SPM Mathematics 2017, Paper 2 (Questions 7 & 8)

Question 7 (5 marks):
During the sport day, students used coupons to buy food and drink. Ali and Larry spent RM31 and RM27 respectively. Ali bought 2 food coupons and 5 drinks coupons while Larry bought 3 coupons and 1 drinks coupon.
Using the matrix method, calculate the price, in RM, of a food coupon and of a drinks coupon.

Solution:
Ali spent RM31. He bought 2 food coupons and 5 drinks coupons.
Larry spent RM27. He bought 3 food coupons and 1 drinks coupons.
x = price of one food coupon
y = price of one drinks coupon

$\begin{array}{l}\left(\begin{array}{l}2\text{ 5}\\ 31\end{array}\right)\left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}31\\ 27\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2\left(1\right)-5\left(3\right)}\left(\begin{array}{l}1-5\\ -3\text{ 2}\end{array}\right)\left(\begin{array}{l}31\\ 27\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{2-15}\left(\begin{array}{l}1\left(31\right)+\left(-5\right)\left(27\right)\\ -3\left(31\right)\text{+}2\left(27\right)\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\frac{1}{-13}\left(\begin{array}{l}-104\\ -39\end{array}\right)\\ \left(\begin{array}{l}x\\ y\end{array}\right)=\left(\begin{array}{l}8\\ 3\end{array}\right)\\ x=8\text{ and }y=3\\ \\ \text{Thus, the price for a food coupon is RM8}\\ \text{and the price for drink coupon is RM3}\text{.}\end{array}$

Question 8 (5 marks):
Diagram 4 shows a parallelogram drawn on a Cartesian plane which represents the locations of Rahman’s house, a cinema, a school and a shop.

It is given that the scale is 1 unit : 1 km.
(a) Calculate the distance, in km, between Rahman’s house and the school.
(b) Find the equation of the straight line that links the school to the cinema.


Solution:
(a)
2y = 3x + 15
When y = 0
3x + 15 = 0
3x = –15
x = –5

Rahman’s house = (–5, 0)
School = (3, 0)

Distance, between Rahman’s house and the school
= 3 – (– 5)
= 8 units
= 8 km

(b)
$\begin{array}{l}2y=3x+15\\ y=\frac{3}{2}x+\frac{15}{2}\\ \text{Thus }m=\frac{3}{2}\\ \text{At point }\left(3,0\right),y_1=m{x}_1+c\\ 0=\frac{3}{2}\left(3\right)+c\\ \frac{9}{2}+c=0\\ c=-\frac{9}{2}\\ \\ \text{Thus, the linear equation is}\\ y=\frac{3}{2}x-\frac{9}{2}\\ 2y=3x-9\end{array}$