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Mathematics

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Question 2: Complete the table

Here is the completed table:

NumberNatural numberWhole numberIntegersRational numberIrrational number
2.1 49\sqrt{49}✓\checkmark✓\checkmark✓\checkmark✓\checkmark
2.2 −643\sqrt[3]{-64}✓\checkmark✓\checkmark
2.3 227\frac{22}{7}✓\checkmark
2.4 −2.4-2.4✓\checkmark
2.5 0×π0 \times \pi✓\checkmark✓\checkmark✓\checkmark

Explanation for each row:

  • 2.1 49\sqrt{49}: 49=7\sqrt{49} = 7. This is a positive whole number, so it is a natural number, whole number, integer, and rational number (71\frac{7}{1}).
  • 2.2 −643\sqrt[3]{-64}: −643=−4\sqrt[3]{-64} = -4. This is a negative whole number, so it is an integer and a rational number (−41\frac{-4}{1}). It is not a natural or whole number.
  • 2.3 227\frac{22}{7}: This is a fraction of two integers, so it is a rational number. It is not a natural number, whole number, or integer.
  • 2.4 −2.4-2.4: This is a terminating decimal, which can be written as a fraction −2410=−125\frac{-24}{10} = \frac{-12}{5}. Therefore, it is a rational number. It is not a natural number, whole number, or integer.
  • 2.5 0×π0 \times \pi: 0×π=00 \times \pi = 0. This is a whole number, an integer, and a rational number (01\frac{0}{1}). It is not a natural number.

Question 3.1: Use prime factorisation to determine LCM and HCF of 540 and 1800.

  1. Find the prime factorisation of 540.

    540=54×10=(2×27)×(2×5)=(2×33)×(2×5)=22×33×51540 = 54 \times 10 = (2 \times 27) \times (2 \times 5) = (2 \times 3^3) \times (2 \times 5) = 2^2 \times 3^3 \times 5^1

  2. Find the prime factorisation of 1800.

    1800=18×100=(2×9)×(102)=(2×32)×(2×5)2=(2×32)×(22×52)=23×32×521800 = 18 \times 100 = (2 \times 9) \times (10^2) = (2 \times 3^2) \times (2 \times 5)^2 = (2 \times 3^2) \times (2^2 \times 5^2) = 2^3 \times 3^2 \times 5^2

  3. Determine the HCF (Highest Common Factor).

    The HCF is the product of the common prime factors raised to the lowest power.

    HCF=2min⁡(2,3)×3min⁡(3,2)×5min⁡(1,2)HCF = 2^{\min(2,3)} \times 3^{\min(3,2)} \times 5^{\min(1,2)} HCF=22×32×51HCF = 2^2 \times 3^2 \times 5^1 HCF=4×9×5HCF = 4 \times 9 \times 5 HCF=36×5HCF = 36 \times 5 HCF=180HCF = 180

  4. Determine the LCM (Lowest Common Multiple).

    The LCM is the product of all prime factors raised to the highest power.

    LCM=2max⁡(2,3)×3max⁡(3,2)×5max⁡(1,2)LCM = 2^{\max(2,3)} \times 3^{\max(3,2)} \times 5^{\max(1,2)} LCM=23×33×52LCM = 2^3 \times 3^3 \times 5^2 LCM=8×27×25LCM = 8 \times 27 \times 25 LCM=216×25LCM = 216 \times 25 LCM=5400LCM = 5400

    The HCF of 540 and 1800 is 180\boxed{\text{180}}.
    The LCM of 540 and 1800 is 5400\boxed{\text{5400}}.

    Question 3.2: Alex and Thomas share 30 sweets. They divide them in the ratio 3:2. How many sweets does Thomas have?

  5. Find the total number of parts in the ratio.

    Total parts =3+2=5= 3 + 2 = 5 parts.

  6. Determine the value of one part.

    Value per part =Total sweetsTotal parts=305=6= \frac{\text{Total sweets}}{\text{Total parts}} = \frac{30}{5} = 6 sweets per part.

  7. Calculate Thomas's share.

    Thomas's share is 2 parts.
    Thomas's sweets =2×6=12= 2 \times 6 = 12 sweets.

    Thomas has 12 sweets\boxed{\text{12 sweets}}.

    Question 3.3: The number of mechanics versus the number of fixed cars in a day is shown in the table below.

    Number of mechanics24126
    Number of cars36189

    Question 3.3.1: Is this an example of direct or inverse proportion?

  8. Check for direct proportion by calculating the ratio Number of mechanicsNumber of cars\frac{\text{Number of mechanics}}{\text{Number of cars}}.

    For the first pair: 2436=23\frac{24}{36} = \frac{2}{3}
    For the second pair: 1218=23\frac{12}{18} = \frac{2}{3}
    For the third pair: 69=23\frac{6}{9} = \frac{2}{3}

    Since the ratio is constant, the relationship is a direct proportion.

    This is an example of direct proportion\boxed{\text{direct proportion}}.

    Question 3.3.2: Find the number of mechanics required to fix 108 cars.

  9. Use the constant ratio from the direct proportion.

    Let MM be the number of mechanics and CC be the number of cars.
    We found that MC=23\frac{M}{C} = \frac{2}{3}.

  10. Substitute the given number of cars (C=108C = 108) into the proportion.

    M108=23\frac{M}{108} = \frac{2}{3}

  11. Solve for MM.

    M=23×108M = \frac{2}{3} \times 108 M=2×1083M = 2 \times \frac{108}{3} M=2×36M = 2 \times 36 M=72M = 72

    The number of mechanics required to fix 108 cars is 72 mechanics\boxed{\text{72 mechanics}}.

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