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ScotlandApplications of MathematicsSyllabus dot point

How do you carry out efficient container packing, and use precedence tables to plan tasks and solve a time-management problem?

Carrying out efficient container packing to fit items into a space, and using precedence tables to plan tasks in order, find the minimum completion time and solve a time-management problem.

A focused answer to the SQA National 5 Applications of Mathematics measurement content on packing and planning, covering efficient container packing to fit items into a space, and using precedence tables to order tasks, find the minimum completion time and solve a time-management problem.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Efficient container packing
  3. Precedence tables and time management
  4. Examples in context
  5. Try this

What this dot point is asking

The SQA wants you to carry out efficient container packing, fitting items into a space and working out how many fit, and to use a precedence table to put tasks in order, find the minimum time to complete them, and solve a time-management problem.

Efficient container packing

Container packing asks how many smaller items fit inside a larger space. The efficient method fits items by lining up each dimension, not by dividing volumes, because items must fit whole.

Dividing the container volume by the box volume can overcount, because it ignores whether the boxes actually fit along each edge, so always work dimension by dimension. It is often worth trying more than one orientation: turning the boxes so a different side lies along the long edge of the container can leave less wasted space and fit more in. The best packing is the orientation that gives the largest whole-number product.

Precedence tables and time management

A precedence table lists each task, how long it takes, and which tasks must be finished before it can start. It is used to plan the order of work and find the shortest time to finish everything.

Examples in context

Packing and scheduling appear in logistics (loading a van or crate), event planning, cooking a meal with several dishes, and project work. Each rests on fitting items dimension by dimension, or ordering tasks so independent ones overlap to save time, the skills here. The efficiency gain from running tasks in parallel, or from choosing the best packing orientation, is exactly what the SQA asks you to find.

A longer chain of tasks is handled the same way: list each task with its predecessors, then track the earliest each can start. The minimum completion time is set by the longest chain of dependent tasks, sometimes called the critical path, because shortening any task off that chain does not finish the project sooner.

Try this

Q1. How many 1010 cm cubes fit in a box 30×20×1030 \times 20 \times 10 cm? [2 marks]

  • Cue. 3×2×1=63 \times 2 \times 1 = 6 cubes.

Q2. Tasks: X (1212 min, first); Y (66 min, after X); Z (99 min, after X), Y and Z independent. Find the minimum time. [3 marks]

  • Cue. 12+9=2112 + 9 = 21 min (Y and Z parallel, longer is Z).

Q3. Boxes 25×25×2025 \times 25 \times 20 cm in a crate 100×50×40100 \times 50 \times 40 cm: how many fit? [2 marks]

  • Cue. 4×2×2=164 \times 2 \times 2 = 16 boxes.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA N5 Apps style3 marksBoxes measuring 3030 cm by 2020 cm by 1515 cm are packed into a crate measuring 9090 cm by 6060 cm by 3030 cm. How many boxes fit if they are all the same way up?
Show worked answer →

Work out how many boxes fit along each dimension, matching the box side to the crate side. Along the 9090 cm: 90÷30=390 \div 30 = 3; along the 6060 cm: 60÷20=360 \div 20 = 3; along the 3030 cm: 30÷15=230 \div 15 = 2 (1 mark for the three divisions). Multiply: 3×3×2=183 \times 3 \times 2 = 18 boxes (1 mark). State the answer: 1818 boxes fit (1 mark). Markers reward dividing each crate dimension by the matching box dimension and multiplying. Counting by volume alone can overcount because boxes may not fit exactly.

SQA N5 Apps style3 marksA precedence table lists tasks A (2020 min, no predecessor), B (1515 min, after A), and C (1010 min, after A). B and C can run at the same time once A is done. Find the minimum time to complete all tasks.
Show worked answer →

Task A must finish first, taking 2020 minutes (1 mark). Once A is done, B and C can run together, so the time for that stage is the longer of the two, 1515 minutes (B), not 15+1015 + 10 (1 mark). Total minimum time: 20+15=3520 + 15 = 35 minutes (1 mark). Markers reward respecting the order from the table, running B and C in parallel, and the total. Adding every task time would give 4545 minutes, which ignores the parallel working.

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