Higher January 2021 Paper 2R Q10
10 A mathematics teacher at a school asked a group of students how far, in kilometres, each student had travelled to get to school that day.
The table gives information about their answers.
| Distance travelled (\(d\) km) | Number of students |
|---|---|
| \(0 \lt d \leqslant 2\) | \(x\) |
| \(2 \lt d \leqslant 4\) | 11 |
| \(4 \lt d \leqslant 6\) | 8 |
| \(6 \lt d \leqslant 8\) | 6 |
| \(8 \lt d \leqslant 10\) | 5 |
The teacher calculated that an estimate for the mean distance travelled by the whole group of students was 4.25 km.
Work out the value of \(x\).
Show your working clearly.
(4)
| Scheme | Marks |
|---|---|
| (11 × 3) + (8 × 5) + (6 × 7) + (5 × 9) (= 160) (= 33 + 40 + 42 + 45 = 160) | M1 |
“160” + \(x\) = 4.25 × (11 + 8 + 6 + 5 + \(x\)) oe or \(\dfrac{\text{“}160\text{”} + x}{\text{“}30\text{”} + x} = 4.25\) or “160” + \(x\) = 4.25 × “30” + 4.25\(x\) | M1 |
| “160” − “127.5” = 4.25\(x\) − \(x\) or 32.5 = 3.25\(x\) | M1 |
| Working required Answer: 10 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: Correct numerical products using midpoints (allowing one error) with intention to add.
May be seen in table.
M1: dep M1 for correct equation ft their 160.
M1: Isolating \(x\) and number terms
A1: dep 1st M1
| Scheme | Marks |
|---|---|
| (11 × 3) + (8 × 5) + (6 × 7) + (5 × 9) (= 33 + 40 + 42 + 45 = 160) | M1 |
| 4.25\(y\) = “160” + [\(y\) – (11 + 8 + 6 + 5)] oe 4.25\(y\) = “160” + \(y\) – 30 | M1 |
| 4.25\(y\) – \(y\) = “160” – 30 or 3.25\(y\) = 130 or \(y\) = 40 | M1 |
| Working required Answer: 10 | A1 |
Notes
M1: Correct numerical products using midpoints (allowing one error) with intention to add.
May be seen in table.
M1: dep M1 for correct equation ft their 160, where \(y\) = total number of pupils
M1: Isolating \(y\) and number terms
or \(y = 40\)
A1: dep 1st M1