2D Area & Perimeter: WJEC GCSE Numeracy Foundation November 2023 Unit 2 Q3

Numeracy Foundation November 2023 Unit 2 Q3

WJECFoundationCurrent spec9 marks2D Area & Perimeter Constructions

3. Siân belongs to a local running club.

(a)

One weekend, Siân plans to run along some local paths.

Siân wants to work out the total distance for each run before she starts.

Siân uses the map below.
The scale of the map is 1 cm represents 0·5 miles.

On screen this diagram may not be shown at actual size. To measure it, use the question paper (PDF) printed at 100%.

Scale: 1 cm represents 0·5 miles

(i) How many miles is it from Aber to Cwm? [2]

(ii) How many miles is it from Aber to Borth and then to Dinas? [2]

(b)

Siân trains at her local rugby pitch, which is shown below.
Siân runs complete laps around the outside of the pitch.

Diagram not drawn to scale

Siân wants to run at least 5000 m.

What is the smallest number of complete laps that Siân needs to run?
You must show all your working. [5]

Mark scheme (a)

Scheme	Marks	Comments
(i) (For Aber to Cwm:) 6 (miles) (5.9 to 6.1)	B2	For B2, FT ‘their distance measured’ × 0.5 correctly evaluated provided the distance is in the range 11.8 to 12.2 cm Award B1 for: distance measured as 12(cm) (±2mm) ‘their stated distance measured outside the range 11.8 to 12.2 cm’ × 0.5 correctly evaluated
(ii) (For Aber to Borth to Dinas:) 4 (miles) (3.8 to 4.2)	B2	FT ‘their distance measured’ × 0.5 correctly evaluated provided the distance is in the range 7.6 to 8.4 cm Award B1 for: Total distance measured as 8(cm) (±4mm) 2 distances seen as 3(cm) (±2mm) AND 5(cm) (±2mm) ‘their stated distance measured outside the range 7.6 to 8.4 cm’ × 0.5 correctly evaluated 1.5 (miles) or 2.8 to 3.2(cm) × 0.5 correctly evaluated (for Aber to Borth) 2.5 (miles) or 4.8 to 5.2 (cm) × 0.5 correctly evaluated (for Borth to Dinas)

Mark scheme (b)

Scheme	Marks	Comments
(Perimeter=) 106 + 68 + 106 + 68 or equivalent 348 (m)	M1 A1
5000 ÷ 348 or 348 × 15 or 348 × 14	M1	Division may be seen as repeated addition/subtraction or repeated trials. Allow 348 × 13 (= 4524) Allow 348 × 16 (= 5568) FT ‘their derived perimeter’ including partial perimeter but not use of 106 or 68 Use of area gains no marks
15 (laps)	A2	Accept a unique embedded answer for A2 or A1 e.g. Award A2 if only 348 × 15 = 5220 seen e.g. Award A1 if only 348 × 14 = 4872 seen Award A1 if both 348 × 15 = 5220 AND 348 × 14 = 4872 seen with no final answer given for the number of laps. Award A1 for 14 or 14.4 or 14.37 or 14.3(6…..) or 14.5 seen. FT 5000 ÷ ‘their 348’ for possible A2 or A1. On FT, award A2 provided they have rounded up to the nearest whole number. Otherwise award A1 for a correctly evaluated answer for 5000 ÷ ‘their 348’ without rounding up. If on FT the answer is a whole number with no rounding required award A1 only. For use of partial perimeter of 174m 5000 ÷ 174 = 29 laps award M0A0M1A2 5000 ÷ 174 = 28.7… award M0A0M1A1 If trials used (apply in line with the MS above): For 174 × 29 = 5046 Award M0A0M1A2 For 174 × 28 = 4872 Award M0A0M1A1 For 174 × 27 (= 4698) Award M0A0M1A0 For 174 × 30 (= 5220) Award M0A0M1A0 Note: if 4872 seen from repeated addition but 28 laps not seen award M0A0M1A0 (need a value for the number of laps or an embedded value of the number of laps)

Corrected by Westie's Workshop

The board's mark scheme printed “partial perimeter of 174cm”. Changed to “partial perimeter of 174m” because the pitch is 106 m by 68 m, so the half-perimeter is 174 m, not 174 cm.

Original: see the board's mark scheme PDF (link above).