# 10.3.5 Index Number Long Questions (Question 5)

Question 5:
Table below 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
(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]

Solution:
(a)
$\begin{array}{l}I=\frac{{P}_{1}}{{P}_{0}}×100\\ I=\frac{{P}_{2014}}{{P}_{2011}}×100\\ 120=\frac{7.20}{x}×100\\ x=6.00\\ \\ y=\frac{5.50}{5.00}×100\\ y=110\end{array}$

(b)
$\begin{array}{l}\text{Composite index for bag in the year}\\ \text{2014 based on the year 2011,}\\ \overline{I}=\frac{\sum IW}{\sum W}\\ \overline{I}=\frac{\left(120\right)\left(7\right)+\left(115\right)\left(3\right)+\left(110\right)\left(6\right)+\left(125\right)\left(8\right)}{7+3+6+8}\\ \overline{I}=\frac{840+345+660+1000}{24}\\ \overline{I}=118.54\end{array}$

(c)
$\begin{array}{l}\overline{I}=\frac{{P}_{2014}}{{P}_{2011}}×100\\ 118.54=\frac{115}{{P}_{2011}}×100\\ {P}_{2011}=97.01\end{array}$

(d)
<$\begin{array}{l}\text{Composite index for the bag in the year}\\ \text{2015 based on the year 2011,}\\ \overline{I}=\frac{\sum IW}{\sum W}\\ =\frac{\left(120×0.9\right)\left(7\right)+\left(115×1.05\right)\left(3\right)+\left(110×1.05\right)\left(6\right)+1000}{24}\\ =\frac{756+362.25+693+1000}{24}\\ =117.14\end{array}$