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Class 12- Applied Mathematics-NCERT Handbook Solutions-Unit-6-Index Numbers and Time Based Data

NCERT Handbook Solutions
Class 12-Applied Mathematics
Exercise 6.11 - Index Numbers & Time Series

Practice Exercise 6.11

Q1
Calculate index numbers from the following data by simple aggregate method taking prices of 1995 as base period.
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Answer: 120

Method: Simple Aggregative Method

\[ P_{01} = \frac{\sum P_1}{\sum P_0} \times 100 \]

Where \( P_0 \) are prices of 1995 (Base) and \( P_1 \) are prices of 2005 (Current).

CommodityPrice 1995 (\( P_0 \))Price 2005 (\( P_1 \))
A8095
B5060
C90100
D3045
Total (\( \sum \))250300

Calculation:

\[ P_{01} = \frac{300}{250} \times 100 = 1.2 \times 100 = 120 \]
Q2
Construct price index number from the following data using:
i) Laspeyre’s Method
ii) Paasche’s Method
iii) Fisher’s Ideal Method
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Answer: Laspeyre's = 146, Paasche's = 149, Fisher's = 147

1. Calculation Table:
Base Year: 2008 (0), Current Year: 2010 (1)

Comm.\( p_0 \)\( q_0 \)\( p_1 \)\( q_1 \)\( p_1 q_0 \)\( p_0 q_0 \)\( p_1 q_1 \)\( p_0 q_1 \)
P284532162010
Q51261072606050
R41551275606048
S21842072368040
Total251172220148

i) Laspeyre’s Method (\( L \)):

\[ L = \frac{\sum p_1 q_0}{\sum p_0 q_0} \times 100 = \frac{251}{172} \times 100 \approx 145.93 \approx 146 \]

ii) Paasche’s Method (\( P \)):

\[ P = \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100 = \frac{220}{148} \times 100 \approx 148.64 \approx 149 \]

iii) Fisher’s Ideal Method (\( F \)):

\[ F = \sqrt{L \times P} = \sqrt{145.93 \times 148.64} = \sqrt{21691.03} \approx 147.27 \approx 147 \]
Q3
Taking 1995 as base year calculate relative index number for the years 1997-2005.
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Answer:

Formula for Price Relative: \( R = \frac{P_1}{P_0} \times 100 \)

YearPrice (\( P \))Calculation (\( \frac{P}{12} \times 100 \))Index Number
199512\( \frac{12}{12} \times 100 \)100
199714\( \frac{14}{12} \times 100 \)117
199913\( \frac{13}{12} \times 100 \)108
200120\( \frac{20}{12} \times 100 \)167
200325\( \frac{25}{12} \times 100 \)208
200521\( \frac{21}{12} \times 100 \)175
Q4
Compute the weighted aggregative index number for the following data.
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Answer: 137 (approx)

Formula: \( I = \frac{\sum P_1 W}{\sum P_0 W} \times 100 \)

VariableWeight (\( W \))Base Price (\( P_0 \))Curr. Price (\( P_1 \))\( P_0 W \)\( P_1 W \)
X6045240300
Y5023100150
Z30123060
Total370510
\[ I = \frac{510}{370} \times 100 = 137.83 \]

Note: The textbook answer key lists 137, likely truncating the decimal.

Q5
Calculate price index number for 2004 taking 1994 as the base year from the following data by simple aggregative method.
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Answer: 164

(Note: The table says 1990 and 2010, but the question text says 1994 and 2004. We use the data in the table.)

Data:

  • \( \sum P_0 \) (Base) = \( 60 + 40 + 100 + 60 + 90 = 350 \)
  • \( \sum P_1 \) (Current) = \( 140 + 60 + 205 + 70 + 100 = 575 \)

Calculation:

\[ I = \frac{\sum P_1}{\sum P_0} \times 100 = \frac{575}{350} \times 100 = 164.28 \approx 164 \]
Q6
Based on the data on the expenses of middle-class families in a certain city, calculate the cost-of-living index during the year 2003 as compared with 1990.
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Answer: 132

Using the Simple Aggregative Method on the expenses:

  • 1990 Expenses (\( P_0 \)): \( 1400 + 200 + 400 + 200 + 250 = 2450 \)
  • 2003 Expenses (\( P_1 \)): \( 1500 + 250 + 750 + 300 + 425 = 3225 \)
\[ \text{Cost of Living Index} = \frac{3225}{2450} \times 100 = 131.63 \approx 132 \]
Q7
From the data given below, obtain the index of retail sales (interpreted as Price Index) in India for years 1995, 1996, 1997.
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Answer:

The relationship between Value, Volume, and Price is: \( \text{Value} = \text{Price} \times \text{Volume} \).

Therefore, \( \text{Price Index} = \frac{\text{Value Index}}{\text{Volume Index}} \times 100 \).

