ML Aggarwal Class 12 Maths Solutions Section C Chapter 05 Index Numbers And Moving Averages

Access free ML Aggarwal Class 12 Maths Solutions Section C Chapter 05 Index Numbers And Moving Averages 2026 below. Students can now access free ML Aggarwal Solutions Solutions for Class 12 Mathematics. These chapter-wise exercises are designed by expert math teachers to help you understand complex formulas and score higher marks in your class tests.

Class 12 Math Section C Chapter 05 Index Numbers And Moving Averages ML Aggarwal Solutions Solutions

Get step-by-step ML Aggarwal Solutions Solutions for Section C Chapter 05 Index Numbers And Moving Averages Class 12 Math below. All answers are updated for the 2026 school curriculum, offering step by step methods to help you solve textbook problems easily.

Section C Chapter 05 Index Numbers And Moving Averages ML Aggarwal Solutions Class 12 Solved Exercises

Index Numbers and Moving Averages

 

5.1 Index Numbers

An index number serves as a ratio between two quantities—such as prices, values, or other economic variables—measured at two different points in time. It enables comparison of how much change has occurred relative to data from a base period or reference period.

Index numbers represent a specialized form of average that tracks shifts in the level of an activity or item, measured either across time, geographic areas, or other features. They are presented as either a ratio or a percentage. For instance, when the consumer price index reaches 175 in 2008 compared to 2001, this signals that consumer prices have climbed by 75% over that seven-year span.

Examining index numbers reveals long-term patterns and trends. By selecting an appropriate time window for calculating index numbers, we can identify seasonal patterns, cyclical fluctuations, unexpected (or unusual) shifts, and long-term directions in any measurable activity—whether sales of ice-cream, school attendance rates, literacy growth in a region, joblessness levels, or production figures for specific companies.

The Wholesale Price Index (WPI) and Consumer Price Index (CPI) rank among the most commonly referenced indicators. Both reflect inflation rates and shifts in living standards. The Consumer Price Index draws from five product categories—Food, Housing (Rent), Household goods, Fuel and light, and Miscellaneous. Each grouping contains numerous individual items—for example, Food encompasses Rice, Wheat, Dal, Milk, and comparable goods.

The key features of index numbers include:

  • They take the form of a ratio or percentage.
  • They function as specialized averages.
  • They track changes in the level of a phenomenon.
  • They gauge the impact of change across a time frame.
  • They measure shifts that cannot be assessed directly—that is, they track relative shifts in an economic activity by measuring the underlying factors that drive that activity.

 

5.1.1 Uses of Index Numbers

Index numbers play a vital role in business and economic work. Their primary applications include:

  1. They serve to assess the pulse of the economy. As a result, index numbers function as indicators of economic health.
  2. They aid in developing appropriate policies and making choices about wages, prices, consumption, and related matters.
  3. They show patterns and inclinations. They serve as markers of inflation or deflation patterns.
  4. They are employed to evaluate the real purchasing strength of money.
  5. They support forecasting of future economic behavior.

 

5.1.2 Classification of Index Numbers

Based on the category of activity they track, index numbers fall into four groups:

  1. (i) Price indexes (ii) Quantity indexes (iii) Value indexes (iv) Special purpose indexes

Price indexes measure shifts in price-related characteristics. The Wholesale price index and Consumer price index serve as two samples of Price indexes.

Quantity indexes measure shifts in some quantity (volume) characteristic. Examples include the index of Industrial production, or the index of scooters sold.

Value indexes track changes in some aspect of value. Special Purpose indexes are developed on an as-needed basis to track particular characteristics.

 

5.1.3 Problems in the Construction of Index Numbers

Several considerations merit attention during the development of index numbers:

  1. (i) Defining the purpose of the Index clearly. No single index works for all purposes. For example, if you are building a consumer price index, exclude wholesale prices, and so forth.
  2. (ii) Selecting base year (or base period) carefully. Pick the reference period—against which relative change will be calculated—with care. It should be reasonably recent, not far back in history. It should represent a typical period - free from unusual circumstances such as wars, floods, disease outbreaks, and so on. In certain situations, the chain base approach may replace a fixed base; for instance, each year's prices link to the previous year rather than to one fixed year.
  3. (iii) Selecting the numbers of items to be included. Since all items cannot be included, focus on items that matter and are representative. Additionally, standardize items so they can be readily recognized after time has passed.
  4. (iv) Selection of price quotations and choice of places. After deciding which items and how many to track, select geographic areas (marketplaces, retail outlets) thoughtfully. This ensures you gather a representative mix of price information.
  5. (v) Choice of an average. Because index numbers are specialized averages, determine which averaging method (arithmetic mean, median, mode, geometric mean, or harmonic mean) to apply during construction. Though geometric mean produces superior outcomes, arithmetic mean is frequently selected to reduce computation burden.
  6. (vi) Selection of appropriate weights. Given that different items are purchased in varying quantities, apply suitable weights to represent how significant each item is relative to others.

