2D Area & Perimeter: WJEC GCSE Numeracy Higher November 2024 Unit 2 Q6

Numeracy Higher November 2024 Unit 2 Q6

WJECHigherCurrent spec7 marks2D Area & Perimeter Upper & Lower Bounds

6. Teifion is going to paint the front of his bungalow.
He will not be painting the roof, door or window.
A diagram of the front of Teifion’s bungalow is shown below.

Front of a bungalow: rectangular wall 610 cm wide and 300 cm high with a roof above; a window 220 cm by 140 cm and a door 100 cm wide and 200 cm high

Diagram not drawn to scale

All the measurements on the diagram are correct to the nearest 10 cm.
A litre of paint will cover 50 000 cm\(^2\), correct to the nearest 10 000 cm\(^2\).

For a good finish, Teifion needs to paint the front of the bungalow 3 times.
Calculate the smallest possible number of litres of paint that Teifion could use to paint the front of his bungalow 3 times. [7]

Smallest possible number of litres of paint that Teifion could use to paint the front of his bungalow 3 times is …………… litres

Mark scheme

Scheme	Marks	Comments
		Accept use of upper bounds e.g. 104.999… with a clear recurring digit. Do not accept use of e.g. 104.99
(Smallest possible area =) 605×295 − 105×205 − 225×145 or equivalent or 178475 − 21525 − 32625 or equivalent or 178475 − 54150	M4	M3 for an appropriate calculation using 3 products with any 2 correct M2 for ‘their 605×295’ − ‘their 105×205’ − ‘their 225×145’ where 600 \(\leqslant\) ‘their 605’ < 610 and 290 \(\leqslant\) ‘their 295’ < 300 and 100 < ‘their 105’ \(\leqslant\) 110 and 200 < ‘their 205’ \(\leqslant\) 210 and 140 < ‘their 145’ \(\leqslant\) 150 and 220 < ‘their 225’ \(\leqslant\) 230 M1 for use of 605×295 OR 105×205 OR 225×145
= 124 325 (cm\(^2\))	A1	CAO May be implied by (3 × 124325 =) 372975
(Least number of litres Teifion could possibly use =) 3 × 124 325 ÷ 55000 or 372975 ÷ 55000	M1	FT ‘their derived 124325’ Allow M1 for 3 × ‘their 124325’ ÷ ‘their 55000’ where 50000 < ‘their 55000’ \(\leqslant\) 60000
= 6.78(13…) or 6.8 (litres)	A1	Allow an answer of 6.7 or 7 from correct working On FT, accept rounding up to nearest whole number On FT, allow truncation to 1 d.p. or better If no marks awarded, and from a misinterpretation of the question, SC3 for an answer of 9.335 or 9.3 or 9.33 or 9.34 or 9.4 or 10 (litres) from \(\dfrac{3 \times (615\times 305 - 95\times 195 - 215\times 135)}{45000}\) or \(\left[\dfrac{3\times(187575-18525-29025)}{45000}\right]\) OR SC2 for any one of the following: \(\dfrac{3 \times (615\times 305 - 95\times 195 - 215\times 135)}{45000}\) or \(\left[\dfrac{3\times(187575-18525-29025)}{45000}\right]\) but not evaluated correctly correct evaluation (unrounded, rounded or truncated) from 3 × (‘their 615×305’ − ‘their 95×195’ − ‘their 215×135’) ÷ ‘their 45000’ where 610 < ‘their 615’ \(\leqslant\) 620 and 300 < ‘their 305’ \(\leqslant\) 310 and 90 \(\leqslant\) ‘their 95’ < 100 and 190 \(\leqslant\) ‘their 195’ < 200 and 210 \(\leqslant\) ‘their 215’ < 220 and 130 \(\leqslant\) ‘their 135’ < 140 and 40000 \(\leqslant\) ‘their 45000’ < 50000 OR SC1 for 3 × (‘their 615×305’ − ‘their 95×195’ − ‘their 215×135’) ÷ ‘their 45000’ using values in the same ranges as SC2, but not evaluated correctly