

    1412=14÷212÷2=76\frac{14}{12} = \frac{14 \div 2}{12 \div 2} = \frac{7}{6}

    The simplified expression is 76\boxed{\frac{7}{6}}.

    b)

  10. Determine the total number of ratio parts.

    Ratio of boys to girls is 7:87:8.
    Total ratio parts =7+8=15= 7 + 8 = 15.

  11. Calculate the value of one ratio part.

    Total learners =60= 60.
    Value of one part =Total learnersTotal ratio parts=6015=4= \frac{\text{Total learners}}{\text{Total ratio parts}} = \frac{60}{15} = 4.

  12. Calculate the number of boys.

    Number of boys =7×4=28= 7 \times 4 = 28.

  13. Calculate the number of girls.

    Number of girls =8×4=32= 8 \times 4 = 32.
    There are 28 boys\boxed{28 \text{ boys}} and 32 girls\boxed{32 \text{ girls}}.

    c)
    Given: radius r=3r = 3 m, height h=5h = 5 m, π=3.142\pi = 3.142.

    i) The volume of the tank

  14. Use the formula for the volume of a cylinder, V=πr2hV = \pi r^2 h.

    V=3.142×(3 m)2×5 mV = 3.142 \times (3 \text{ m})^2 \times 5 \text{ m} V=3.142×9 m2×5 mV = 3.142 \times 9 \text{ m}^2 \times 5 \text{ m} V=3.142×45 m3V = 3.142 \times 45 \text{ m}^3 V=141.39 m3V = 141.39 \text{ m}^3

    The volume of the tank is 141.39 m3\boxed{141.39 \text{ m}^3}.

    ii) The surface area of the tank

  15. Use the formula for the surface area of a closed cylinder, A=2πr(r+h)A = 2\pi r (r+h).

    A=2×3.142×3 m×(3 m+5 m)A = 2 \times 3.142 \times 3 \text{ m} \times (3 \text{ m} + 5 \text{ m}) A=2×3.142×3 m×8 mA = 2 \times 3.142 \times 3 \text{ m} \times 8 \text{ m} A=6.284×24 m2A = 6.284 \times 24 \text{ m}^2 A=150.816 m2A = 150.816 \text{ m}^2

    The surface area of the tank is 150.816 m2\boxed{150.816 \text{ m}^2}.

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