 

5.1.4 Methods of Construction of Index Numbers

When just one item is involved and its two separate values are available at two separate times (or locations, and so forth), the index number becomes simply the quotient of two figures, shown as a percentage. For instance, if 2 lac cars were registered in 1990 and 10 lac cars in 2000, the (quantity) index equals \( \frac{10 \text{ lac}}{2 \text{ lac}} \times 100 = 500 \). Similarly, if in Mumbai commercial space rent is Rs. 1 per sq. foot per month and in New York it is Rs. 2.50 per sq. foot per month, the index of rental of New York relative to Mumbai is \( \frac{2.50}{1.00} \times 100 = 250 \).

When dealing with multiple items rather than one, prices for multiple goods are supplied for both the present year and the base year. At times, separate weights or amounts are provided for those items as well. A variety of approaches exist for computing index numbers in these cases.

Index number methods:

Unweighted methods:

(i) Simple aggregative method

When \( \Sigma p_1 \) represents the combined total of current-year prices and \( \Sigma p_0 \) represents the combined total of base-year prices, the price index for the current year is:

\[ P_{01} = \frac{\Sigma p_1}{\Sigma p_0} \times 100 \]

(ii) Simple average of price relatives method

Price Relative refers to the current year price divided by the base year price, stated as a percentage - that is, Price Relative = \( \frac{p_1}{p_0} \times 100 \). As an example, if a colour TV cost Rs. 12000 in 1995 and Rs. 18000 in 2008, the price relative is \( \frac{18000}{12000} \times 100 = 150 \).

When multiple items are considered, first compute the price relative for each, then calculate their simple average to find the index number. The formula for determining price index via this method is:

\[ P_{01} = \frac{\Sigma \left(\frac{p_1}{p_0} \times 100\right)}{N} \]
where N is the number of items.

To reduce calculation effort, this form is sometimes used:

\[ P_{01} = \frac{\Sigma \left(\frac{p_1}{p_0}\right) \times 100}{N} \text{ or } \frac{1}{N} \Sigma \left(\frac{p_1}{p_0} \times 100\right) \]

Weighted methods (or Arithmetic Mean method):

(iii) Weighted aggregate method

When base prices and current prices are given together with the weights or amounts for each item, the index number based on weighted aggregates is:

\[ P_{01} = \frac{\Sigma p_1 w}{\Sigma p_0 w} \times 100 \]

(iv) Weighted average of price relatives method

This represents the frequently chosen method for constructing consumer or wholesale price index when base and current prices along with weights or amounts are supplied. Weighted average of price relatives is given by:

\[ P_{01} = \frac{\Sigma \left(\frac{p_1}{p_0} \times 100\right) \times w}{\Sigma w} \]
or

\[ P_{01} = \frac{\Sigma Iw}{\Sigma w} \]
where \( I = \frac{p_1}{p_0} \times 100 \), the price relative.

 

ILLUSTRATIVE EXAMPLES

 

Example 1. Find by simple aggregate method, the index number from the following data:

CommodityBase Price (Rs.)Current Price (Rs.)
Rice3035
Wheat2225
Fish5464
Potato2025
Coal1518

Solution: We build the table shown below:
CommodityBase Price (Rs.) p₀Current Price (Rs.) p₁
Rice3035
Wheat2225
Fish5464
Potato2025
Coal1518
TotalΣp₀ = 141Σp₁ = 167

Therefore, the required index number using the simple aggregate method is:
\[ P_{01} = \frac{\Sigma p_1}{\Sigma p_0} \times 100 = \frac{167}{141} \times 100 = 118.44 \]
Consequently, there is an average rise of roughly 18.44% in the cost of these items.

Exam Tip: Keep column totals organized in a clear table format and double-check your summation before computing the final ratio - arithmetic mistakes here are common.

 

Example 2. In the above example, calculate price relative of Fish and Coal.
Solution: Price relative of fish = \( \frac{\text{current price of fish}}{\text{base price of fish}} \times 100 = \frac{64}{54} \times 100 = 118.5 \)
Price relative of coal = \( \frac{\text{current price of coal}}{\text{base price of coal}} \times 100 = \frac{18}{15} \times 100 = 120 \)

Exam Tip: Price relatives show the percentage change for each individual commodity - verify your division is accurate by checking that larger numbers yield larger price relatives.

 

Example 3. For data in example 1, calculate price index using the price relative method.
Solution: We build the following table:

CommodityBase Price (Rs.) p₀Current Price (Rs.) p₁Price relative p₁/p₀ × 100
Rice3035116.67
Wheat2225113.64
Fish5464118.52
Potato2025125
Coal1518120
TotalΣ p₁/p₀ × 100 = 593.83

The required index number is the simple average of price relatives:
\[ P_{01} = \frac{1}{N} \Sigma \left(\frac{p_1}{p_0} \times 100\right) = \frac{593.83}{5} = 118.77 \]
Observe that the simple aggregate method in example 1 gave us 118.44, which differs slightly from this result.

Exam Tip: Different methods yield slightly different results - always identify which method the question asks for and follow it precisely.

