NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Volume and capacity of prisms

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A library in Adelaide stores rectangular prism-shaped wooden crates for bookshelves. Each crate has internal dimensions of 91 cm in length, 41 cm in width, and 30 cm in height.

The crates are to be filled with archival boxes that each occupy a volume of 1.2 litres.

What is the exact number of archival boxes that can fit inside one crate, assuming no space is wasted between boxes? (Note: 1 litre = 1000 cm³)

Concept illustrations

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A rectangular prism. A prism of length 4 cm, width 3 cm and height 2 cm has volume 24 cm³ = 24 mL. The perspective drawing is schematic.4 cm3 cm2 cm
A prism of length 4 cm, width 3 cm and height 2 cm has volume 24 cm³ = 24 mL. The perspective drawing is schematic.
Square the length conversion factor. A 1 m by 1 m square is 100 cm by 100 cm. Its area is 10,000 cm². The 10-by-10 grid shows 100 squares, each 10 cm by 10 cm.1 m = 100 cm1 m100 cm
A 1 m by 1 m square is 100 cm by 100 cm. Its area is 10,000 cm². The 10-by-10 grid shows 100 squares, each 10 cm by 10 cm.

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Explanation

Worked solution and answer (click to reveal)

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Worked solution

  1. The crate is a rectangular prism with internal dimensions: length = 91 cm, width = 41 cm, height = 30 cm.
  2. The volume of a rectangular prism is calculated using the formula: volume = length × width × height.
  3. Substitute the given dimensions: volume = 91 cm × 41 cm × 30 cm.
  4. First, compute 91 × 41: 91 × 40 = 3640, and 91 × 1 = 91; so 3640 + 91 = 3731.
  5. Now multiply 3731 × 30: 3731 × 3 = 11,193, then ×10 = 111,930 cm³.
  6. The volume of one archival box is 1.2 litres. Since 1 litre = 1000 cm³, convert 1.2 litres to cm³: 1.2 × 1000 = 1200 cm³.
  7. To find how many boxes fit, divide the crate's volume by the box's volume: 111,930 cm³ ÷ 1200 cm³ per box.
  8. Perform the division: 111,930 ÷ 1200. First, simplify by dividing both numerator and denominator by 30: 111,930 ÷ 30 = 3731, 1200 ÷ 30 = 40. So now: 3731 ÷ 40.
  9. Now divide 3731 by 40: 40 × 93 = 3720, and 3731 − 3720 = 11. So 3731 ÷ 40 = 93 + 11/40 = 93.275.
  10. Since boxes must be whole and no space is wasted, only complete boxes fit. Therefore, the maximum number of boxes is 93.
  11. Check: 93 boxes × 1200 cm³ = 111,600 cm³. The crate has 111,930 cm³, so 111,930 − 111,600 = 330 cm³ remains, which is less than one box (1200 cm³), so 93 is correct.
  12. The exact number of archival boxes that can fit is 93.

Reference answer: 93

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