Foundation June 2023 Paper 2 Q14
14 Packets of sweets are put into boxes.
Each packet is a cuboid, 80 mm by 60 mm by 20 mm.
Each box is a cuboid, 72 cm by 48 cm by 24 cm.
Work out the greatest number of packets that can be put into each box. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 864 | B1 | (indep) for process to convert to common units, eg \(72 \times 10\ (= 720)\) or \(48 \times 10\ (= 480)\) or \(24 \times 10\ (= 240)\) or \(80 \div 10\ (= 8)\) or \(60 \div 10\ (= 6)\) or \(20 \div 10\ (= 2)\) or or \(\text{``}96000\text{''} \div 10^3\ (= 96)\) or \(\text{``}82944\text{''} \times 10^3\ (= 82944000)\) or \(\text{``}0.864\text{''} \times 10^3\) |
| P1 | for using volume, eg \(80 \times 60 \times 20\ (= 96000)\) or \(72 \times 48 \times 24\ (= 82944)\) or for start of process to find number of packets using one dimension, eg \(\text{``}720\text{''} \div 80\ (= 9)\) or \(\text{``}480\text{''} \div 60\ (= 8)\) or \(\text{``}240\text{''} \div 20\ (= 12)\) or \(48 \div \text{``}8\text{''}\ (= 6)\) or \(72 \div \text{``}2\text{''}\ (= 36)\) or \(24 \div \text{``}6\text{''}\ (= 4)\) | |
| P1 | for full process with or without unit conversion, eg \(\text{``}82\,944\,000\text{''} \div \text{``}96000\text{''}\) or \(\text{``}82944\text{''} \div \text{``}96\text{''}\) or for \(\text{``}9\text{''} \times \text{``}8\text{''} \times \text{``}12\text{''}\) or \(\text{``}36\text{''} \times \text{``}6\text{''} \times \text{``}4\text{''}\) or \(\text{``}0.9\text{''} \times \text{``}0.8\text{''} \times \text{``}1.2\text{''}\ (= 0.864)\) | |
| A1 | cao |
Additional guidance
This mark can be awarded at any stage
One correct conversion for their method is enough for the award of this mark
Working may be seen on diagram.
May be implied by correctly dividing the areas of corresponding faces
\(9 \times 24 \times 4\)
\(12 \times 24 \times 3\)
\(12 \times 6 \times 12\)
\(36 \times 8 \times 3\)
Note sight of digits 864 with decimal point and/or extra zeros scores P2