 

Example 4. Let us assume that with prices given in example 1, a Bengali family buys quantities of rice, wheat, fish, potato and coal in the ratio 3:1:3:2:2. Find weighted aggregate price index.
Solution: We build the table shown below:

CommodityBase Price (Rs.) p₀Current Price (Rs.) p₁Weight wp₀wp₁w
Rice3035390105
Wheat222512225
Fish54643162192
Potato202524050
Coal151823036
Total344408

The required index number using weighted aggregates is:
\[ P_{01} = \frac{\Sigma p_1 w}{\Sigma p_0 w} \times 100 = \frac{408}{344} \times 100 = 118.60 \]

Exam Tip: Weights represent the relative importance of each item - ensure you multiply base and current prices separately by their weights before summing.

 

Example 5. With above data (as in example 4), calculate price index using weighted average of price relatives.
Solution: Construct the table below:

CommodityBase Price (Rs.) p₀Current Price (Rs.) p₁Weight wPrice relative I = p₁/p₀ × 100Iw
Rice30353116.67350
Wheat22251113.64113.64
Fish54643118.52355.56
Potato20252125250
Coal15182120240
TotalΣw = 11ΣIw = 1309.2

Therefore, the index using weighted average of price relatives is:
\[ P_{01} = \frac{\Sigma Iw}{\Sigma w} = \frac{1309.2}{11} = 119.02 \]

Exam Tip: Multiply each price relative by its weight, then divide the sum of these products by the sum of weights - this accounts for how frequently each item is purchased.

 

Example 6. With data from Example 1, consider the case of a Punjabi family which uses more wheat than rice or fish. Calculate price index using weighted aggregate as well as using weighted average of price relatives, assuming that weights are 10, 50, 10, 20, 20.
Solution: Using weighted aggregate, the required price index = \( \frac{\Sigma p_1 w}{\Sigma p_0 w} \times 100 \)

CommodityBase Price (Rs.) p₀Current Price (Rs.) p₁Weight wp₀wp₁w
Rice303510300350
Wheat22255011001250
Fish546410540640
Potato202520400500
Coal151820300360
Total26403100

Hence, the price index using weighted aggregates is:
\[ P_{01} = \frac{\Sigma p_1 w}{\Sigma p_0 w} \times 100 = \frac{3100}{2640} \times 100 = 117.42 \]
Now, to calculate the price index using weighted average of price relatives, construct the following table:
CommodityBase Price (Rs.) p₀Current Price (Rs.) p₁Weight wPrice relative I = p₁/p₀ × 100Iw
Rice303510116.671166.7
Wheat222550113.645682
Fish546410118.521185.2
Potato2025201252500
Coal1518201202400
Total11012933.9

Hence, the price index using weighted average of price relatives is:
\[ P_{01} = \frac{\Sigma Iw}{\Sigma w} = \frac{12933.9}{110} = 117.58 \]
Comparing these findings with examples 4 and 5 shows that the Bengali family has experienced a bigger strain than the Punjabi family. Note that wheat prices have risen less compared to rice and fish prices. The Bengali family consumes more fish and rice relative to wheat (check the weights), whereas the Punjabi family consumes more wheat than rice and fish (check the weights). This illustrates the influence weights have on the price index.

Exam Tip: Weights fundamentally shift how much each item affects the final index - a family consuming more of an item with a larger price rise will experience a higher index than one consuming less of it.

 

Example 7. Construct the index number for 1991 taking 1990 as the base year by simple average of price relatives method:

CommodityABCDE
Price in 1990 (Rs.)1008016022040
Price in 1991 (Rs.)14012018024040

Solution: Construct the table as below:
CommodityPrice in 1990 (Rs.) p₀Price in 1991 (Rs.) p₁Price relative p₁/p₀ × 100
A100140140
B80120150
C160180112.5
D220240109.1
E4040100
Total= 611.6

Hence, the required price index using simple average of price relatives is:
\[ P_{01} = \frac{1}{N} \Sigma \left(\frac{p_1}{p_0} \times 100\right) = \frac{611.6}{5} = 122.32 \]

Exam Tip: When one item shows no change (like commodity E), its price relative equals 100 - include it in your total as it contributes to the overall index.

 

Example 8. The price index for the following data for the year 2011 taking 2001 as the base year was 127. The simple average of price relatives method was used. Find the value of x:

ItemsABCDEF
Price (Rs. per unit) in year 2001807050201825
Price (Rs. per unit) in year 201110087.506122x32.50

Solution: Construct the table as below:
ItemsPrice (Rs. per unit) in year 2001 p₀Price (Rs. per unit) in year 2011 p₁Price relative p₁/p₀ × 100
A80100\( \frac{100}{80} \times 100 = 125 \)
B7087.50\( \frac{87.50}{70} \times 100 = 125 \)
C5061\( \frac{61}{50} \times 100 = 122 \)
D2022\( \frac{22}{20} \times 100 = 110 \)
E18x\( \frac{x}{18} \times 100 = \frac{50x}{9} \)
F2532.50\( \frac{32.50}{25} \times 100 = 130 \)
Total\( 612 + \frac{50x}{9} \)

Here, N = total number of items = 6.
Using simple average of price relative method, price index = \( \frac{1}{N} \Sigma \left(\frac{p_1}{p_0} \times 100\right) = \frac{612 + \frac{50x}{9}}{6} = 127 \) (given)

\[ \implies 612 + \frac{50x}{9} = 6 \times 127 \implies \frac{50x}{9} = 762 - 612 \]

\[ \implies \frac{50x}{9} = 150 \implies x = 27 \]

Hence, the value of x = 27.

