SPM Mathematics 2025, Paper 2 (Question 17)


Question 17:
Ali owns a campsite.
(a) Diagram 11 shows the campsite which is a rectangular field, ABCD, made up of a durian orchard, APQ and a camping ground, PQBCD. P is the midpoint of AD and Q is the midpoint of AB.

Calculate the area, in m2, of the camping ground. [3 marks]


(b) Ali notices that there is a rectangular swampy area next to the camping ground and he wants to put a fence around it. The swamp has the length of (4x + 13) metres and the width of (3x) metres and has an area of 36 m2.
If the cost of installing the fence is RM14.50 per metre, calculate the total cost of the fence.
[4 marks]

(c) A small tent takes up a rectangular area which is 5 m long and 4 m wide.
A large tent takes up a rectangular area which is 6 m long and 5 m wide.
The camping ground can accommodate x small tents and y large tents.
The number of large tents is at most twice the number of small tents.
This can be written as y ≤ 2 x.


(i) The total area that all the tents take up is not more than 1500 m2 and this can be represented by a linear inequality, ax + by 1500.
State the values of a and of b.

(ii) In the answer space, draw and shade the region that satisfies the system of linear inequalities, x ≥ 0, y ≥ 0, y 2x and ax + by 1500.

(iii) Hence, determine the maximum number of small tent if the number of large tent is 20.
[4 marks]




(d) Zuki and his friends want to hold a family day.
Diagram 12 shows two types of camping packages.
Table 8 shows information of four families who join the family day.



Zuki assumes that each family will rent a tent.
Based on the packages provided, which package will save more?
Justify your answer by providing numerical values.
[4 marks]


Answer:
(a)
$$ \begin{aligned} &\text { Area of the camping ground }\\ &\begin{aligned} & =(82 \times 60) \mathrm{m}^2-\left(\frac{1}{2} \times 41 \times 30\right) \mathrm{m}^2 \\ & =4305 \mathrm{~m}^2 \end{aligned} \end{aligned} $$

(b)
$$ \begin{aligned} \text { Area } & =(4 x+13)(3 x) \\ 36 & =(4 x+13)(3 x) \\ 36 & =12 x^2+39 x \\ 0 & =12 x^2+39 x-36 \\ 0 & =(4 x-3)(x+4) \end{aligned} $$
$$ \begin{aligned} &x=\frac{3}{4},-4 \text { not accepted } \text { ) }\\ &\text { Cost of installing the fence }\\ &\begin{aligned} & =[(4 x+13) \mathrm{m}+(4 x+13) \mathrm{m}+3 x \mathrm{~m}+3 x \mathrm{~m}] \times \text { RM14.50 } \\ & =\left[4\left(\frac{3}{4}\right)+13+4\left(\frac{3}{4}\right)+13+3\left(\frac{3}{4}\right)+3\left(\frac{3}{4}\right)\right] \times \text { RM14.50 } \\ & =\text { RM529.25 } \end{aligned} \end{aligned} $$


(c)(i) 
$$ \begin{aligned} a= \text { area of a small tent } & =5 \times 4 \\ & =20 \end{aligned} $$
$$ \begin{aligned} b=\text { area of a big tent } & =6 \times 5 \\ & =30 \end{aligned} $$
$$ a=20, b=30 $$

(c)(ii)


Straight line 20x + 30y = 1500 drawn correctly. (gained 1 mark)
The region shaded correctly. (gained 1 mark)

(c)(iii) Maximum number of small tent = 45


(d)
$$ \begin{aligned} &\text { / Price for package } A\\ &\begin{aligned} & =(10 \times \mathrm{RM} 55)+(8 \times \mathrm{RM} 30) \\ & =\mathrm{RM} 550+\mathrm{RM} 240 \\ & =\mathrm{RM} 790 \end{aligned} \end{aligned} $$
$$ \begin{aligned} & \text { Price for package } B\\ &\begin{aligned} & =\text { RM60 + RM40 + RM60 + RM40 + RM150 + RM550 } \\ & =\text { RM900 } \end{aligned} \end{aligned} $$
Package A = RM790, Package B = RM 900
Package A with the price of RM790 will save more.

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