Numeracy Higher November 2016 Unit 2 Q5
5. Rhodri has carried out an experiment to measure the diameters of 20 spherical dust particles, in microns.
Here are his results.
| Diameter, \(d\) (microns) | Frequency |
|---|---|
| \(1 \leqslant d \lt 2\) | 2 |
| \(2 \leqslant d \lt 4\) | 6 |
| \(4 \leqslant d \lt 5\) | 8 |
| \(5 \leqslant d \lt 9\) | 4 |
Rhodri is told,
‘The ratio of dust particles with diameters less than 4 microns to those with diameters greater than or equal to 4 microns is 7 : 8.’
He finds this fact is true when he considers all 45 dust particles.
How many of the extra 25 dust particles have a diameter of less than 4 microns?
You must show your working. [3]
The height of the cell is 2 microns.
The circumference of the cell is 5 microns.
Calculate the volume of the cell he sees under the microscope.
Give your answer in microns3, correct to 1 significant figure. [5]
Volume is ……………… microns3
| Scheme | Marks | Comments |
|---|---|---|
| (i) Mid points : 1.5, 3, 4.5, 7 | B1 | |
| 1.5 × 2 + 3 × 6 + 4.5 × 8 + 7 × 4 (= 3 + 18 + 36 + 28 = 85) | M1 | FT ‘their mid points’ provided each one lies within the appropriate group, including bounds |
| ÷ 20 4.25 (microns) | m1 A1 | Accept 4.3 from correct working, i.e. 85 ÷ 20 seen in working Do not accept 4.2 unless 4.25 or 85 ÷ 20 seen in working |
| (ii) 45 dust particles means 3×7 : 3×8 21 : 24 or 21 in total equivalent (A further) 13 (dust particles) | M1 m1 A1 | Accept 7×45/(7+8) : 8×45/(7+8) Allow M1, m1, A0 for sight of 8 + 13 = 21 Alternative: Trial & improvement, e.g. 18 : 27 (is 2 : 3 incorrect) 19 : 26 (incorrect) 20 : 25 (is 4 : 5 incorrect) 21 : 24 (is 7 : 8 correct!!) M1 for sight from the above list: a trial with correct simplification shown AND --- either for a second trial with correct simplification shown --- or the second trial has clearly been dismissed m1 Selection of 21 : 24 A1 (a further) 13 (dust particles) |
| Scheme | Marks | Comments |
|---|---|---|
| (Circumference) 5 = 2 × π × r or 5 = π × d Radius of the cylinder \(\dfrac{5}{2\pi}\) | M1 A1 | (5/2π = 0.79577…) |
| Volume π × (5/2π)2 × 2 4 (microns3) | m1 A2 | FT ‘their r’ provided M1 awarded provided ‘their r’ ≠ 5/π A1 for 25/2π or 3.9(…) or 4.0 (microns3) |