Exam Tip: When a variable is missing, set up the equation using the index formula, then solve algebraically - be careful with fractions when multiplying out.

 

Example 9. Calculate the index number for 2005 with 2000 as the base year by weighted aggregate method:

CommodityPrice (in Rs.) in the year 2000Price (in Rs.) in the year 2005Weights
A14018010
B4005507
C1002506
D1251508
E2003004
(I.S.C. 2007)
Solution: Construct the table as below:
CommodityBase Price (Rs.) in 2000, p₀Current Price (Rs.) in 2005, p₁Weight wp₀wp₁w
A1401801014001800
B400550728003850
C10025066001500
D125150810001200
E20030048001200
Total66009550

Using weighted aggregate method, index number = \( \frac{\Sigma p_1 w}{\Sigma p_0 w} \times 100 = \frac{9550}{6600} \times 100 = 144.696 = 144.7 \) (approximately).

Exam Tip: Round your final answer appropriately - check the question to see if it specifies decimal places, and follow that guidance consistently.

 

Example 10. Calculate the index number for the year 2006 with 1996 as the base year by the weighted average of price relative method from the following data:

CommodityABCDE
Weight402552010
Price (Rs. per unit) year 199632.0080.001.0010.244.00
Price (Rs. per unit) year 200640.00120.001.0015.363.00
(I.S.C. 2009)
Solution: We construct the table as below:
CommodityWeightBase year 1996 p₀year 2006 p₁Price relative I = p₁/p₀ × 100Iw
A4032.0040.001255000
B2580.00120.001503750
C51.001.00100500
D2010.2415.361503000
E104.003.0075750
Total10013000

Using weighted average of price relative method, index number = \( \frac{\Sigma Iw}{\Sigma w} = \frac{13000}{100} = 130 \).

Exam Tip: When total weights equal 100, the calculation simplifies dramatically - the sum of Iw values becomes the index directly.

 

Example 11. The price relatives and weights of a set of commodities are given below:

CommodityABCD
Price Relative125120127119
Weightx2xyy + 3

If the sum of weights is 40 and the index for the set is 122, find the numerical values of x and y.
Solution: The data may be entered in a table like this:
CommodityWeight wPrice relative IIw
Ax125125x
B2x120240x
Cy127127y
Dy + 3119119y + 357
TotalΣw = 3x + 2y + 3ΣIw = 365x + 246y + 357

Since it is given that sum of weights is 40, we get:
3x + 2y + 3 = 40
\[ \implies 3x + 2y = 37 \quad \ldots(i) \]

Since the index number is 122, we get:
\[ \frac{\Sigma Iw}{\Sigma w} = 122 \implies \frac{365x + 246y + 357}{40} = 122 \]

\[ \implies 365x + 246y + 357 = 4880 - 357 = 4523 \quad \ldots(ii) \]

To solve (i) and (ii), multiply (i) by 123:
\[ 369x + 246y = 4551 \quad \ldots(iii) \]

Subtracting (ii) from (iii), we get:
4x = 28 \( \implies \) x = 7

Putting x = 7 in (i), we get y = 8.

Exam Tip: When weights are given as algebraic expressions, set up two equations - one from the total of weights and one from the index formula - then solve simultaneously.

 

Example 12. The wholesale price index (or price relative) of rice in 2002 compared to 2000 is 130. If the cost of rice was Rs. 12 per kg in 2000, calculate the cost in 2002.
Solution: Let the cost of rice be Rs. p per kg in 2002.
Then, by given:
\[ 130 = \frac{p}{12} \times 100 \]

\[ \implies p = \frac{130 \times 12}{100} = 15.60 \]

Hence, the price of rice in 2002 is Rs. 15.60 per kg.

Exam Tip: Rearrange the price index formula to solve for the unknown price - isolate the variable before substituting numbers.

 

Example 13. During a certain period, the cost of living index number goes from 110 to 200 and the salary of a worker is also raised from Rs. 325 to Rs. 500. Does the worker really gains or loses, and by how much amount in real terms?
Solution: Real wage = \( \frac{\text{Actual wage}}{\text{Cost of living index}} \times 100 \)

So real wage of Rs. 325 = \( \frac{325}{110} \times 100 = \) Rs. 295.45

and real wage of Rs. 500 = \( \frac{500}{200} \times 100 = \) Rs. 250

So the worker actually loses i.e., Rs. (295.45 - 250) = Rs. 45.45 in real terms.

