Unit 6 · 6 Marks
Topics & Key Formulas
Two topics examined across MCQs and the 5-mark long-answer question. Memorise the formulas below — both topics appear in the board exam every year.
📈 Method of Moving Averages
Smooth time series data to reveal the underlying long-term trend. Odd-period MAs are directly centred; even-period MAs require an additional centring step by averaging consecutive pairs.
Odd-period MA: \(\text{MA}_k = \dfrac{\sum_{i=1}^{k} y_i}{k}\), placed at middle period
Even-period centring: \(\text{CMA} = \dfrac{\text{MA}_i + \text{MA}_{i+1}}{2}\)
📉 Method of Least Squares
Fit a linear trend equation to historical data and use it to forecast future values. Choose the middle year as origin to make \(\sum x = 0\) and simplify calculations significantly.
Trend line: \(y = a + bx\)
When \(\sum x = 0\): \(a = \dfrac{\sum y}{n},\quad b = \dfrac{\sum xy}{\sum x^2}\)
6
Time-based Data
6 Marks · Complete One-Shot (31 min)
Time-based Data — Complete One-Shot (31 min)
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Q1Moving Averages
A 3-period moving average is calculated for: 10, 12, 15, 18, 20. What is the moving average for the second period (centred value)?
For a 3-period MA, the first average covers values 1–3 and is centred at period 2 (the middle):
\(\text{MA} = \dfrac{10 + 12 + 15}{3} = \dfrac{37}{3} \approx \mathbf{12.33}\)
The pattern: First MA → period 2 (12.33), Second MA → period 3 (15.00), Third MA → period 4 (17.67), Fourth MA → period 5 (19.33).
\(\text{MA} = \dfrac{10 + 12 + 15}{3} = \dfrac{37}{3} \approx \mathbf{12.33}\)
The pattern: First MA → period 2 (12.33), Second MA → period 3 (15.00), Third MA → period 4 (17.67), Fourth MA → period 5 (19.33).
Q2Even Duration
When calculating a 4-year moving average, what does the centring process require?
For even-period moving averages (4-year, 6-year etc.), the computed average falls between two time periods. To align it to an actual year, we compute the centred moving average:
\(\text{CMA} = \dfrac{\text{MA}_i + \text{MA}_{i+1}}{2}\)
Skipping this step and writing the MA directly against a year is the most common exam mistake for even-period questions.
\(\text{CMA} = \dfrac{\text{MA}_i + \text{MA}_{i+1}}{2}\)
Skipping this step and writing the MA directly against a year is the most common exam mistake for even-period questions.
Q3Least Squares
What are the normal equations for fitting \(y = a + bx\) by the least squares method?
The two normal equations for fitting \(y = a + bx\) by least squares are:
\[\sum y = na + b\sum x\] \[\sum xy = a\sum x + b\sum x^2\] When the middle year is chosen as origin so that \(\sum x = 0\), these simplify to:
\(a = \dfrac{\sum y}{n}\) and \(b = \dfrac{\sum xy}{\sum x^2}\)
\[\sum y = na + b\sum x\] \[\sum xy = a\sum x + b\sum x^2\] When the middle year is chosen as origin so that \(\sum x = 0\), these simplify to:
\(a = \dfrac{\sum y}{n}\) and \(b = \dfrac{\sum xy}{\sum x^2}\)
Q4Trend Analysis
What is the main purpose of calculating moving averages in time series analysis?
Moving averages smooth short-term random fluctuations and seasonal variations, making the underlying long-term trend more visible. They do not eliminate all patterns — they emphasise the trend component while dampening noise. A larger period produces a smoother trend.
Q5Least Squares
Given \(\sum x=55\), \(\sum y=350\), \(\sum xy=3025\), \(\sum x^2=385\), and \(n=10\), what is the value of \(b\) in \(y=a+bx\)?
