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What is the length of a ruler in metres if it is 100 centimetres long?

Mathematics

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8 parts
10

1 m\text{1 m}

11

2300 kg\text{2300 kg}

12

2415000 mL\text{2415000 mL}

13

18.25 kg\text{18.25 kg}

14

7805 mL or 7 L 805 mL\text{7805 mL or 7 L 805 mL}

15

4000 kilograms\text{4000 kilograms}

16

4466 mL\text{4466 mL}

17

209320 g\text{209320 g}

Step-by-step solution

18 steps

Part 10· 2 steps

The length of a ruler is 100 centimetres. What is this length in metres?

  1. Recall the conversion factor.

    There are 100 centimetres in 1 metre.

    1 m=100 cm1 \text{ m} = 100 \text{ cm}
  2. Convert centimetres to metres.

    To convert 100 centimetres to metres, divide by 100.

    100 cm×1 m100 cm=100100 m100 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{100}{100} \text{ m} =1 m= 1 \text{ m}

    The length of the ruler in metres is 1 m\boxed{\text{1 m}}.

Part 11

2300 kg\text{2300 kg}

A container weighs 2 tonnes and 300 kilograms. Convert this mass into kilograms.

  1. Recall the conversion factor for tonnes to kilograms.

    There are 1000 kilograms in 1 tonne.

    1 tonne=1000 kg1 \text{ tonne} = 1000 \text{ kg}
  2. Convert tonnes to kilograms.

    2 tonnes=2×1000 kg=2000 kg2 \text{ tonnes} = 2 \times 1000 \text{ kg} = 2000 \text{ kg}
  3. Add the remaining kilograms.

    Total mass = 2000 kg+300 kg2000 \text{ kg} + 300 \text{ kg}

    =2300 kg= 2300 \text{ kg}

    The total mass of the container in kilograms is 2300 kg\boxed{\text{2300 kg}}.

Part 12

2415000 mL\text{2415000 mL}

A factory produces 345 litres of sunflower oil every day. How many millilitres of oil will the factory produce in one week?

  1. Recall the conversion factor for litres to millilitres.

    There are 1000 millilitres in 1 litre.

    1 L=1000 mL1 \text{ L} = 1000 \text{ mL}
  2. Calculate daily production in millilitres.

    345 L=345×1000 mL=345000 mL345 \text{ L} = 345 \times 1000 \text{ mL} = 345000 \text{ mL}
  3. Calculate weekly production.

    There are 7 days in a week.
    Weekly production = Daily production in mL ×\times 7 days

    =345000 mL/day×7 days= 345000 \text{ mL/day} \times 7 \text{ days} =2415000 mL= 2415000 \text{ mL}

    The factory will produce 2415000 mL\boxed{\text{2415000 mL}} of oil in one week.

Part 13

18.25 kg\text{18.25 kg}

A maize sack weighs 18250 grams. Convert this mass into kilograms.

  1. Recall the conversion factor for grams to kilograms.

    There are 1000 grams in 1 kilogram.

    1 kg=1000 g1 \text{ kg} = 1000 \text{ g}
  2. Convert grams to kilograms.

    To convert grams to kilograms, divide by 1000.

    18250 g×1 kg1000 g=182501000 kg18250 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = \frac{18250}{1000} \text{ kg} =18.25 kg= 18.25 \text{ kg}

    The mass of the maize sack in kilograms is 18.25 kg\boxed{\text{18.25 kg}}.

Part 14

7805 mL or 7 L 805 mL\text{7805 mL or 7 L 805 mL}

Two cows produce milk as follows: the first cow produces 2 litres and 353 millilitres and the second cow produces 5 litres and 452 millilitres. Find the total volume of milk produced by the two cows.

  1. Convert all volumes to a common unit (millilitres).

    Recall that 1 L=1000 mL1 \text{ L} = 1000 \text{ mL}.

