Higher June 2025 Paper 2 Q5
5
(a) The length of a fence panel is 160 cm to the nearest 10 cm
Complete the error interval. [2 marks]
\(\ldots\ldots\ldots\ \text{cm} \leqslant \text{length} \lt \ldots\ldots\ldots\ \text{cm}\)
(b) A different fence panel measures 2 metres to the nearest 10 cm
Kim says that the total length of three of these fence panels must be less than 6.2 m
Show that Kim is correct. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 155 cm \(\leqslant\) length \(\lt\) 165 cm | B2 | oe B1 155 or 165 in correct position SC1 165 cm \(\leqslant\) length \(\lt\) 155 cm |
Additional guidance
Accept \(164.\dot{9}\) for 165
Accept eg 155.0 for 155
| Answer | Mark | Comments |
|---|---|---|
| 2.05 or \(6.2 \div 3\) or \(2.0\dot{6}\) | M1 | oe eg in cm |
| \(2.05 \times 3\) and 6.15 or \(205 \times 3\) and 6.15 or \(205 \times 3\) and 615 and 620 or \(2.0\dot{6}\) and 2.05 or \(206.\dot{6}\) and 205 | A1 | accept eg \(2.05 + 2.05 + 2.05\) for \(2.05 \times 3\) |
Additional guidance
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Accept rounding or truncating of \(2.0\dot{6}\) or \(206.\dot{6}\) to at least 4 sf Accept rounding or truncating of \(2.0\dot{6}\) to 3 sf only if \(6.2 \div 3\) is seen Accept rounding or truncating of \(206.\dot{6}\) to 3 sf only if \(620 \div 3\) is seen | |
| Accept \(2.04\dot{9}\) for 2.05 | |
| Ignore any reference to lower bounds | |
| Ignore reference to units | |
| 2.05 or 205 may be embedded eg \(2.05 + 2.06 + 2.09 = 6.2\) with no further work or explanation | M1A0 |
| \(2.06 + 2.06 + 2.08 = 6.2\) with no further work or explanation \(2.06 + 2.06 + 2.06 = 6.18\) with a clear reference to 2.05 and a full explanation | M0A0 M1A1 |