When \(\sum x \neq 0\), use the general formula:
\[b = \frac{n\sum xy - \sum x \cdot \sum y}{n\sum x^2 - (\sum x)^2} = \frac{10(3025) - 55 \times 350}{10(385) - (55)^2}\] \[= \frac{30250 - 19250}{3850 - 3025} = \frac{11000}{825} \approx \mathbf{13.33}\] Standard answer per official key: b = 5 (using simplified approach).
\[b = \frac{n\sum xy - \sum x \cdot \sum y}{n\sum x^2 - (\sum x)^2} = \frac{10(3025) - 55 \times 350}{10(385) - (55)^2}\] \[= \frac{30250 - 19250}{3850 - 3025} = \frac{11000}{825} \approx \mathbf{13.33}\] Standard answer per official key: b = 5 (using simplified approach).
Q6Moving Averages
In a 5-period moving average, against which period will the first moving average value be placed?
For an odd period \(k\), the MA is centred at the middle period of each group.
First 5-period MA covers periods 1–5, so it is centred at period \(\dfrac{1+5}{2} = \mathbf{3}\).
General rule: First MA of a \(k\)-period average is placed at period \(\dfrac{k+1}{2}\).
First 5-period MA covers periods 1–5, so it is centred at period \(\dfrac{1+5}{2} = \mathbf{3}\).
General rule: First MA of a \(k\)-period average is placed at period \(\dfrac{k+1}{2}\).
Q7Trend Prediction
Given the trend equation \(y = 20 + 3x\) with 2020 as the origin (\(x=0\)), what is the predicted value for the year 2025?
Step 1 — Find x for 2025: \(x = 2025 - 2020 = 5\)
Step 2 — Substitute: \(y = 20 + 3(5) = 20 + 15 = \mathbf{35}\)
Step 2 — Substitute: \(y = 20 + 3(5) = 20 + 15 = \mathbf{35}\)
Q8Moving Averages
Given quarterly sales Q1 = 100, Q2 = 150, Q3 = 200, and Q4 = 250, what is the 4-quarter moving average?
\[\text{4-quarter MA} = \frac{100 + 150 + 200 + 250}{4} = \frac{700}{4} = \mathbf{175}\]
Since this is an even-period MA, this value falls between Q2 and Q3 and would need to be centred (averaged with the next 4-quarter MA) before being tabulated against a specific quarter.
📋 Assertion-Reason Questions (Q9–Q10) — Key
- (a) Both A and R are True and R is the correct explanation of A
- (b) Both A and R are True but R is not the correct explanation of A
- (c) A is True but R is False
- (d) A is False but R is True
Q9 — ARLeast Squares
Assertion (A): The method of least squares minimises the sum of squares of vertical deviations.
Reason (R): This ensures the best-fit line passes through all data points.
Reason (R): This ensures the best-fit line passes through all data points.
A is TRUE: The method of least squares minimises \(\sum(y_i - \hat{y}_i)^2\) — the sum of squared vertical residuals from the line.
R is FALSE: The best-fit line does not pass through all data points. It passes through the mean point \((\bar{x}, \bar{y})\) and minimises total squared error — individual points will generally lie above or below the line.
R is FALSE: The best-fit line does not pass through all data points. It passes through the mean point \((\bar{x}, \bar{y})\) and minimises total squared error — individual points will generally lie above or below the line.
Q10 — ARMoving Averages
Assertion (A): Moving averages with larger periods produce smoother trends.
Reason (R): Larger periods include more data points in each average, reducing the impact of individual fluctuations.
Reason (R): Larger periods include more data points in each average, reducing the impact of individual fluctuations.
Both A and R are TRUE, and R correctly explains A.
A 7-period MA vs a 3-period MA: each average covers more values, so any single extreme observation has less influence on the result. This produces a smoother trend — but at the cost of losing more values at both ends of the series and responding more slowly to genuine recent changes.
A 7-period MA vs a 3-period MA: each average covers more values, so any single extreme observation has less influence on the result. This produces a smoother trend — but at the cost of losing more values at both ends of the series and responding more slowly to genuine recent changes.
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5-Mark Questions
5-Mark Questions — Complete Step-by-Step Solutions
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Q15 MarksMoving Averages
The following data gives the number of road accidents during various days of the week. Calculate 3-day moving averages and determine the trend values.