YearSales Volume IndexSales Value IndexCalculated Index
1995101105\( \frac{105}{101} \times 100 = 104 \)
1996113108\( \frac{108}{113} \times 100 = 96 \)
1997106124\( \frac{124}{106} \times 100 = 117 \)
Q8
Calculate the price index number for the following data using weighted aggregative method.
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Answer: 153

Comm.Weight (\( W \))Base (\( P_0 \))Curr (\( P_1 \))\( P_0 W \)\( P_1 W \)
P149012012601680
Q201017200340
R35406014002100
S1550937501395
Total36105515
\[ \text{Index} = \frac{5515}{3610} \times 100 = 152.77 \approx 153 \]
Q9
Based on the given data, check whether i) Paasche’s formula and, ii) Fisher’s formula will satisfy the time reversal test.
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Answer: Paasche's - No; Fisher's - Yes

Time Reversal Test: Requires \( P_{01} \times P_{10} = 1 \).

1. Paasche's Formula:

\[ P_{01} = \frac{\sum p_1 q_1}{\sum p_0 q_1} \quad \text{and} \quad P_{10} = \frac{\sum p_0 q_0}{\sum p_1 q_0} \]

Calculating using table data:

  • \( \sum p_1 q_1 = 6(15)+4(20)+10(4) = 90+80+40 = 210 \)
  • \( \sum p_0 q_1 = 4(15)+6(20)+8(4) = 60+120+32 = 212 \)
  • \( \sum p_0 q_0 = 4(10)+6(15)+8(5) = 40+90+40 = 170 \)
  • \( \sum p_1 q_0 = 6(10)+4(15)+10(5) = 60+60+50 = 170 \)
\[ P_{01} \times P_{10} = \frac{210}{212} \times \frac{170}{170} = 0.99 \neq 1 \]

Therefore, Paasche's formula does not satisfy the test.

2. Fisher's Formula:

Fisher's Index is mathematically constructed to satisfy the Time Reversal Test. The product of Fisher's index for 0 to 1 and 1 to 0 always results in unity.

\[ \sqrt{\frac{\sum p_1 q_0}{\sum p_0 q_0} \times \frac{\sum p_1 q_1}{\sum p_0 q_1}} \times \sqrt{\frac{\sum p_0 q_1}{\sum p_1 q_1} \times \frac{\sum p_0 q_0}{\sum p_1 q_0}} = 1 \]

Therefore, Fisher's formula satisfies the test.

Q10
The annual rainfall (in cm) was recorded for Cherrapunji, Meghalaya. Determine the trend of rainfall by 3-year moving averages.
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Note: The data values provided in the question image (1.2, 1.9, 2.0...) differ from the data used to generate the answer key (which yields averages around 2.55). The solution below is calculated correctly based on the Question Data provided.
YearRainfall3-Year Total3-Year Moving Avg
20011.2--
20021.95.11.70
20032.05.31.77
20041.45.51.83
20052.14.81.60
20061.35.21.73
20071.84.21.40
20081.14.21.40
20091.3--
Q11
Compute the seasonal indices by 4-year moving averages from the given data of production of paper.
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Method: Centered 4-Year Moving Average

YearValue4-Year Total4-Year AvgCentered Avg
19802450---
19811470---
1982215078701967.51812.5
1983180066301657.51712.5
1984121071101777.51791.25
1985195079601990.01897.5
1986230092302307.52138.75
1987250099602490.02393.75
19882480---
19892680---
Q12
Given below is the data of workers welfare expenses in steel industries during 2001 - 2005. Use method of least squares to find the best fit for a straight-line trend and compute expected sale trend for year 2006.
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Answer: \( Y_t = 275 + 81.5X \); 2006 Prediction: 519.5

Solution:
Center year: 2003 (so \( X=0 \) at 2003). \( n=5 \).

YearExp (\( Y \))\( X \)\( X^2 \)\( XY \)
2001160-24-320
2002185-11-185
2003220000
200430011300
2005510241020
Sum1375010815

Constants:

\[ a = \frac{\sum Y}{n} = \frac{1375}{5} = 275 \] \[ b = \frac{\sum XY}{\sum X^2} = \frac{815}{10} = 81.5 \]

Trend Equation: \( Y_c = 275 + 81.5X \)

Prediction for 2006:
For 2006, \( X = 3 \).
\( Y_{2006} = 275 + 81.5(3) = 275 + 244.5 = 519.5 \)

Q13
Fit a straight-line trend by method of least squares for the following data and also find the trend value for year 1998.
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Answer: \( Y_t = 234 + 1.6X \); 1998 Prediction: 245.2

Solution:
Data has \( n=6 \) (even years). We take deviations from the mid-point (between 1994 and 1995) and multiply by 2 to get integer \( X \) values.

YearProd (\( Y \))\( X \)\( X^2 \)\( XY \)
1992210-525-1050
1993225-39-675
1994275-11-275
199522011220
199624039720
19972355251175
Sum1405070115

Calculations:

\[ a = \frac{\sum Y}{n} = \frac{1405}{6} \approx 234.16 \quad (\text{Rounded to } 234 \text{ in key}) \] \[ b = \frac{\sum XY}{\sum X^2} = \frac{115}{70} \approx 1.64 \quad (\text{Rounded to } 1.6 \text{ in key}) \]

Trend Equation: \( Y_c = 234 + 1.6X \)

Prediction for 1998:
For 1998, \( X = 7 \).
\( Y_{1998} = 234 + 1.6(7) = 234 + 11.2 = 245.2 \) tons.

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