Question 5:
Table 3 shows the competitions joined by a group of students.
(a) Complete the Venn diagram in the answer space.
[2 marks]
(b) Hence, find
(i) the number of students who joined one competition only, [1 mark]
(ii) the number of students who did not join both the colouring and cooking competitions. [2 marks]
Solution:
(a)
(b)(i)
Number of students who joined one competition only
= 12 + 6 + 11
= 29
(b)(ii)
The number of students who did not join both the colouring and cooking competitions
= 12 + 4 + 6 + 10 + 11
= 43
Table 3 shows the competitions joined by a group of students.
(a) Complete the Venn diagram in the answer space.
[2 marks]
(b) Hence, find
(i) the number of students who joined one competition only, [1 mark]
(ii) the number of students who did not join both the colouring and cooking competitions. [2 marks]
Solution:
(a)
(b)(i)
Number of students who joined one competition only
= 12 + 6 + 11
= 29
(b)(ii)
The number of students who did not join both the colouring and cooking competitions= 12 + 4 + 6 + 10 + 11
= 43
Question 6:
Diagram 4 shows a speed-time graph for a car.
(a) Describe the motion of the car between the 20th second and the 25th second. [1 mark]
(b) Calculate the total distance, in m , traveled by the car during acceleration and deceleration. [3 marks]
Solution:
(a) The car was experiencing constant speed between the 20th second and the 25th second.
(b) The car was experiencing acceleration from 10th – 20th second and deceleration from 25th – 50th second.
For speed-time graph, total distance is area under the graph,
$$ \begin{aligned} & =\text { Area of trapezium }+ \text { Area of triangle } \\ & =\frac{1}{2}(17+13)(10)+\frac{1}{2}(25)(17) \\ & =150+212.5 \\ & =362.5 \mathrm{~m} \end{aligned} $$
Diagram 4 shows a speed-time graph for a car.
(a) Describe the motion of the car between the 20th second and the 25th second. [1 mark]
(b) Calculate the total distance, in m , traveled by the car during acceleration and deceleration. [3 marks]
Solution:
(a) The car was experiencing constant speed between the 20th second and the 25th second.
(b) The car was experiencing acceleration from 10th – 20th second and deceleration from 25th – 50th second.
For speed-time graph, total distance is area under the graph,
$$ \begin{aligned} & =\text { Area of trapezium }+ \text { Area of triangle } \\ & =\frac{1}{2}(17+13)(10)+\frac{1}{2}(25)(17) \\ & =150+212.5 \\ & =362.5 \mathrm{~m} \end{aligned} $$