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} (\begin{array}{l} 25 \\ 3 1 \end{array}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 31 \\ 27 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2 (1) - 5 (3)} (\begin{array}{l} 1 - 5 \\ - 32 \end{array}) (\begin{array}{l} 31 \\ 27 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2 - 15} (\begin{array}{l} 1 (31) + (- 5) (27) \\ - 3 (31) + 2 (27) \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 13} (\begin{array}{l} - 104 \\ - 39 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 8 \\ 3 \end{array}) \\ x = 8 and y = 3 \\ Thus, the price for a food coupon is RM8 \\ and the price for drink coupon is RM3 . \end{array}

$\begin{array}{l} (\begin{array}{l} 25 \\ 3 1 \end{array}) (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 31 \\ 27 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2 (1) - 5 (3)} (\begin{array}{l} 1 - 5 \\ - 32 \end{array}) (\begin{array}{l} 31 \\ 27 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{2 - 15} (\begin{array}{l} 1 (31) + (- 5) (27) \\ - 3 (31) + 2 (27) \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = \frac{1}{- 13} (\begin{array}{l} - 104 \\ - 39 \end{array}) \\ (\begin{array}{l} x \\ y \end{array}) = (\begin{array}{l} 8 \\ 3 \end{array}) \\ x = 8 and y = 3 \\ Thus, the price for a food coupon is RM8 \\ and the price for drink coupon is RM3 . \end{array}$

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} 2 y = 3 x + 15 \\ y = \frac{3}{2} x + \frac{15}{2} \\ Thus m = \frac{3}{2} \\ At point (3, 0), y_{1} = m x_{1} + c \\ 0 = \frac{3}{2} (3) + c \\ \frac{9}{2} + c = 0 \\ c = - \frac{9}{2} \\ Thus, the linear equation is \\ y = \frac{3}{2} x - \frac{9}{2} \\ 2 y = 3 x - 9 \end{array}

$\begin{array}{l} 2 y = 3 x + 15 \\ y = \frac{3}{2} x + \frac{15}{2} \\ Thus m = \frac{3}{2} \\ At point (3, 0), y_{1} = m x_{1} + c \\ 0 = \frac{3}{2} (3) + c \\ \frac{9}{2} + c = 0 \\ c = - \frac{9}{2} \\ Thus, the linear equation is \\ y = \frac{3}{2} x - \frac{9}{2} \\ 2 y = 3 x - 9 \end{array}$

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