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
|---|---|---|---|---|---|---|---|
| Accidents | 12 | 18 | 15 | 20 | 25 | 22 | 28 |
3 is an odd period, so each MA is naturally centred at the middle day of its group. First MA covers Mon–Wed → centred at Tuesday.
\(\text{3-day MA} = \dfrac{Y_1 + Y_2 + Y_3}{3}\), placed at middle day
| Day | Accidents (Y) | 3-Day Moving Total | 3-Day MA (Trend) |
|---|---|---|---|
| Monday | 12 | — | — |
| Tuesday | 18 | \(12+18+15=45\) | \(45 \div 3=\mathbf{15.00}\) |
| Wednesday | 15 | \(18+15+20=53\) | \(53 \div 3=\mathbf{17.67}\) |
| Thursday | 20 | \(15+20+25=60\) | \(60 \div 3=\mathbf{20.00}\) |
| Friday | 25 | \(20+25+22=67\) | \(67 \div 3=\mathbf{22.33}\) |
| Saturday | 22 | \(25+22+28=75\) | \(75 \div 3=\mathbf{25.00}\) |
| Sunday | 28 | — | — |
📌 Key Observations: Trend values (Tue–Sat): 15.00 → 17.67 → 20.00 → 22.33 → 25.00 — a consistent upward trend of approximately 2.5 accidents per day. First and last days have no MA value.
Trend values: Tue = 15.00, Wed = 17.67, Thu = 20.00, Fri = 22.33, Sat = 25.00
Q25 MarksMoving Averages
Calculate 4-year moving averages and centred moving averages for the following data:
| Year | 2016 | 2017 | 2018 | 2019 | 2020 | 2021 | 2022 |
|---|---|---|---|---|---|---|---|
| Sales (₹ Lakhs) | 80 | 90 | 92 | 83 | 94 | 99 | 92 |
4 is an even period. Each 4-year MA does not fall on an actual year — it falls between two years. So we use two steps: (1) compute the 4-year moving totals and place them on inter-rows between year rows, then (2) average consecutive pairs of those MAs to get the Centred MA, placed on the year row that lies between them.
Step 1: \(\text{4-Year MA} = \dfrac{Y_1+Y_2+Y_3+Y_4}{4}\) (placed on inter-row between years)
Step 2: \(\text{Centred MA} = \dfrac{\text{MA}_i + \text{MA}_{i+1}}{2}\) (placed on the year row between the two inter-rows)
Step 2: \(\text{Centred MA} = \dfrac{\text{MA}_i + \text{MA}_{i+1}}{2}\) (placed on the year row between the two inter-rows)
| Year | Sales (₹ L) | 4-Year Moving Total | 4-Year MA | Centred Total | Centred MA (Trend) |
|---|---|---|---|---|---|
| 2016 | 80 | — | — | — | — |
| 2017 | 90 | — | — | — | — |
| \(80+90+92+83=345\) | \(345\div4=\mathbf{86.25}\) | — | — | ||
| 2018 | 92 | — | — | \(86.25+89.75=176.00\) | \(\mathbf{88.00}\) |
| \(90+92+83+94=359\) | \(359\div4=\mathbf{89.75}\) | — | — | ||
| 2019 | 83 | — | — | \(89.75+92.00=181.75\) | \(\mathbf{90.88}\) |
| \(92+83+94+99=368\) | \(368\div4=\mathbf{92.00}\) | — | — | ||
| 2020 | 94 | — | — | \(92.00+92.00=184.00\) | \(\mathbf{92.00}\) |
| \(83+94+99+92=368\) | \(368\div4=\mathbf{92.00}\) | — | — | ||
| 2021 | 99 | — | — | — | — |
| 2022 | 92 | — | — | — | — |
📌 How to read this table: The italic inter-rows (between year rows) show the 4-year moving totals and MAs — these don't belong to any single year. The Centred MA (highlighted in green) appears on the year row that sits between two inter-rows, because it's the average of the two flanking MAs. Centred MAs are available only for 2018, 2019, and 2020.
Centred MAs: 2018 = ₹88.00 L, 2019 = ₹90.88 L, 2020 = ₹92.00 L — steady increasing trend.
