NavGanitMateby NavGreen Compute, AustraliaBased on Victorian Curriculum Mathematics v2.0
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Volume and capacity of prisms

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A library in Melbourne is designing a new rectangular prism-shaped storage unit for archival documents. The internal dimensions of the unit are 130 cm in length, 56 cm in width, and 45 cm in height.

The unit is to be filled completely with archival boxes, each with a volume of 1.35 litres.

What is the exact number of archival boxes that can fit inside the storage unit, assuming no space is wasted between boxes? (Note: 1 litre = 1000 cm^3)

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 storage unit is a rectangular prism with internal dimensions: length = 130 cm, width = 56 cm, height = 45 cm.
  2. The volume of a rectangular prism is calculated using the formula: volume = length × width × height.
  3. Substitute the given values: volume = 130 cm × 56 cm × 45 cm.
  4. First, compute 130 × 56: 130 × 56 = (100 × 56) + (30 × 56) = 5600 + 1680 = 7280 cm².
  5. Now multiply 7280 cm² by 45 cm: 7280 × 45.
  6. Break 45 into 40 + 5: 7280 × 40 = 291,200; 7280 × 5 = 36,400. Add: 291,200 + 36,400 = 327,600 cm³.
  7. The volume of one archival box is 1.35 litres. Convert litres to cm³ using the given conversion: 1 litre = 1000 cm³.
  8. So, 1.35 litres = 1.35 × 1000 = 1350 cm³.
  9. To find how many boxes fit, divide the total volume of the unit by the volume of one box: 327,600 cm³ ÷ 1350 cm³ per box.
  10. Perform the division: 327,600 ÷ 1350. First, simplify by dividing both numerator and denominator by 30: 327,600 ÷ 30 = 10,920; 1350 ÷ 30 = 45. So, 10,920 ÷ 45.
  11. Now divide 10,920 by 45: 45 × 240 = 10,800; 10,920 − 10,800 = 120. 120 ÷ 45 = 2.666..., so 240 + 2.666... = 242.666... But since we need an exact integer count and no space is wasted, check if 1350 × 242 = 327,600.
  12. Calculate 1350 × 242: 1350 × 200 = 270,000; 1350 × 40 = 54,000; 1350 × 2 = 2,700. Add: 270,000 + 54,000 = 324,000; 324,000 + 2,700 = 326,700. This is less than 327,600. Try 1350 × 243: 326,700 + 1350 = 328,050, which exceeds 327,600. So 242 boxes give 326,700 cm³, leaving 900 cm³ unused. But 327,600 ÷ 1350 = 242.666... which is not an integer. However, the problem states 'assuming no space is wasted between boxes', so the number must be an integer. Therefore, only 242 boxes can fit exactly without exceeding volume. But 242 × 1350 = 326,700, and 327,600 − 326,700 = 900 cm³ unused. Since 900 < 1350, a 243rd box cannot fit. Thus, the maximum number of boxes is 242.

Reference answer: 242

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