Exam Tip: Real wages account for inflation - a nominal salary increase may actually represent a real wage decrease if inflation has risen faster than the salary increase.

 

Exercise 5.1

 

1. Fill in the blanks:

  1. (i) Index numbers are __________ types of ratio.
  2. (ii) Index numbers are barometers of __________.
  3. (iii) Quantity indexes measure changes in __________ characteristic, compared to __________ period.
  4. (iv) Weighted indexes are __________ to unweighted indexes.
  5. (v) Base period is the period of __________ activities.
  6. (vi) If the price index is 132, it means that price has increased by __________ compared to base period.
  7. (vii) If the price index is 88, it means that price has decreased by __________ compared to base period.

 

2. (i) Price relative of coal is 125 in 2001 compared to 2000. If the coal cost Rs. 8 per kg in 2000, find its cost in 2001.

(ii) Price relative of TV set is 90 in 2001 compared to 2000. If a TV set cost Rs. 9000 in 2000, find its cost in 2001.

(iii) Price relative of sugar is 110 in 2002 compared to 2001. If sugar costs Rs. 16.50 per kg in 2002, what did it cost in 2001?

(iv) Price relative of maize is 80 in 2002 compared to 2001. If maize costs Rs. 12 per kg in 2002, what did it cost in 2001?

 

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

CommoditiesPrice in Rs. in the year 1995Price in Rs. in the year 2005
A45
B6057
C3642

Using 1995 as the base year, calculate a simple aggregate price index for 2005. (I.S.C. 2008)

 

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

CommoditiesPrice in 1989Price in 1990
Butter2021
Cheese1612
Milk33
Eggs2.802.80

(I.S.C 2003)

 

5. From the following data, compute price index by using simple average of price relatives:

Commodities and unitPrice in 1989 (Rs.)Price in 1990 (Rs.)
Butter (kg)20.0021.00
Cheese (kg)15.0014.00
Milk (kg)3.003.00
Bread (l)2.802.80
Eggs (Doz.)6.008.00
Ghee (1 tin)250.00260.00

(I.S.C. 2004)

 

6. Consider the following data:

ItemsUnitsPrice (in Rs.) In 1994 (p₀)Price (in Rs.) In 1998 (p₁)
Wheat1 kg5.607.20
Rice1 kg17.2024.80
Pulses1 kg36.0044.00
Milk1 l24.0030.00
Clothing1 m199.00130.00

Using 1994 as the base year, calculate the index for 1998 correct up to one decimal using (a) simple aggregate method (b) simple average of relatives method.

 

7. Taking 2003 as the base year, with an index number 100, calculate an index number for 2007, based on

(i) simple aggregate (ii) price relatives

derived from the table given below:

CommodityABCD
Price per unit in 200320102540
Price per unit in 200724203040

 

8. Calculate a cost of living index from the following table of prices and weights.

WeightPrice Index
Food35108.5
Rent9102.6
Clothes1097.0
Fuel7100.9
Miscellaneous39103.7

 

9. Construct Index number for following data:

CommodityButterBreadTeaBacon
Relative Index181116110152
Weight41237

 

10. Find the consumer index number for the year 2010 using year 2000 as the base year by using method of weighted aggregates:

CommodityABCDE
2000 Price per unit (Rs.)16400-505-122
2010 Price per unit (Rs.)20600-506-251-50
Weights402552010

(I.S.C. 2012)

 

11. Based on year 1988 as base, the index numbers for 1988, 1989, 1990, 1991 and 1992 are 100, 110, 120, 200 and 400. Now taking 1992 as base year, calculate index numbers for years 1988, 1989, 1990, 1991 and 1992.

 

12. The price quotations of four different commodities for 2001 and 2009 are as given below. Calculate the index number for 2009 with 2001 as the base year by using weighted average of price relative method.

CommodityWeightPrice (in Rs.) 2009Price (in Rs.) 2001
A109.004.00
B494.405.00
C369.006.00
D43.602.00

(I.S.C. 2011)

 

13. Calculate the index number for the year 1979 with 1970 as base from the following data using weighted average of price relatives:

CommodityweightsPrice (in Rs.) 1970Price (in Rs.) 1979
A222.506.20
B483.304.40
C176.2512.75
D130.650.90

(I.S.C. 2005)

 

19. Find the consumer price index for 1994 on the base of 1988, from the following data, using the method of weighted relatives:

ItemFoodRentClothingFuelMiscellaneous
Price in 1988 (in Rs.)20010015050100
Price in 1994 (in Rs.)280200120100200
Weight3020201020

(I.S.C. 1997)

 

20. The following table shows the prices per unit in 1980 and 1984 with weights of commodities A, B, C, D:

CommodityWeightsPrice per unit in 1980Price per unit in 1984
A202530
B252030
C155070
D40510

Taking 1980 as base year with index number 100, calculate the index number of 1984 based on weighted average of price relatives. (I.S.C. 2000)

 

21. Taking 1975 as the base year, with an index number 100, calculate an index number for 1979, based on weighted average of price relatives from the table given below:

CommodityABCD
Weight30152530
Price per unit in 19752010540
Price per unit in 197924203040

(I.S.C. 2002)

 

5.2 Moving Averages

Imagine this data: monthly ice cream sales over one year; yearly rainfall across the last 20 years; weekly price index for 52 weeks. Data collected at regular intervals over time is called a time series. Usually these intervals are the same length. Frequently, analyzing a time series—whether for short-term or long-term patterns—is important. The long-term direction, known as secular trend, typically gets calculated using a regression line:

\[ y - \bar{y} = b_{yx} (x - \bar{x}) \]

Three additional types of variation are significant:

  1. Seasonal variation. For instance, soft drink and ice cream sales jump higher in summer compared to winter. Crockery purchases spike during holiday seasons (Diwali, Christmas, and so forth) versus other months.
  2. Cyclical variation. You may have read about the rise and fall of the Roman empire. Fashion publications discuss the rise and fall of hemlines. Stock markets go up, down, up, down like a yoyo. The challenge is we lack certainty about cycle length (or we'd all be billionaires!), yet such cyclical movements show up across many time series.
  3. Irregular variations. When mustard oil faced a sudden ban, Soya oil sales climbed sharply and abnormally. When elections were called, printing press revenue jumped unexpectedly. When floods struck, crop yields dropped sharply and abnormally. These jumps in values stem from some unusual event.

This examination suggests that for analysis or forecasting, simple regression lines are often insufficient. What's fundamentally necessary is "smoothening of curve".

 

5.2.1 Purpose of Moving Averages

Moving averages get used to handle cyclical shifts by reducing swings from cyclical patterns in time series. Cyclical swings get reduced by calculating the mean of values across a set span of successive years (months or weeks, and so forth). The span itself—measured in years, months, weeks, and so on—is based on how long each cycle lasts in the time series. The duration covered while taking these averages is called the period of the cycle.

 

5.2.2 Method for Finding Moving Averages

The average of values across a span of years (months or weeks, and so forth) is calculated and placed at the midpoint of the span. When the span chosen matches the duration of one cycle (or two cycles, or several cycles), this causes cycles to vanish.

For an annual time series x₁, x₂, x₃, …, xₙ:

(i) 3-yearly moving averages:

\[ \frac{x_1 + x_2 + x_3}{3}, \frac{x_2 + x_3 + x_4}{3}, \frac{x_3 + x_4 + x_5}{3}, \ldots \]
which are placed against years 2, 3, 4, … respectively.

(ii) 5-yearly moving averages:

\[ \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5}, \frac{x_2 + x_3 + x_4 + x_5 + x_6}{5}, \ldots \]
which are placed against years 3, 4, … respectively.

(iii) 4-yearly moving averages:

\[ \frac{x_1 + x_2 + x_3 + x_4}{4}, \frac{x_2 + x_3 + x_4 + x_5}{4}, \ldots \]
which are placed against years 2.5, 3.5, … respectively. Furthermore, to sync the time axis between the moving averages and the raw data, we average pairs of consecutive moving averages. The first and second moving average, averaged together, is placed against \( \frac{2.5 + 3.5}{2} = \) 3rd year; the second and third moving average, averaged together, is placed against \( \frac{3.5 + 4.5}{2} = \) 4th year, and so on.

This is called 4-yearly centred moving average.

Note. When the span is even, a centred moving average calculation is needed.

The following examples make the above concept clear.

 

ILLUSTRATIVE EXAMPLES
Example 1. (i) Obtain the three year moving averages for the following series of observations.

Year19951996199719981999200020012002
Annual Sales (In 0000 Rs.)3.64.34.33.44.45.43.42.4

(ii) Obtain the five year moving average.
(iii) Construct also the 4-year centred moving average.
Solution: (i) The first 3-year moving average is \( \frac{3.6 + 4.3 + 4.3}{3} = \frac{12.2}{3} = 4.067 \), and is placed against 2nd year i.e. 1996; the second 3-year moving average is \( \frac{4.3 + 4.3 + 3.4}{3} = \frac{12.0}{3} = 4.0 \), and is placed against 3rd year i.e. 1997, and so on. Thus, we have:
Calculation of 3-year moving averages:
YearAnnual sale3-year moving total3-year moving average
19953.6-1/3 -
19964.312.24.067
19974.312.04.00
19983.412.14.03
19994.413.24.40
20005.413.24.40
20013.411.23.73
20022.4--

(ii) The first 5-yearly moving average is \( \frac{3.6 + 4.3 + 4.3 + 3.4 + 4.4}{5} = \frac{20.0}{5} = 4.00 \), and is placed against 3rd year i.e. 1997. The second 5-yearly moving average is \( \frac{4.3 + 4.3 + 3.4 + 4.4 + 5.4}{5} = \frac{21.8}{5} = 4.36 \), and is placed against 4th year i.e. 1998, and so on. Thus, we have:
Calculation of 5-year moving averages:
YearAnnual sale5-year moving total5-year moving average
19953.6--
19964.3-1/5 -
19974.320.04.00
19983.421.84.36
19994.420.94.18
20005.419.03.80
20013.4--
20022.4--

(iii) With 4-year moving averages, the first stage—averaging 4 values—places them between years. To synchronize with the time axis of the original data, we average each pair of successive moving averages. Thus, we produce the following table:

Exam Tip: When working with moving averages, carefully track which year each average aligns with - for odd-period averages, align with the middle year; for even-period averages, use centred moving averages to align properly with the original data timeline.