    First cow's production:

    2 L=2×1000 mL=2000 mL2 \text{ L} = 2 \times 1000 \text{ mL} = 2000 \text{ mL}

    Total for first cow = 2000 mL+353 mL=2353 mL2000 \text{ mL} + 353 \text{ mL} = 2353 \text{ mL}

    Second cow's production:

    5 L=5×1000 mL=5000 mL5 \text{ L} = 5 \times 1000 \text{ mL} = 5000 \text{ mL}

    Total for second cow = 5000 mL+452 mL=5452 mL5000 \text{ mL} + 452 \text{ mL} = 5452 \text{ mL}

  2. Add the total volumes from both cows.

    Total volume = 2353 mL+5452 mL2353 \text{ mL} + 5452 \text{ mL}

    =7805 mL= 7805 \text{ mL}

    Alternatively, we can add litres and millilitres separately:
    Total litres = 2 L+5 L=7 L2 \text{ L} + 5 \text{ L} = 7 \text{ L}
    Total millilitres = 353 mL+452 mL=805 mL353 \text{ mL} + 452 \text{ mL} = 805 \text{ mL}
    Total volume = 7 L 805 mL7 \text{ L } 805 \text{ mL}

    The total volume of milk produced by the two cows is 7805 mL or 7 L 805 mL\boxed{\text{7805 mL or 7 L 805 mL}}.

Part 15

4000 kilograms\text{4000 kilograms}

Which measurement is the largest among 7500 grams, 3 tonnes, 4000 kilograms, and 2350 milligrams? Give reasons for your answer.

  1. Convert all measurements to a common unit, such as kilograms (kg).

    Recall the conversion factors:
    1 kg=1000 g1 \text{ kg} = 1000 \text{ g}
    1 tonne=1000 kg1 \text{ tonne} = 1000 \text{ kg}
    1 g=1000 mg1 \text{ g} = 1000 \text{ mg} (so 1 kg=1000×1000 mg=1000000 mg1 \text{ kg} = 1000 \times 1000 \text{ mg} = 1000000 \text{ mg})

    • 7500 grams (g) to kilograms (kg):

      7500 g×1 kg1000 g=7.5 kg7500 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 7.5 \text{ kg}
    • 3 tonnes to kilograms (kg):

      3 tonnes×1000 kg1 tonne=3000 kg3 \text{ tonnes} \times \frac{1000 \text{ kg}}{1 \text{ tonne}} = 3000 \text{ kg}
    • 4000 kilograms (kg): This is already in kilograms.

      4000 kg4000 \text{ kg}
    • 2350 milligrams (mg) to kilograms (kg):

      2350 mg×1 g1000 mg×1 kg1000 g=23501000000 kg=0.00235 kg2350 \text{ mg} \times \frac{1 \text{ g}}{1000 \text{ mg}} \times \frac{1 \text{ kg}}{1000 \text{ g}} = \frac{2350}{1000000} \text{ kg} = 0.00235 \text{ kg}
  2. Compare the values in kilograms.

    We have:
    7.5 kg7.5 \text{ kg}
    3000 kg3000 \text{ kg}
    4000 kg4000 \text{ kg}
    0.00235 kg0.00235 \text{ kg}

  3. Identify the largest measurement.

    Comparing these values, 4000 kg4000 \text{ kg} is the largest.

    The largest measurement is 4000 kilograms\boxed{\text{4000 kilograms}}.
    Reason: When all measurements are converted to a common unit (kilograms), 4000 kg is the largest value (7.5 kg7.5 \text{ kg}, 3000 kg3000 \text{ kg}, 4000 kg4000 \text{ kg}, 0.00235 kg0.00235 \text{ kg}).

Part 16

4466 mL\text{4466 mL}

A bucket of water has a volume of 10 litres. If 5534 millilitres of water have been added to the bucket, how many millilitres of water are needed to fill the bucket?

  1. Convert the total volume of the bucket to millilitres.

    Recall that 1 L=1000 mL1 \text{ L} = 1000 \text{ mL}.
    Total bucket volume = 10 L=10×1000 mL=10000 mL10 \text{ L} = 10 \times 1000 \text{ mL} = 10000 \text{ mL}

  2. Subtract the amount of water already added from the total volume.

    Water needed = Total bucket volume - Water added

    =10000 mL−5534 mL= 10000 \text{ mL} - 5534 \text{ mL} =4466 mL= 4466 \text{ mL}

    4466 mL\boxed{\text{4466 mL}} of water are needed to fill the bucket.

Part 17

209320 g\text{209320 g}

A trader purchased products with the following masses: 76320 grams of flour, 61000 grams of rice, and 72000 grams of salt. What is the total mass of the products purchased?

  1. Add the masses of all the products.

    All masses are already in grams, so we can add them directly.
    Total mass = Mass of flour + Mass of rice + Mass of salt

    =76320 g+61000 g+72000 g= 76320 \text{ g} + 61000 \text{ g} + 72000 \text{ g} =209320 g= 209320 \text{ g}

    The total mass of the products purchased is 209320 g\boxed{\text{209320 g}}.

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