Basic Trigonometry: WJEC GCSE Numeracy Intermediate November 2016 Unit 2 Q9

Numeracy Intermediate November 2016 Unit 2 Q9

WJECIntermediateCurrent spec14 marksBasic Trigonometry Compound Measures Percentages

9. Gwenda enjoys road running.

(a) She keeps a record of her run each day this week.

Day	Sun	Mon	Tues	Wed	Thurs	Fri	Sat
Distance	4·6 km	5·4 km	2·2 km	6·2 km	7·2 km	2·2 km	3·4 km
Time	26 mins	31 mins	12 mins	35 mins	40 mins	14 mins	22 mins

Last week, her average speed for the week was 9·6 kilometres per hour.
Calculate Gwenda’s percentage improvement in her average speed from last week to this week.
You must show all your working. [6]

Percentage improvement is …………………… %

(b) The diagram shows the cross-section of one part of her run.

Right-angled triangle: horizontal 1600 metres, vertical 200 metres

Diagram not drawn to scale

Calculate the angle of elevation of the road. [3]

(c)

Uneven road from A up to B, with a dashed straight line from A to B; the vertical rise to B is 300 metres, with a right angle at the bottom

Diagram not drawn to scale

Gwenda runs on another section of uneven road from A to B.
The rise in this section of the road is 300 metres.
The angle of elevation of B from A is 10°.

(i) Calculate an estimate of how far Gwenda has run.
State any assumption you have made. [4]

Assumption: …………………………………………

(ii) What is the impact of your assumption on your answer? [1]

Mark scheme (a)

Scheme	Marks	Comments
Sight of 31.2 and 3 or 180	B1
\(\dfrac{31.2}{3}\) OR (9.6 km/h =) 0.16 (km/min) with \(\dfrac{31.2}{180}\)	M2	Allow 31.2/7 ÷ 3/7 FT \(\dfrac{\text{‘their total distance’}}{\text{‘their total time in hours’}}\) M1 for \(\dfrac{\text{'their total distance'}}{\text{'their total time in minutes'}}\) allow 31.2/7 ÷ 180/7
10.4 (km/h) OR 0.17(3.. km/min)	A1	FT from 1 arithmetic error in calculating either 31.2 or 3, i.e. one of these values needs to be correct Do not FT from denominator in minutes unless 0.16 (km/min) seen Allow a final answer from a correct method that rounds to 10.4, e.g. 10.3(54..km/h) from PA (Note: 31.2 ÷ 7 = 4.45714… 3 ÷ 7 = 0.42857… 180 ÷ 7 = 25.71428…) If no marks so far, allow SC1 for evaluating ‘a distance ÷ time in hours’ correctly (Sun to Sat : 10.615.., 10.45, 11, 10.6286, 10.8, 9.4286, 9.2727…)
% improvement 100 × (10.4 – 9.6) ÷ 9.6 or 100 × 10.4 ÷ 9.6 - 100 or equivalent	M1	FT ‘their 10.4’ provided it is >9.6 OR FT ‘their 0.17(3…)’ provided it is >0.16
8(.333…%)	A1	(Note: use of 10.354… leads to 7.85%) If previously M0, A0 for % improvement, allow SC1 for an answer of 108%, or similar from FT

Scheme	Marks	Comments
tan elevation = \(\dfrac{200}{1600}\) or equivalent	M1
Angle of elevation is 7(.125….°)	A2	A1 for tan^-1 0.125 or tan^-1 (200/1600)

Mark scheme (c)

Scheme	Marks	Comments
(i) run = \(\dfrac{300}{\sin 10^\circ}\)	M2	M1 for sin10° = 300/run
1727(.631… m) or 1728(m)	A1	ISW Accept reasonable estimates (rounding or truncation) following correct working, e.g. 1700, 1730, 1750
Assumption, e.g. ‘road is straight’, ‘used a right-angled triangle’, ‘the road is smooth’, ‘Gwenda runs in a straight line’	E1	Depends on a previous attempt to use right-angled triangle trigonometry or Pythagoras’ theorem Accept ‘Gwenda doesn’t zigzag up the hill’
(ii) Impact, e.g. ‘run could be longer', 'it is an under estimate', ‘bumps could make it longer’	E1	Independent of (c)(i) Allow ‘it is inaccurate’ Do not accept 'shorter' alone However, accept ‘shorter than the actual length’