Numeracy Higher November 2017 Unit 2 Q4
4. Bethan has a plan of her rectangular lawn, which she has labelled ABCD.
She wants to cut out a triangular flowerbed from her lawn, labelled GHD.
Bethan decides that AG : GD should be 1 : 2 and that DH = HC.
She has made a sketch shown below.
Diagram not drawn to scale
(a) Calculate the length of GH. [4]
(b) The flowerbed, GHD, is to have a flexible edging strip placed around its perimeter.
The edging strip costs £3.50 per metre and can only be bought in strips of complete metres.
The edging strip costs £3.50 per metre and can only be bought in strips of complete metres.
- How much will the edging strip cost Bethan?
- What length of strip will be left over?
Give your answer in centimetres. [4]
Cost £ ……………
…………… cm left over
| Scheme | Marks | Comments |
|---|---|---|
| DG = 3.2 (m) and DH = 3.4 (m) | B1 | May be seen on the diagram |
| (GH2 =) 3.22 + 3.42 | M1 | FT ‘their 3.2’ and ‘their 3.4’ provided they are ≠4.8 and ≠ 6.8(m) |
| (GH)2 = 21.8 or (GH =) √21.8 | M1 | FT ‘their 3.2’ (DG) and ‘their 3.4’ (DH) including use of 4.8 and 6.8(m) |
| 4.7(m) or 4.67(m) or 4.66(9…m) or 4.6(m) | A1 | Allow FT from M0, M1 including use of 4.8 and 6.8(m) to give 8.3(2…m) (i.e. B0, M0, M1, A1) FT from M1, M0 for the correctly evaluated square root of ‘their 21.8’ provided ‘their answer’ > 3.4 (m) |
Corrected by Westie's Workshop
- The board's mark scheme printed “provided ‘their answer’ > 3.4 (cm)”. Changed to “3.4 (m)” because all the lengths in this question are in metres.
Original: see the board's mark scheme PDF (link above).
| Scheme | Marks | Comments |
|---|---|---|
| (Perimeter) 4.669… + 3.2 + 3.4 | M1 | (11.2m, 11.269..m, 11.27m or 11.3m) FT ‘their derived 4.669…’ (from (a)) + ‘their DG <4.8’ + ‘their DH <6.8’, however if no response in (a) accept ‘their GH’ if clearly stated provided >3.3 but <8.4 (m) |
| Cost 12 × 3.50 | M1 | FT ‘their derived perimeter’ provided:
|
| (£)42 | A1 | Do not FT further for premature rounding of lengths to find ‘their perimeter’, no further marks (Otherwise FT) |
| Appropriate for the perimeter 70(cm) or 73.(095….cm) or 74(cm) or 80 (cm) left over | B1 | Strict FT 100 × (‘their 12’ – ‘their correctly evaluated derived perimeter<12’), which leads to left over bit ≥ 0 e.g. 74(cm) from a perimeter 11.26m Answer must be in cm Accept use of rounded or truncated answers for ‘their derived perimeter’ A fully correct FT for rounding lengths up prematurely, e.g. if 4.7m used: 4.7 is 5 strips, 3.4 and 3.2 is 4 strips each, gives 13 m, so 13 × £3.50 = (£)45.5(0) with 170(cm) left over, this is awarded M0, M1, A0, B0 |