 

Question 5. The following data relate to the pay of workers employed at a factory.

Type of workerRate of pay (Rs/hour)Average number of hours worked per week - JuneAverage number of hours worked per week - NovemberNumber of workers employed
Skilled6.503738.526
Semi-skilled5.1038.539.514
Unskilled3.5040.54042

Calculate a weighted aggregate index of average weekly pay for November (June = 100), using the number of workers employed as a weighting factor.
Answer: First, find the average weekly pay for each worker type in both June and November by multiplying the hourly rate by the average hours worked per week. For Skilled workers: June = 6.50 × 37 = 240.5; November = 6.50 × 38.5 = 250.25. For Semi-skilled workers: June = 5.10 × 38.5 = 196.35; November = 5.10 × 39.5 = 201.45. For Unskilled workers: June = 3.50 × 40.5 = 141.75; November = 3.50 × 40 = 140.

Now apply the weighted aggregate index formula using the number of workers as weights:

Weighted Aggregate Index = (Sum of November pay × weights) / (Sum of June pay × weights) × 100

Sum of November pay × weights = (250.25 × 26) + (201.45 × 14) + (140 × 42) = 6506.5 + 2820.3 + 5880 = 15206.8

Sum of June pay × weights = (240.5 × 26) + (196.35 × 14) + (141.75 × 42) = 6253 + 2748.9 + 5953.5 = 14955.4

Weighted Aggregate Index = (15206.8 / 14955.4) × 100 = 101.68 (approximately 101.7)

In simple words: A weighted index shows how pay changed from June to November. We multiply each worker type's pay by how many workers there are, then compare November's total to June's total.

Exam Tip: Always ensure that the weights (number of workers) are applied consistently to both the base period and current period values. Double-check your multiplication of rates by hours before applying weights.

 

Question 6. Calculate the 5 yearly moving averages of the number of students in a college from the following data and plot them on a graph paper.

Year1981198219831984198519861987198819891990
No. of students332317357392402405510427405438

(I.S.C. 2012)
Answer: The 5-yearly moving average is calculated by taking the average of 5 consecutive years of data, then shifting one year forward and repeating. Beginning with years 1981-1985: (332 + 317 + 357 + 392 + 402) / 5 = 1800 / 5 = 360. Next, years 1982-1986: (317 + 357 + 392 + 402 + 405) / 5 = 1873 / 5 = 374.6. Then, years 1983-1987: (357 + 392 + 402 + 405 + 510) / 5 = 2066 / 5 = 413.2. Next, years 1984-1988: (392 + 402 + 405 + 510 + 427) / 5 = 2136 / 5 = 427.2. Then, years 1985-1989: (402 + 405 + 510 + 427 + 405) / 5 = 2149 / 5 = 429.8. Finally, years 1986-1990: (405 + 510 + 427 + 405 + 438) / 5 = 2185 / 5 = 437.

The 5-yearly moving averages are: 360, 374.6, 413.2, 427.2, 429.8, 437. These values are plotted on a graph paper and connected by a line, which is positioned at the middle year of each 5-year block. The graph reveals an overall upward trend in student enrollment across the period, with some fluctuations but consistent growth especially between 1983 and 1988.

In simple words: A moving average smooths out changes by looking at groups of 5 years together. Each new average shifts forward by one year, which helps you see the general direction and trend more clearly.

Exam Tip: Plot the moving average points at the middle year of each 5-year period (i.e., years 1983, 1984, 1985, 1986, 1987, 1988). Connect these points with a smooth line to show the trend clearly.

 

Question 7. The number of traffic offences committed in a certain city over a period of 3 years is given in the following table.

PeriodJan.-MarchApril-JuneJuly-Sept.Oct.-Dec.
200074564869
200183524981
200294604879

Draw a graph illustrating these figures. Calculate suitable moving averages and plot them on the same graph. Comment on the result.
Answer: First, plot the original data as points on a graph with quarters on the horizontal axis (Jan.-March 2000, April-June 2000, etc.) and the number of offences on the vertical axis. The raw data shows considerable fluctuation within and across years, with peaks in the first quarter (Jan.-March) of each year rising from 74 to 83 to 94, and valleys in the third quarter (July-Sept.), which remains relatively stable around 48-49 offences.

