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What is the volume of the water tank in cubic metres? Three bells ring at intervals of 3 minutes, 6 minutes and 9 minutes respectively. After how long will they ring again together?

Mathematics

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Answer

2496 m32496 \, \text{m}^3

18 minutes18 \, \text{minutes}

155 cm2155 \, \text{cm}^2

Step-by-step solution

11 steps

: The tank below was used to harvest water in Embakasi South Constituency. What is the volume of the water tank in cubic metres?

  1. Identify the dimensions of the rectangular tank from the image.

    Length (LL) = 24 m24 \, \text{m}
    Width (WW) = 13 m13 \, \text{m}
    Height (HH) = 8 m8 \, \text{m}

  2. Use the formula for the volume of a rectangular prism.

    V=L×W×HV = L \times W \times H
  3. Substitute the values and calculate the volume.

    V=24 m×13 m×8 mV = 24 \, \text{m} \times 13 \, \text{m} \times 8 \, \text{m} V=312 m2×8 mV = 312 \, \text{m}^2 \times 8 \, \text{m} V=2496 m3V = 2496 \, \text{m}^3

    The volume of the water tank is 2496 m3\boxed{2496 \, \text{m}^3}.

    : Three bells ring at intervals of 3 minutes, 6 minutes and 9 minutes respectively. After how long will they ring again together?

  4. To find when the bells will ring together again, calculate the Least Common Multiple (LCM) of their ringing intervals

    3, 6, and 9 minutes.

  5. Find the prime factorization of each number.

    3=313 = 3^1 6=21×316 = 2^1 \times 3^1 9=329 = 3^2
  6. Determine the LCM by taking the highest power of all prime factors present.

    LCM(3,6,9)=21×32\text{LCM}(3, 6, 9) = 2^1 \times 3^2 LCM(3,6,9)=2×9\text{LCM}(3, 6, 9) = 2 \times 9 LCM(3,6,9)=18\text{LCM}(3, 6, 9) = 18

    The bells will ring together again after 18 minutes\boxed{18 \, \text{minutes}}.

    : A rectangular garden in the shape below was fenced leaving a pathway. Calculate the shaded area.

  7. Identify the dimensions of the outer rectangle.

    Outer Length (LoL_o) = 15 cm15 \, \text{cm}
    Outer Width (WoW_o) = 13 cm13 \, \text{cm}

  8. Calculate the area of the outer rectangle.

    Areaouter=Lo×Wo\text{Area}_{outer} = L_o \times W_o Areaouter=15 cm×13 cm\text{Area}_{outer} = 15 \, \text{cm} \times 13 \, \text{cm} Areaouter=195 cm2\text{Area}_{outer} = 195 \, \text{cm}^2
  9. Identify the dimensions of the inner (unshaded) rectangle.

    Inner Length (LiL_i) = 10 cm10 \, \text{cm}
    Inner Width (WiW_i) = 4 cm4 \, \text{cm}

  10. Calculate the area of the inner rectangle.

    Areainner=Li×Wi\text{Area}_{inner} = L_i \times W_i Areainner=10 cm×4 cm\text{Area}_{inner} = 10 \, \text{cm} \times 4 \, \text{cm} Areainner=40 cm2\text{Area}_{inner} = 40 \, \text{cm}^2
  11. Calculate the shaded area by subtracting the inner area from the outer area.

    Shaded Area=Areaouter−Areainner\text{Shaded Area} = \text{Area}_{outer} - \text{Area}_{inner} Shaded Area=195 cm2−40 cm2\text{Shaded Area} = 195 \, \text{cm}^2 - 40 \, \text{cm}^2 Shaded Area=155 cm2\text{Shaded Area} = 155 \, \text{cm}^2

    The shaded area is 155 cm2\boxed{155 \, \text{cm}^2}.

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