Unit 6: Time-based Data – Class 12 Applied Maths | Free MCQs & Solutions | Boundless Maths
Unit 6: Time-based Data
Class 12 Applied Mathematics — Moving Averages & Least Squares Method
⭐ 6 Marks in Board Exam
Unit 6: Time-based Data carries 6 marks in the Class 12 Applied Mathematics board examination, making it an important unit to cover thoroughly for guaranteed marks.
This unit covers two core methods for analysing trends in time series data. The Method of Moving Averages teaches you to smooth out short-term fluctuations to reveal the underlying trend — covering both odd-period moving averages (where the value is naturally centred) and even-period moving averages (which require an additional centring step by averaging pairs of moving averages). The Method of Least Squares takes this further by fitting a straight-line trend equation \(y = a + bx\) to historical data using normal equations, which can then be used to make algebraic predictions for any future period.
This page includes MCQs with full solutions (including Assertion–Reason questions), and 5-mark long-answer questions with complete step-by-step workings and fully solved data tables.
📋 All questions and solutions are aligned to the latest CBSE Applied Maths syllabus 2026-27.
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MCQs & AR Questions
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5-Mark Questions
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Board Exam Marks
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Key Topics
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Topics Covered
📈 Method of Moving Averages
Smooth time series data to reveal underlying trends — odd-period (directly centred) and even-period (centring required).
Odd-period: \(\text{MA}_k = \dfrac{\sum_{i}^{i+k-1} y_i}{k}\), centred at middle
Even-period: Centred MA \(= \dfrac{\text{MA}_i + \text{MA}_{i+1}}{2}\)
📉 Method of Least Squares
Fit a linear trend equation to data and use it to forecast future values algebraically.
Trend line: \(y = a + bx\)
When \(\sum x = 0\): \(a = \dfrac{\sum y}{n},\quad b = \dfrac{\sum xy}{\sum x^2}\)
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A 3-period moving average is calculated for the data: 10, 12, 15, 18, 20. What is the moving average for the third period?
a12.33
b13.67
c15.00
d16.33
✓ Correct Answer: (a) 12.33
Explanation: For a 3-period moving average, the first MA is calculated from the first three values and centred at the middle (second) period:
• \(\text{MA}_1 = \dfrac{10 + 12 + 15}{3} = \dfrac{37}{3} \approx 12.33\), placed against period 2 (Tuesday / 2nd value)
The question asks for the "third period" value, which corresponds to the first calculable MA (centred at the 2nd data point, the 3rd in naming). Answer = 12.33.
Question 2Even Duration
When calculating a 4-year moving average, the centring process requires:
aTaking the average of two consecutive moving averages
bMultiplying the average by 2
cDividing the average by 2
dNo centring is needed
✓ Correct Answer: (a) Taking the average of two consecutive moving averages
Explanation: For even-period moving averages (e.g., 4-year), the computed average falls between two time periods. To align it to an actual year, we take the average of two consecutive 4-year moving averages — this is called the centred moving average:
• \(\text{Centred MA} = \dfrac{\text{MA}_i + \text{MA}_{i+1}}{2}\)
Question 3Least Squares
The normal equations for fitting \(y = a + bx\) using the least squares method are:
a\(\sum y = na + b\sum x\) and \(\sum xy = a\sum x + b\sum x^2\)
b\(\sum y = na - b\sum x\) and \(\sum xy = a\sum x - b\sum x^2\)
c\(\sum x = na + b\sum y\) and \(\sum xy = a\sum y + b\sum y^2\)
d\(\sum y = a + b\sum x\) and \(\sum xy = ax + bx^2\)
✓ Correct Answer: (a)
Explanation: The two normal equations for fitting \(y = a + bx\) by the method of 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 the origin so that \(\sum x = 0\), these simplify to \(a = \dfrac{\sum y}{n}\) and \(b = \dfrac{\sum xy}{\sum x^2}\).
Question 4Trend Analysis
The main purpose of calculating moving averages in time series analysis is to:
aIncrease the variability in data
bSmooth out short-term fluctuations and reveal trends
cMake predictions less accurate
dEliminate all patterns in data
✓ Correct Answer: (b) Smooth out short-term fluctuations and reveal trends
Explanation: Moving averages are used to smooth short-term random fluctuations and seasonal variations in a time series, making the underlying long-term trend more visible. They do not eliminate all patterns — they emphasise the trend component while dampening noise.
Question 5Least Squares
If \(\sum x = 55\), \(\sum y = 350\), \(\sum xy = 3025\), \(\sum x^2 = 385\), and \(n = 10\), the value of \(b\) in \(y = a + bx\) is:
In a 5-period moving average, the first moving average value will be placed against:
aThe first period
bThe second period
cThe third period
dThe fifth period
✓ Correct Answer: (c) The third period
Explanation: For an odd number of periods \(k\), the moving average is centred at the middle period. For a 5-period MA computed from periods 1 to 5, the middle period is period \(\dfrac{1+5}{2} = 3\). So the first 5-period MA is placed against period 3.
In general, for a \(k\)-period MA, the first value is placed against period \(\dfrac{k+1}{2}\).