Using a 4-quarterly moving average (since data is quarterly), calculate averages by taking each group of 4 consecutive quarters: (74 + 56 + 48 + 69) / 4 = 61.75 for the first moving average (positioned at the middle of the first year). Next, (56 + 48 + 69 + 83) / 4 = 64, then (48 + 69 + 83 + 52) / 4 = 63, then (69 + 83 + 52 + 49) / 4 = 63.25, then (83 + 52 + 49 + 81) / 4 = 66.25, then (52 + 49 + 81 + 94) / 4 = 69, then (49 + 81 + 94 + 60) / 4 = 71, then (81 + 94 + 60 + 48) / 4 = 70.75, and finally (94 + 60 + 48 + 79) / 4 = 70.25. Plot these moving averages on the same graph and connect them with a line.

The moving average line smooths out the short-term variations present in the raw data, revealing a clearer upward trend overall. Traffic offences generally increase from around 61-64 offences per quarter in 2000 to approximately 70 offences per quarter by late 2002. However, the fourth quarter (Oct.-Dec.) consistently shows higher offence numbers across all three years compared to other quarters, suggesting seasonal or cyclical patterns in traffic violations. The first quarter also tends to record higher offences than the middle quarters, indicating potential seasonal factors - possibly related to weather, holiday travel, or increased vehicle movement - that influence traffic violation rates.

In simple words: The moving average removes the ups and downs, showing the real trend. Traffic offences go up over time, and there are more offences in the first and last quarters of each year.

Exam Tip: Always use a 4-quarterly moving average for quarterly data to eliminate seasonal effects. Mention both the overall trend and any seasonal patterns you observe - this comprehensive analysis scores full marks.

 

Exercise 5.1

1. (i) specialised (ii) economic activity (iii) quantity (or volume), base (or fixed) (iv) preferred (v) normal (vi) 32% (vii) 12%
2. (i) Rs 10 per kg (ii) Rs 8100 (iii) Rs 15 per kg (iv) Rs 15 per kg
3. 104
4. 95
5. 105.9
6. (i) 129.1 (ii) 130.0
7. (i) 114 (ii) 135
8. 104.4
9. 135
10. 138.39
11. 25.0, 27.5, 30.0, 50.0, 100
12. 128.10
13. 171.24
14. 164.05
15. 137.27
16. 171.23
17. x = 80, y = 70
18. x = 40, y = 53
19. 158
20. 162.5
21. 246

 

Exercise 5.2

1. 3.667, 5.333, 6.667, 8.333, 10.333
2. 47, 55, 52.33, 60.33, 64.33
3. 4.67, 5.33, 6, 7, 8, 8.33, 9
4. 2.33, 5.67, 8.33, 16.67, 27.67, 27.67, 29.67, 20.67, 20, 10.33, 5.33, 2
5. The 3-day moving averages are 2.3, 5.6, 12.3, 19.6, 31.0, 34.3, 26.3, 20.0, 11.3, 5.3, 2.0. This pattern shows a steady increase (arrival of fresh cases every day) and then decrease (control over epidemic).
6. 470, 484.6, 503.2, 515.2, 517.8, 523.6
7. 368, 381.6, 390.4, 402.2, 419.8, 439.8

Free study material for Mathematics

Download ML Aggarwal Solutions Solutions for Class 12 Math PDF

You can easily download the complete chapter-wise PDF for ML Aggarwal Class 12 Maths Solutions Section C Chapter 05 Index Numbers And Moving Averages on Studiestoday.com. Our expert-curated ML Aggarwal Solutions Solutions for Class 12 Mathematics are fully optimized for quick revision before your upcoming weekly tests and terminal exams.

Explore More Study Resources for Class 12 Math

Beyond these ML Aggarwal Solutions chapters, you can access free online mock tests, printable sample papers, syllabus details, and short revision notes for the 2026 academic session across our platform.

FAQs

Are these ML Aggarwal Solutions Solutions for Class 12 updated for the 2026 session?

Yes, all solved questions and step-by-step exercises provided on this page are updated based on the latest 2026 edition of the ML Aggarwal Solutions textbook matching the current school curriculum

Can I download Section C Chapter 05 Index Numbers And Moving Averages solutions in PDF format for free on Studiestoday?

Absolutely. You can easily download printable PDF versions of <strong>ML Aggarwal Class 12 Maths Solutions Section C Chapter 05 Index Numbers And Moving Averages</strong> entirely for free. Simply click the download button on our portal to save it for offline study

Who prepared these ML Aggarwal Solutions Class Class 12 Solutions?

These chapter-wise answers for Class 12 Mathematics have been meticulously solved and verified by expert math teachers who specialize in the ML Aggarwal Solutions curriculum

Will practicing ML Aggarwal Solutions Class 12 Math problems help me score better in exams?

Yes, practicing these exercises thoroughly will significantly improve your foundational concepts. The step-by-step layout helps you understand how formulas are applied, ensuring you score top marks in your Class 12 tests and school examinations.

How should I use these ML Aggarwal Solutions solutions for Section C Chapter 05 Index Numbers And Moving Averages?

We highly recommend trying to solve the Section C Chapter 05 Index Numbers And Moving Averages textbook questions on your own first. Use these expert solutions to double-check your calculations, rectify mistakes, and learn faster shortcuts for complex math problems.