 # SPM Additional Mathematics (Model Test Paper)

SPM Additional Mathematics (Model Test Paper)

Section C
[20 marks]
Answer any two questions from this section.

Question 12
Diagram 6 shows a quadrilateral KLMN. Calculate
(a)KML, [2 marks]
(b) the length, in cm, of KM,   [3 marks]
(c) area, in cm2, of triangle KMN,   [3 marks]
(d) a triangle K’L’M’ has the same measurements as those given for triangle KLM, that is K’L’= 12.4 cm, L’M’= 9.5 cm and ∠L’K’M’= 43.2o, which is different in shape to triangle KLM.
(i) Sketch the triangle K’L’M’,
(ii) State the size of ∠K’M’L’. [2 marks]

(a)
$\begin{array}{l}\frac{\mathrm{sin}\angle KML}{12.4}=\frac{\mathrm{sin}{43.2}^{o}}{9.5}\\ \mathrm{sin}\angle KML=\frac{\mathrm{sin}{43.2}^{o}}{9.5}×12.4\\ \mathrm{sin}\angle KML=0.8935\\ \angle KML={63.32}^{o}\end{array}$

(b)
∠KLM = 180o – 43.2o – 63.32o = 73.48o
KM2 = 9.52 + 12.42 – 2(9.5)(12.4) cos 73.48o
KM2 = 244.01 – 66.99
KM2 = 177.02
KM = 13.30 cm

(c)
KM2 = MN2+ KN2– 2(MN)(KN) cos ∠KNM
13.302 = 5.42 + 9.92– 2(5.4)(9.9) cos ∠KNM
176.89 = 127.17 – 106.92 cos ∠KNM
$\begin{array}{l}\mathrm{cos}\angle KNM=\frac{127.17-176.89}{106.92}\\ \angle KNM={117.71}^{o}\end{array}$

Area of triangle KMN
$\begin{array}{l}=\frac{1}{2}\left(5.4\right)\left(9.9\right)\mathrm{sin}{117.71}^{o}\\ =23.66\text{}c{m}^{2}\end{array}$

(d)(i) (d)(ii)
∠K’M’L’ = 180o – 63.32o = 116.68o

Question 13
Table 2 shows the prices, the price indices and the proportion of four materials, ABC and D used in the production of a type of bag.

 Material Price (RM) for the year Price index in the year 2014 based on the year 2011 Proportion 2011 2014 A x 7.20 120 7 B 8.00 9.20 115 3 C 5.00 5.50 y 6 D 3.00 3.75 125 8
Table 2

(a) Calculate the value of x and of y. [2 Marks]
(b)   Find the composite index for the bag in the year 2014 based on the year 2011. [3 Marks]
(c) Given the cost for the production of the bag in the year 2014 is RM 115, find the corresponding cost in the year 2011. [2 Marks]
(d)   From the year 2014 to year 2015, the price indices of material B and C increase by 5%, material A decrease by 10% and material D remains unchanged.
Calculate the composite index in the year 2015 based on the year 2011. [3 Marks]

$\begin{array}{l}I=\frac{{P}_{1}}{{P}_{0}}×100\\ I=\frac{{P}_{2014}}{{P}_{2011}}×100\\ 120=\frac{7.20}{x}×100\\ x=6.00\\ \\ y=\frac{5.50}{5.00}×100\\ y=110\end{array}$
$\begin{array}{l}\text{Composite index for bag in the year}\\ \text{2014 based on the year 2011,}\\ \overline{I}=\frac{\sum IW}{\sum W}\\ \overline{I}=\frac{\left(120\right)\left(7\right)+\left(115\right)\left(3\right)+\left(110\right)\left(6\right)+\left(125\right)\left(8\right)}{7+3+6+8}\\ \overline{I}=\frac{840+345+660+1000}{24}\\ \overline{I}=118.54\end{array}$
$\begin{array}{l}\overline{I}=\frac{{P}_{2014}}{{P}_{2011}}×100\\ 118.54=\frac{115}{{P}_{2011}}×100\\ {P}_{2011}=97.01\end{array}$
<$\begin{array}{l}\text{Composite index for the bag in the year}\\ \text{2015 based on the year 2011,}\\ \overline{I}=\frac{\sum IW}{\sum W}\\ =\frac{\left(120×0.9\right)\left(7\right)+\left(115×1.05\right)\left(3\right)+\left(110×1.05\right)\left(6\right)+1000}{24}\\ =\frac{756+362.25+693+1000}{24}\\ =117.14\end{array}$