Higher June 2023 Paper 1R Q11
11 The diagram shows a block of iron in the shape of a cuboid.
Diagram NOT accurately drawn
The block has length \(w\) cm, width 5 cm and height 4 cm
The density of iron is 7.8 g/cm3
The mass of the block is 1950 g
Work out the value of \(w\)
(3)
| Scheme | Marks |
|---|---|
| \((V =)\;\dfrac{1950}{7.8}\;(= 250)\) or \(7.8 = \dfrac{1950}{w \times 5 \times 4}\) or \(7.8 = \dfrac{1950}{w \times 20}\) | M1 |
eg \(w = \dfrac{1950}{7.8 \times 5 \times 4}\) or \(20w = \dfrac{1950}{7.8}\) or \(20w\) = “250” or 4 × 5 × \(w\) = “250” OR eg \(\dfrac{1950}{5 \times 4 \times 7.8}\) or 1950 ÷ (20 × 7.8) or 1950 ÷ 156 or “250” ÷ 20 | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 12.5 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: for correct method to find volume using mass ÷ density or a correct equation with correct expression for volume
(may be embedded in another calculation)
M1: for a fully correct equation in \(w\)
or
a fully correct calculation to find the value of \(w\) (may be labelled eg \(x\) or \(L\))