SPM Mathematics 2018, Paper 2 (Questions 11 & 12)

Question 11 (6 marks):
Diagram 8 shows sectors OABC and ODE with the common centre O respectively.

Diagram 8

Calculate
(a) the perimeter, in cm, of the whole diagram,
(b) the area, in cm², of the shaded region.

Solution:
(a)
$\begin{array}{l}\text{Perimeter of the whole diagram}\\ =OP+PE+\text{Arc }DE+DC\\ +\text{Arc }CBA+AO\\ =10+5+\left(\frac{{50}^o}{{360}^o}\times 2\times \frac{22}{7}\times 15\right)+5\\ +\left(\frac{{230}^o}{{360}^o}\times 2\times \frac{22}{7}\times 10\right)+10\\ =30+\frac{275}{21}+\frac{2530}{63}\\ =\frac{5245}{63}\\ =83.254\text{ cm}\end{array}$


(b)
$\begin{array}{l}\text{Area of the shaded region}\\ =\left(\frac{{230}^o}{{360}^o}\times \frac{22}{7}\times {10}^2\right)\\ +\left[\left(\frac{{50}^o}{{360}^o}\times \frac{22}{7}\times {15}^2\right)-\left(\frac{{50}^o}{{360}^o}\times \frac{22}{7}\times {10}^2\right)\right]\\ =\frac{12650}{63}+\left(\frac{1375}{14}-\frac{2750}{63}\right)\\ =\frac{3575}{14}\\ =255.36{\text{ cm}}^2\end{array}$

Question 12 (12 marks):
(a) Complete Table 2 in the answer space, for the equation $y=\frac{30}{x}$  by writing down the values of y when x = 2 and x = 5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of $y=\frac{30}{x}\text{ for }0\le x\le 7.$  

(c) From the graph in 12(b), find
(i) the value of y when x = 2.6,
(ii) the value of x when y = 17.5.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation $\frac{30}{x}=-5x+30\text{ for }0\le x\le 7.$
State the values of x.

Answer:
Table 2


Solution:
(a)



(b)



(c) From the graph
(i) When x = 2.6; y = 11.5
(ii) When y = 17.5; x = 1.7

(d)
$\begin{array}{l}\text{Given,}\frac{30}{x}=-5x+30\\ y=-5x+30\\ \text{From the graph; }x=1.25\text{ and }4.75\end{array}$