OP Malhotra Class 11 Maths Solutions Chapter 30 Index Numbers Chapter Test

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S Chand Class 11 ICSE Maths Solutions Chapter 30 Index Numbers Chapter Test

Question 1.
Construct the consumer price index number for 1990 , taking 1989 as the base year and using simple average of price relative method for the following data :

Solution:
We construct the table of values as under:

using simple average of price relatives method
required index number = P₀₁ = \(\frac{1}{\mathrm{~N}} \Sigma\left(\frac{\mathrm{P}_1}{\mathrm{P}_0} \times 100\right)\) = \(\frac{1}{4} \times 380\) = 95

Question 2.
A small industrial concern used three raw materials A, B and C in the manufacturing process, the prices of the materials was as shown below :

Using 1995 as the base year, calculate a simple aggregate price index for 2005.
Solution:
We construct the following table as under :

By simple aggregate method, we have
required price index for 2015 = P₀₁ = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\) = \(\frac{104}{100} \times 100\) = 104

Question 3.
Find the consumer price index number for the year 2010 as the base year by using method of weighted aggregates.

Solution:
We construct the table of values as under:

Thus by weighted aggregate method,
required index number = \(\frac{\Sigma \mathrm{P}_1 w}{\Sigma \mathrm{P}_0 w} \times 100\) = \(\frac{2442.5}{1764.9} \times 100\) = 138.39

Question 4.
The price of six different commodities for years 2009 and year 2011 are as follows :

The index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x and y if the total price in 2009 is ₹ 360.
Solution:
We construct the table of values given as under :

Since total price in 2009 = ₹ 360 ⇒ 250 + x = 360 ⇒ x = 110
Using simple aggregate method, index number = \(\frac{\Sigma \mathrm{P}_1}{\Sigma \mathrm{P}_0} \times 100\)
⇒ 125 = \(\frac{390+y}{360} \times 100\)
\(\frac{125 \times 36}{10}\) = 390 + y
⇒ 390 + y = 450
⇒ y = 60