Higher June 2017 Paper 3 Q7
7
(a) The length of a pipe is 6 metres to the nearest metre.
Complete the error interval for the length of the pipe. [2 marks]
Answer \(\;\ldots\ldots\ldots\;\) m \(\;\leqslant\;\) length \(\;\lt\;\) \(\;\ldots\ldots\ldots\;\) m
(b) The length of a different pipe is 4 metres to the nearest metre.
Olly says,
“The total length of the two pipes is 11 metres to the nearest metre.”
Give an example to show that he could be correct. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| 5.5 in the correct position | B1 | oe |
| 6.5 in the correct position | B1 | oe |
Additional guidance
| 5.50 or \(5\dfrac{1}{2}\) or \(\dfrac{11}{2}\) | B1 |
| 6.50 or \(6\dfrac{1}{2}\) or \(\dfrac{13}{2}\) | B1 |
| Answer | Mark | Comments |
|---|---|---|
| One correctly evaluated trial using (6, 6.5] + (4, 4.5) or (6, 6.5) + (4, 4.5] or two values in the ranges given that work if correctly evaluated | M1 | eg 6.3 + 4.1 = 10.4 eg 6.4, 4.2 |
| One correctly evaluated trial using (6, 6.5) + (4, 4.5) with an answer that rounds to 11 | A1 | eg 6.4 + 4.2 = 10.6 Ignore fw |
Additional guidance
| 6.4 + 4.4 = 10.8 (= 11) do not need to show 11 | M1A1 |
| 6.4999 + 4.4999 = 10.9998 | M1A1 |
| 6.5 + 4.4 = 10.9 | M1A0 |
| 4.5 + 6.2 = 10.7 | M1A0 |
| 6 + 4 = 10 | M0 |
| 6.5 + 4.5 = 11 | M0 |
| \(6.4\dot{9} + 4.4\dot{9} = 11\) | M0 |