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Express 48 and 60 as a product of their prime factors. Hence, find the LCM and GCD of 48 and 60.

Mathematics

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3 parts
b
12
240
c

70 sweets70 \text{ sweets}

d

96π cm296\pi \text{ cm}^2

Step-by-step solution

9 steps

Here are the solutions to the questions from the image:

1. a) Express 48 and 60 as a product of their prime factors.

Step 1: Find the prime factors of 48.

48=2×2448 = 2 \times 24 =2×2×12= 2 \times 2 \times 12 =2×2×2×6= 2 \times 2 \times 2 \times 6 =2×2×2×2×3= 2 \times 2 \times 2 \times 2 \times 3 =24×31= 2^4 \times 3^1

Step 2: Find the prime factors of 60.

60=2×3060 = 2 \times 30 =2×2×15= 2 \times 2 \times 15 =2×2×3×5= 2 \times 2 \times 3 \times 5 =22×31×51= 2^2 \times 3^1 \times 5^1

The prime factorization of 48 is 24×3\boxed{2^4 \times 3} and for 60 is 22×3×5\boxed{2^2 \times 3 \times 5}.

Part b· 3 steps

Hence, find the LCM and GCD of 48 and 60.

  1. Use the prime factorizations from part (a).

    48=24×31×5048 = 2^4 \times 3^1 \times 5^0
    60=22×31×5160 = 2^2 \times 3^1 \times 5^1

  2. Find the Greatest Common Divisor (GCD).

    For GCD, take the lowest power of each common prime factor.
    Common prime factors are 2 and 3.
    Lowest power of 2 is 222^2.
    Lowest power of 3 is 313^1.

    GCD(48,60)=22×31=4×3=12GCD(48, 60) = 2^2 \times 3^1 = 4 \times 3 = 12
  3. Find the Least Common Multiple (LCM).

    For LCM, take the highest power of all prime factors present in either number.
    Highest power of 2 is 242^4.
    Highest power of 3 is 313^1.
    Highest power of 5 is 515^1.

    LCM(48,60)=24×31×51=16×3×5=240LCM(48, 60) = 2^4 \times 3^1 \times 5^1 = 16 \times 3 \times 5 = 240

    The GCD of 48 and 60 is 12\boxed{12} and the LCM is 240\boxed{240}.

Part c

70 sweets70 \text{ sweets}

A packet of sweets is shared between Amina, Pauline and Sharon in the ratio 2:3:5. If Pauline received 42 sweets, how many did Sharon receive?

  1. Determine the value of one ratio part.

    The ratio for Amina : Pauline : Sharon is 2:3:52:3:5.
    Pauline's share corresponds to 3 parts of the ratio.
    Given that Pauline received 42 sweets, we can find the value of one part:

    3 parts=42 sweets3 \text{ parts} = 42 \text{ sweets} 1 part=423=14 sweets1 \text{ part} = \frac{42}{3} = 14 \text{ sweets}
  2. Calculate Sharon's share.

    Sharon's share corresponds to 5 parts of the ratio.

    Sharon’s sweets=5 parts×14 sweets/part\text{Sharon's sweets} = 5 \text{ parts} \times 14 \text{ sweets/part} Sharon’s sweets=70 sweets\text{Sharon's sweets} = 70 \text{ sweets}

    Sharon received 70 sweets\boxed{70 \text{ sweets}}.

Part d

96π cm296\pi \text{ cm}^2

A right circular cone has a base diameter of 12 cm and a height of 8 cm. Find the total surface area.

  1. Determine the radius and calculate the slant height.

    The base diameter is 12 cm12 \text{ cm}, so the radius r=122=6 cmr = \frac{12}{2} = 6 \text{ cm}.
    The height h=8 cmh = 8 \text{ cm}.
    The slant height ll can be found using the Pythagorean theorem:

    l=r2+h2l = \sqrt{r^2 + h^2} l=62+82l = \sqrt{6^2 + 8^2} l=36+64l = \sqrt{36 + 64} l=100l = \sqrt{100} l=10 cml = 10 \text{ cm}
  2. Calculate the area of the base.

    The area of the circular base is Abase=πr2A_{base} = \pi r^2.

    Abase=π×(6)2A_{base} = \pi \times (6)^2 Abase=36π cm2A_{base} = 36\pi \text{ cm}^2
  3. Calculate the lateral surface area.

    The lateral surface area is Alateral=πrlA_{lateral} = \pi r l.

    Alateral=π×6×10A_{lateral} = \pi \times 6 \times 10 Alateral=60π cm2A_{lateral} = 60\pi \text{ cm}^2
  4. Calculate the total surface area.

    The total surface area (TSA) is the sum of the base area and the lateral surface area.

    TSA=Abase+AlateralTSA = A_{base} + A_{lateral} TSA=36π+60πTSA = 36\pi + 60\pi TSA=96π cm2TSA = 96\pi \text{ cm}^2

    The total surface area is 96π cm2\boxed{96\pi \text{ cm}^2}.

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