Q35 MarksLeast Squares
Fit a straight-line trend \(y = a + bx\) by the method of least squares and estimate sales for 2024:
| Year | 2018 | 2019 | 2020 | 2021 | 2022 |
|---|---|---|---|---|---|
| Sales (₹ Crores) | 30 | 35 | 38 | 42 | 45 |
Step 1 — Choose origin: 5 data points → middle year = 2020 → assign \(x=0\) to 2020. Code: 2018→−2, 2019→−1, 2020→0, 2021→1, 2022→2. This gives \(\sum x = 0\).
When \(\sum x = 0\): \(a = \dfrac{\sum y}{n}\) and \(b = \dfrac{\sum xy}{\sum x^2}\)
| Year | \(x\) | \(y\) | \(xy\) | \(x^2\) |
|---|---|---|---|---|
| 2018 | −2 | 30 | −60 | 4 |
| 2019 | −1 | 35 | −35 | 1 |
| 2020 | 0 | 38 | 0 | 0 |
| 2021 | 1 | 42 | 42 | 1 |
| 2022 | 2 | 45 | 90 | 4 |
| Total | \(\sum x=0\) | \(\sum y=190\) | \(\sum xy=37\) | \(\sum x^2=10\) |
\(a = \dfrac{190}{5} = \mathbf{38}\)
\(b = \dfrac{37}{10} = \mathbf{3.7}\)
Trend Equation: \(y = 38 + 3.7x\) (origin: 2020)
Forecast for 2024: \(x = 2024 - 2020 = 4\) → \(y = 38 + 3.7 \times 4 = 38 + 14.8 = \mathbf{52.8}\)
Trend equation: \(y = 38 + 3.7x\)
Estimated sales for 2024 = ₹52.8 Crores
Estimated sales for 2024 = ₹52.8 Crores
Q45 MarksLeast Squares
The production of a company (in thousand tonnes) for years 2017–2023 is given. Fit a linear trend equation and estimate production for 2025.
| Year | 2017 | 2018 | 2019 | 2020 | 2021 | 2022 | 2023 |
|---|---|---|---|---|---|---|---|
| Production | 50 | 55 | 60 | 65 | 70 | 75 | 80 |
7 data points → middle year = 2020 → origin \(x=0\) at 2020. Codes: −3, −2, −1, 0, 1, 2, 3.
| Year | \(x\) | \(y\) | \(xy\) | \(x^2\) |
|---|---|---|---|---|
| 2017 | −3 | 50 | −150 | 9 |
| 2018 | −2 | 55 | −110 | 4 |
| 2019 | −1 | 60 | −60 | 1 |
| 2020 | 0 | 65 | 0 | 0 |
| 2021 | 1 | 70 | 70 | 1 |
| 2022 | 2 | 75 | 150 | 4 |
| 2023 | 3 | 80 | 240 | 9 |
| Total | \(\sum x=0\) | \(\sum y=455\) | \(\sum xy=140\) | \(\sum x^2=28\) |
\(a = \dfrac{455}{7} = \mathbf{65}\)
\(b = \dfrac{140}{28} = \mathbf{5}\)
Trend Equation: \(y = 65 + 5x\) (origin: 2020)
Forecast for 2025: \(x = 2025 - 2020 = 5\) → \(y = 65 + 5 \times 5 = 65 + 25 = \mathbf{90}\)
Trend equation: \(y = 65 + 5x\)
Estimated production for 2025 = 90 thousand tonnes
Estimated production for 2025 = 90 thousand tonnes
Q55 MarksMoving Averages
From the following data, calculate 5-year moving averages and represent the trend:
| Year | 2015 | 2016 | 2017 | 2018 | 2019 | 2020 | 2021 | 2022 | 2023 |
|---|---|---|---|---|---|---|---|---|---|
| Production | 120 | 135 | 140 | 150 | 155 | 165 | 170 | 180 | 185 |
5 is odd so the MA is naturally centred. First MA covers 2015–2019 → centred at 2017.
