Numeracy Intermediate June 2019 Unit 2 Q9
9. Mr Jakob notices a crack in a vertical wall which stands on horizontal ground.
(a) 
Mr Jakob fixes two temporary supports against the wall, as shown in the diagram below.
Diagram not drawn to scale
(i) 
Calculate the length of Support 1. [3]
Diagram not drawn to scale
(ii)
The length of Support 2 is 2·6 m.
Calculate the angle between the horizontal ground and Support 2. [3]
(b)
Mr Jakob gets a quote of £516 for rebuilding his wall.
The quote includes:
- 8 hours’ labour costs at £22.50 per hour,
- a 20% discount off the cost of the bricks.
Calculate the cost of the bricks before the discount. [3]
| Scheme | Marks | Comments | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| (i) (Support12 =) 0.92 + 1.12 | M1 | Scale drawings are not accepted | ||||||||||||
| Support12 = 2.02 or (Support1 =) √2.02 | A1 | Do not accept rounded to 2, unless final answer is 1.42(1... m)** | ||||||||||||
| (Support 1 =) 1.4(2... m) | A1 | FT from M1 for the correctly evaluated square root of ‘their 2.02’ provided ‘their answer’ > 1.1 (m) **Note, award as follows:
| ||||||||||||
| (ii) sin base angle = \(\dfrac{1.1 + 0.8}{2.6}\) | M1 | OR alternative full method using Pythagoras’ theorem then cos or tan | ||||||||||||
| sin-1 \(\dfrac{1.1 + 0.8}{2.6}\) or sin-1 0.73(0769....) | m1 | OR FT correct statement for ‘their inverse trig ratio’ | ||||||||||||
| (Base angle =) 46.95(...°) or 47(°) | A1 | Allow 46.88(...°) or 46.9(°) ISW unless subtracted from 90° If no marks, award SC1 for an answer of 50.7(°) or 51(°) from working with Support 1 |
| Scheme | Marks | Comments |
|---|---|---|
| (Discount cost of bricks) (516 – 8 × 22.5(0) =) (£) 336 | B1 | |
| 100 × 336 ÷ 80 or 100 × \(\dfrac{336}{80}\) | M1 | FT 'their 516 – 8 × 22.5(0)’ provided ≠ 516 and ≠ 180 for M1 and possible A1 |
| (£) 420 | A1 | If no marks, award SC2 for (516 ÷ 0.8 - 180 =) (£) 465 OR SC1 for (516 ÷ 0.8 =) (£) 645 or (100 × 180 ÷ 80 =) (£) 225 |