Index Numbers and Time Based Data — NCERT Solutions

Correct Answer: ii) Consumer price index

The Consumer Price Index (CPI), also called the cost-of-living index, is specifically designed to compare the cost of maintaining a given standard of living across different time periods or different cities/regions.

Given:
Base year = 1995, Current year = 2005.

Formula (Simple Aggregative Method):
$$P_{01} = \frac{\sum p_1}{\sum p_0} \times 100$$

Calculation:
$$\sum p_0 = 80 + 50 + 90 + 30 = 250$$
$$\sum p_1 = 95 + 60 + 100 + 45 = 300$$

$$P_{01} = \frac{300}{250} \times 100 = \boxed{120}$$

Interpretation: Prices in 2005 are 20% higher than in 1995.

Given: Base year = 2008 ($p_0, q_0$), Current year = 2010 ($p_1, q_1$).

Step 1: Prepare the calculation table.

i) Laspeyres' Index:
$$P_L = \frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100 = \frac{251}{172} \times 100 \approx \boxed{146}$$

ii) Paasche's Index:
$$P_P = \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100 = \frac{220}{148} \times 100 \approx \boxed{149}$$

iii) Fisher's Ideal Index:
$$P_F = \sqrt{P_L \times P_P} = \sqrt{146 \times 149} = \sqrt{21754} \approx \boxed{147}$$

Given: Base year = 1995, $p_0 = 12$.

Formula (Price Relative):
$$I = \frac{p_n}{p_0} \times 100$$

Calculations:

Result: The relative index numbers are 100, 117, 108, 167, 208, 175 for the respective years.

Given: Current year prices ($p_1$), Base year prices ($p_0$), Weights ($W$).

Formula:
$$P_{01} = \frac{\sum p_1 W}{\sum p_0 W} \times 100$$

Calculation table:

$$P_{01} = \frac{510}{370} \times 100 = 137.84 \approx \boxed{137}$$

Given: Base year = 1990, Current year = 2010.

Formula:
$$P_{01} = \frac{\sum p_1}{\sum p_0} \times 100$$

Calculation:
$$\sum p_0 = 60 + 40 + 100 + 60 + 90 = 350$$
$$\sum p_1 = 140 + 60 + 205 + 70 + 100 = 575$$

$$P_{01} = \frac{575}{350} \times 100 = 164.28 \approx \boxed{164}$$

Interpretation: Prices in 2010 are approximately 64% higher than in 1990.

Given: Base year = 1990 ($p_0$), Current year = 2003 ($p_1$).

Formula (Simple Aggregative Method for Cost of Living Index):
$$\text{CLI} = \frac{\sum p_1}{\sum p_0} \times 100$$

Calculation:
$$\sum p_0 = 1400 + 200 + 400 + 200 + 250 = 2450$$
$$\sum p_1 = 1500 + 250 + 750 + 300 + 425 = 3225$$

$$\text{CLI} = \frac{3225}{2450} \times 100 = 131.63 \approx \boxed{132}$$

Interpretation: The cost of living in 2003 is approximately 32% higher than in 1990.

Given: Index of sales volume and index of sales value for each year.

Concept:
$$\text{Value Index} = \text{Price Index} \times \text{Volume Index}$$
$$\Rightarrow \text{Price Index} = \frac{\text{Value Index}}{\text{Volume Index}} \times 100$$

Calculations:

Year 1995:
$$\text{Price Index} = \frac{105}{101} \times 100 = 103.96 \approx \mathbf{104}$$

Year 1996:
$$\text{Price Index} = \frac{108}{113} \times 100 = 95.58 \approx \mathbf{96}$$

Year 1997:
$$\text{Price Index} = \frac{124}{106} \times 100 = 116.98 \approx \mathbf{117}$$

Result:

Given: Weights ($W$), Base year prices ($p_0$), Current year prices ($p_1$).

Formula:
$$P_{01} = \frac{\sum p_1 W}{\sum p_0 W} \times 100$$

Calculation table:

$$P_{01} = \frac{5515}{3610} \times 100 = 152.77 \approx \boxed{153}$$

Time Reversal Test: An index satisfies the time reversal test if $P_{01} \times P_{10} = 1$.

Let base year = 0 ($p_0, q_0$) and current year = 1 ($p_1, q_1$).