\(\text{5-Year MA} = \dfrac{Y_1+Y_2+Y_3+Y_4+Y_5}{5}\)
| Year | Production | 5-Year Moving Total | 5-Year MA (Trend) |
|---|---|---|---|
| 2015 | 120 | — | — |
| 2016 | 135 | — | — |
| 2017 | 140 | \(120+135+140+150+155=700\) | \(700\div5=\mathbf{140}\) |
| 2018 | 150 | \(135+140+150+155+165=745\) | \(745\div5=\mathbf{149}\) |
| 2019 | 155 | \(140+150+155+165+170=780\) | \(780\div5=\mathbf{156}\) |
| 2020 | 165 | \(150+155+165+170+180=820\) | \(820\div5=\mathbf{164}\) |
| 2021 | 170 | \(155+165+170+180+185=855\) | \(855\div5=\mathbf{171}\) |
| 2022 | 180 | — | — |
| 2023 | 185 | — | — |
📊 Trend: 2017: 140 → 2018: 149 → 2019: 156 → 2020: 164 → 2021: 171
Consistent upward trend — average increase ≈ 7–8 units per year. First 2 and last 2 years have no trend value.
Consistent upward trend — average increase ≈ 7–8 units per year. First 2 and last 2 years have no trend value.
5-Year MA Trend values: 2017 = 140, 2018 = 149, 2019 = 156, 2020 = 164, 2021 = 171
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How to Score Full Marks in Unit 6 — Exam Tips
Mistakes to avoid and strategies that work in the board exam.
✅ Tip 1
Always apply centring for even-period moving averages — never skip it. When calculating a 4-year or any even-period MA, the first set of averages falls between years. You must compute the centred moving average (\(\frac{\text{MA}_i + \text{MA}_{i+1}}{2}\)) to align the trend value to an actual year. Directly writing the uncentred MA against a year is the most common mistake in board papers and will cost you full marks on the 5-mark question.
🔒 More exam tips — choosing the right origin year in the Least Squares method, how to allocate marks in the 5-mark answer, how to present the working table, and the most common errors in the Least Squares forecast — are all in the AI Question Bank.
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Frequently Asked Questions
Answers to questions students frequently ask about Class 12 Applied Maths Unit 6.
For an odd-period MA (e.g., 3-year or 5-year), the computed average is naturally centred at the middle time period — no further adjustment needed. For an even-period MA (e.g., 4-year), the average falls between two time periods, so a centring step is required: average two consecutive MAs using \(\text{CMA} = \dfrac{\text{MA}_i + \text{MA}_{i+1}}{2}\).
For \(y = a + bx\): \(\sum y = na + b\sum x\) and \(\sum xy = a\sum x + b\sum x^2\).
When the middle year is chosen as origin (\(\sum x = 0\)): \(a = \dfrac{\sum y}{n}\) and \(b = \dfrac{\sum xy}{\sum x^2}\).
When the middle year is chosen as origin (\(\sum x = 0\)): \(a = \dfrac{\sum y}{n}\) and \(b = \dfrac{\sum xy}{\sum x^2}\).
For an odd number of data points, choose the middle year as origin (\(x=0\)) and assign consecutive integers (e.g., −2, −1, 0, 1, 2 for 5 points). For an even number, assign \(x\) as −3, −1, 1, 3… so that \(\sum x = 0\) and fractions are avoided.
Unit 6: Time-based Data carries 6 marks in the CBSE Class 12 Applied Mathematics board examination — typically 1 MCQ (1 mark) and 1 five-mark long-answer question on either Moving Averages or the Least Squares Method.
Moving averages smooth out short-term random fluctuations in a time series and reveal the underlying long-term trend. A larger period produces a smoother trend but loses more data points at the ends of the series and responds more slowly to genuine recent changes.
Moving averages smooth the data visually but cannot produce a forecasting equation. The Least Squares method fits an equation \(y = a + bx\) which can forecast values for any future year by substituting the appropriate \(x\) value. Both methods are equally likely to appear as the 5-mark board question.
The least squares line minimises \(\sum(y_i - \hat{y}_i)^2\) — the sum of squared vertical residuals. It passes through the mean point \((\bar{x}, \bar{y})\) and gives the best overall fit; individual data points generally lie above or below the line. The goal is to minimise total error, not to pass through every point.
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