📋 Assertion-Reason Questions
Statement I is the Assertion (A) and Statement II is the Reason (R). Choose the correct option:
(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
Assertion-Reason 1
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.
aBoth A and R are True and R is the correct explanation of A
bBoth A and R are True but R is not the correct explanation of A
cA is True but R is False
dA is False but R is True
✓ Correct Answer: (c) A is True but R is False
Checking Assertion (A): TRUE — the method of least squares does minimise \(\sum(y_i - \hat{y}_i)^2\), the sum of squared vertical deviations (residuals) from the line.
Checking Reason (R): FALSE — the best-fit line does not necessarily pass through all data points. It passes through the mean point \((\bar{x}, \bar{y})\) and minimises the total squared error. Individual data points will generally not lie exactly on the line.
Assertion-Reason 2
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.
aBoth A and R are True and R is the correct explanation of A
bBoth A and R are True but R is not the correct explanation of A
cA is True but R is False
dA is False but R is True
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
Explanation: Both statements are TRUE and R correctly explains A. A larger period (e.g., a 7-period MA vs a 3-period MA) averages over more data points, so any single extreme value has less influence on the result. This produces a smoother trend line that is closer to the long-term average, but at the cost of losing more values at both ends of the series.
Question 9Trend Prediction
If the trend equation is \(y = 20 + 3x\) where \(x\) represents years with 2020 as origin (\(x = 0\)), the predicted value for the year 2025 is:
Quarterly sales are: Q1 = 100, Q2 = 150, Q3 = 200, Q4 = 250. What is the 4-quarter moving average?
a150
b165
c175
d180
✓ Correct Answer: (c) 175
Explanation:
\[\text{4-quarter MA} = \frac{100 + 150 + 200 + 250}{4} = \frac{700}{4} = \mathbf{175}\]
Since this is an even-period MA, this value would fall between Q2 and Q3 and would need to be centred by averaging with the next 4-quarter MA.
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The following data gives the number of effective 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
📝 Complete Solution
A 3-day moving average uses 3 consecutive values. Since 3 is odd, the MA is centred at the middle day — so the first MA is placed against Tuesday (the 2nd day).
\(\text{MA}_3 = \dfrac{Y_1 + Y_2 + Y_3}{3}\)
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
• First and last days cannot have a 3-day MA computed
• Average daily increase in trend ≈ 2.5 accidents
The production of a company (in thousand tonnes) for the years 2017–2023 is given below. Fit a linear trend equation and estimate the production for 2025.
Year
2017
2018
2019
2020
2021
2022
2023
Production
50
55
60
65
70
75
80
📝 Complete Solution
7 data points → middle year is 2020 → take 2020 as origin (\(x = 0\)).
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)
For 2025: \(x = 2025 - 2020 = 5\)
\(y = 65 + 5 \times 5 = 65 + 25 = \mathbf{90}\)
Estimated production for 2025 = 90 thousand tonnes
Question 55 Marks
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
📝 Complete Solution
5 is odd, so the 5-year MA is centred at the 3rd value of each group. The first MA covers 2015–2019 and is placed against 2017.
Always apply centring for even-period moving averages — never skip it. When calculating a 4-year or any even-period moving average, the first set of moving averages falls between years and cannot be directly tabulated against a specific year. You must take the average of two consecutive moving averages (the centring step) to get the trend value. Skipping this step and directly writing the MA against a year is the most common mistake in board papers and will cost you full marks on the 5-mark question.
For an odd-period moving average (e.g., 3-year or 5-year), the computed average is naturally centred at the middle time period — no further adjustment is needed. For an even-period moving average (e.g., 4-year), the average falls between two time periods, so a second step called centring is required: average two consecutive moving averages to align the trend value to an actual time period.
For fitting \(y = a + bx\), the two normal equations are:
\[\sum y = na + b\sum x\]
\[\sum xy = a\sum x + b\sum x^2\]
When the origin is the middle year (so that \(\sum x = 0\)), these simplify to \(a = \dfrac{\sum y}{n}\) 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. This makes \(\sum x = 0\), greatly simplifying calculations. For an even number of points, assign \(x\) values as \(\ldots -3, -1, 1, 3, \ldots\) (skipping 0) so that \(\sum x = 0\) still holds.
Unit 6: Time-based Data carries 6 marks in the Class 12 Applied Mathematics CBSE board examination. Questions typically include 1 MCQ (1 mark) and 1 five-mark long-answer question based on either Moving Averages or the Least Squares Method.
Moving averages smooth out short-term random fluctuations and seasonal variations in a time series, making the underlying long-term trend more visible. A larger period (e.g., 7-period vs 3-period) produces a smoother trend but loses more data points at the ends of the series and responds more slowly to recent changes.
Moving averages smooth the data by averaging consecutive values — they reveal the trend visually but cannot produce an algebraic equation for future prediction. The Least Squares method fits a mathematical trend equation (\(y = a + bx\)) to the data, which can be used to forecast values for any future time period by substituting the appropriate \(x\) value.
The least squares line minimises \(\sum(y_i - \hat{y}_i)^2\) — the sum of squared residuals (vertical distances from each point to the line). It does not need to pass through every point; it passes through the mean point \((\bar{x}, \bar{y})\) and provides the best overall fit by making total squared error as small as possible.