Step 1: Compute required sums.

i) Paasche's Formula — Time Reversal Test:

$$P_{01}^{\text{Paasche}} = \frac{\sum p_1 q_1}{\sum p_0 q_1} = \frac{210}{212}$$

For $P_{10}^{\text{Paasche}}$, swap 0 and 1 (i.e., current becomes base and base becomes current):
$$P_{10}^{\text{Paasche}} = \frac{\sum p_0 q_0}{\sum p_1 q_0} = \frac{170}{170} = 1$$

$$P_{01} \times P_{10} = \frac{210}{212} \times 1 = \frac{210}{212} \neq 1$$

Conclusion: Paasche's formula does NOT satisfy the time reversal test.

ii) Fisher's Formula — Time Reversal Test:

$$P_{01}^{\text{Fisher}} = \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} = \sqrt{\frac{170}{170} \times \frac{210}{212}}$$

$$P_{10}^{\text{Fisher}} = \sqrt{\frac{\sum p_0 q_1}{\sum p_1 q_1} \times \frac{\sum p_0 q_0}{\sum p_1 q_0}} = \sqrt{\frac{212}{210} \times \frac{170}{170}}$$

$$P_{01} \times P_{10} = \sqrt{\frac{170}{170} \times \frac{210}{212}} \times \sqrt{\frac{212}{210} \times \frac{170}{170}}$$
$$= \sqrt{\frac{170 \times 210 \times 212 \times 170}{170 \times 212 \times 210 \times 170}} = \sqrt{1} = 1$$

Conclusion: Fisher's formula DOES satisfy the time reversal test.

Given: Annual rainfall data from 2001 to 2009.

Method: 3-year moving averages — each average is the mean of 3 consecutive years.

Formula: $\text{3-year MA} = \frac{y_t + y_{t+1} + y_{t+2}}{3}$, centred at year $t+1$.

Calculation table:

Result: The 3-year moving averages (trend values) are: 1.70, 1.77, 1.83, 1.60, 1.73, 1.40, 1.40 for years 2002 through 2008 respectively. The trend shows slight fluctuation with no strong upward or downward movement.

Given: Annual production data from 1980 to 1989.

Method: 4-year moving averages.

Step 1: Compute 4-year moving totals (sum of 4 consecutive years).

Step 2: Compute 4-year moving averages (divide by 4).

Step 3: Centre the averages (average of two consecutive 4-year averages to align with actual years).

Result: The centred 4-year moving averages (trend values) for years 1983–1988 are: 1812.5, 1717.5, 1796.25, 1902.5, 2148.75, 2398.75 respectively. These represent the trend component of the time series.

Given: $n = 5$ years. Let $A = 2003$ (middle year) so that $\sum x = 0$.

Step 1: Set up the table.

Step 2: Calculate $a$ and $b$.

Since $\sum x = 0$:
$$a = \frac{\sum y}{n} = \frac{1375}{5} = 275$$
$$b = \frac{\sum xy}{\sum x^2} = \frac{815}{10} = 81.5$$

Step 3: Equation of straight-line trend:
$$\boxed{y = 275 + 81.5x}$$

Step 4: i) Trend values:

Step 5: iii) Expected trend for 2006:
$$x_{2006} = 2006 - 2003 = 3$$
$$\hat{y}_{2006} = 275 + 81.5(3) = 275 + 244.5 = \boxed{₹519.5 \text{ lakh}}$$

Given: $n = 6$ (even number of years). Use coding: $x = \frac{\text{Year} - 1994.5}{0.5}$, so that $\sum x = 0$.

Step 1: Set up the table.

Step 2: Calculate $a$ and $b$.

Since $\sum x = 0$:
$$a = \frac{\sum y}{n} = \frac{1405}{6} = 234.17$$
$$b = \frac{\sum xy}{\sum x^2} = \frac{115}{70} = 1.643$$

Step 3: Equation of straight-line trend:
$$\hat{y} = 234.17 + 1.643x \quad \text{where } x = \frac{\text{Year} - 1994.5}{0.5}$$

Step 4: Trend value for 1998:
$$x_{1998} = \frac{1998 - 1994.5}{0.5} = \frac{3.5}{0.5} = 7$$
$$\hat{y}_{1998} = 234.17 + 1.643 \times 7 = 234.17 + 11.50 = \boxed{245.67 \text{ tons}}$$