(Long Questions) – Question 2

Posted on April 22, 2020 by user

Question 2:

Table shows the prices, the price indices and weightage for four types of stationery A, B, C and D.

(a) Find the value of

(i) x, (ii) y, (iii) z.

(b) The composite index for the price of the stationery in the year 2009 based on the year 2008 is 118.5. Calculate the value of m.

(c) The total expenditure for the stationery in the year 2008 is RM350. Calculate the corresponding total expenditure in the year 2009.

(d) The price index for B in the year 2010 based on the year 2008 is 132. Calculate the price index for B in the year 2010 based on the year 2009.

Solution:
(a)(i)

\begin{array}{l} I = \frac{P_{1}}{P_{0}} \times 100 \\ x = \frac{P_{2009}}{P_{2008}} \times 100 \\ x = \frac{2}{2.5} \times 100 \\ x = 80 \end{array}

(a)(ii)

\begin{array}{l} 135 = \frac{P_{2009}}{2} \times 100 \\ 135 = \frac{y}{2} \times 100 \\ y = R M 2.70 \end{array}

(a)(iii)

\begin{array}{l} 116 = \frac{5.80}{P_{2008}} \times 100 \\ 116 = \frac{5.80}{z} \times 100 \\ z = RM5 \end{array}

(b)

\begin{array}{l} Composite index for items in the year 2009 based on the year 2008, \bar{I} = 118.5 \\ \bar{I} = \frac{\sum I W}{\sum W} \\ 118.5 = \frac{(80) (2) + (120) (3) + (135) (m) + (116) (4)}{2 + 3 + m + 4} \\ 118.5 = \frac{984 + 135 m}{9 + m} \\ 1066.5 + 118.5 m = 984 + 135 m \\ 16.5 m = 82.5 \\ m = 5 \end{array}

(c)

\begin{array}{l} \bar{I} = \frac{P_{2009}}{P_{2008}} \times 100 \\ 118.5 = \frac{P_{2009}}{350} \times 100 \\ ∴ Total expenditure in the year 2009, P_{2009} = R M 414.75. \end{array}

(d)

\begin{array}{l} For stationery B, \\ It is given \frac{P_{2010}}{P_{2008}} \times 100 = 132 \to \frac{P_{2010}}{P_{2008}} = \frac{132}{100} \\ and \frac{P_{2009}}{P_{2008}} \times 100 = 120 \to \frac{P_{2009}}{P_{2008}} = \frac{120}{100} \\ Price index for stationery B in the year 2010 based on the year 2009, \\ I_{10 / 09} = \frac{P_{2010}}{P_{2009}} \times 100 \\ I_{10 / 09} = \frac{P_{2010}}{P_{2008}} \times \frac{P_{2008}}{P_{2009}} \times 100 \\ I_{10 / 09} = \frac{132}{100} \times \frac{100}{120} \times 100 \\ I_{10 / 09} = 110 \end{array}