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A motorist started a journey at 8:30 am, travelling at 120 km/hr and covered a distance of 240 km. Calculate the time taken and finish time. Given that a=4, b=2 and c=3. Find the value of abc and ac/b

Mathematics

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Answers

5 parts
a

2 hours\text{2 hours}

b

10:30 am\text{10:30 am}

b(iii)
16
11, 13, 15

Step-by-step solution

18 steps

Here are the solutions to the problems:

23. A motorist started a journey at 8:30 am, travelling at 120 km/hr and covered a distance of 240 km.

Part a· 3 steps

Calculate the time he took.

  1. Identify the given values and the formula.

    Speed=120 km/hr\text{Speed} = 120 \text{ km/hr} Distance=240 km\text{Distance} = 240 \text{ km} Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}
  2. Substitute the values into the formula.

    Time=240 km120 km/hr\text{Time} = \frac{240 \text{ km}}{120 \text{ km/hr}}
  3. Calculate the time.

    Time=2 hours\text{Time} = 2 \text{ hours}

    The time he took is 2 hours\boxed{\text{2 hours}}.

Part b

10:30 am\text{10:30 am}

At what time did he finish the journey?

  1. Identify the start time and the duration of the journey.

    Start time=8:30 am\text{Start time} = 8:30 \text{ am} Duration=2 hours\text{Duration} = 2 \text{ hours}
  2. Add the duration to the start time.

    8:30 am+2 hours=10:30 am8:30 \text{ am} + 2 \text{ hours} = 10:30 \text{ am}

    He finished the journey at 10:30 am\boxed{\text{10:30 am}}.

    24. Given that a=4a=4, b=2b=2 and c=3c=3. Find the value of:

Part b(i)24

abcabc

  1. Substitute the given values into the expression.

    abc=4×2×3abc = 4 \times 2 \times 3
  2. Multiply the numbers.

    abc=8×3abc = 8 \times 3 abc=24abc = 24

    The value of abcabc is 24\boxed{24}.

Part b(ii)6

acb\frac{ac}{b}

  1. Substitute the given values into the expression.

    acb=4×32\frac{ac}{b} = \frac{4 \times 3}{2}
  2. Multiply the numbers in the numerator.

    acb=122\frac{ac}{b} = \frac{12}{2}
  3. Divide the numbers.

    acb=6\frac{ac}{b} = 6

    The value of acb\frac{ac}{b} is 6\boxed{6}.

Part b(iii)16

a2a^2

  1. Substitute the given value of aa into the expression.

    a2=42a^2 = 4^2
  2. Calculate the square.

    a2=4×4a^2 = 4 \times 4 a2=16a^2 = 16

    The value of a2a^2 is 16\boxed{16}.

    25. The sum of three consecutive odd numbers is 39. Find the numbers.

  3. Let the first odd number be xx.

    The next consecutive odd number is x+2x+2.
    The third consecutive odd number is x+4x+4.

  4. Set up the equation based on the given sum.

    x+(x+2)+(x+4)=39x + (x+2) + (x+4) = 39
  5. Combine like terms.

    3x+6=393x + 6 = 39
  6. Subtract 6 from both sides of the equation.

    3x=39−63x = 39 - 6 3x=333x = 33
  7. Divide by 3 to solve for xx.

    x=333x = \frac{33}{3} x=11x = 11
  8. Find the three consecutive odd numbers.

    First number: x=11x = 11
    Second number: x+2=11+2=13x+2 = 11+2 = 13
    Third number: x+4=11+4=15x+4 = 11+4 = 15
    The three consecutive odd numbers are 11,13,15\boxed{11, 13, 15}.

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