When observed over a long period of time, a time series data can predict trends that can forecast increase, decrease, or stagnation of a variable under consideration. The table below shows the sale of an item in a district during 1996–2001:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]When calculating a straight-line trend using least squares:
- Assign coded time values (\( t \)) to simplify calculations.
- Use the formulas \( a = \frac{\sum y}{N} \) and \( b = \frac{\sum (t \cdot y)}{\sum t^2} \).
Step 1: Assign coded time variables (\( t \)) for convenience:
Let \( t = -2, -1, 0, 1, 2, 3 \) for the years 1996 to 2001, respectively.
Step 2: Tabulate the data:

Step 3: Use the least squares formula for straight-line trend:
\[ y = a + bt, \]
where:
\[ a = \frac{\sum y}{N}, \quad b = \frac{\sum (t \cdot y)}{\sum t^2}. \]
Substitute the values:
\[ a = \frac{\sum y}{N} = \frac{35.6}{6} = 5.93, \quad b = \frac{\sum (t \cdot y)}{\sum t^2} = \frac{22.4}{19} = 1.18. \]
Step 4: Write the equation:
\[ y = 5.93 + 1.18t. \]
Step 1: Use the trend equation \( y = 5.93 + 1.18t \).
Step 2: Compute the trend values for \( t = -2, -1, 0, 1, 2, 3 \):
\[ \begin{array}{|c|c|c|} \hline \textbf{Year} & t & \textbf{Trend Value (y)} \\ \hline 1996 & -2 & 5.93 + 1.18(-2) = 3.57 \\ 1997 & -1 & 5.93 + 1.18(-1) = 4.75 \\ 1998 & 0 & 5.93 + 1.18(0) = 5.93 \\ 1999 & 1 & 5.93 + 1.18(1) = 7.11 \\ 2000 & 2 & 5.93 + 1.18(2) = 8.29 \\ 2001 & 3 & 5.93 + 1.18(3) = 9.47 \\ \hline \end{array} \]Step 3: Compute the trend value for 2002 (\( t = 4 \)):
\[ y = 5.93 + 1.18(4) = 10.65. \]Final Answer:
Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]Step 1: Assign \( t \): Let \( t = -3, -2, -1, 0, 1, 2, 3 \) for the years 2004 to 2010.
Step 2: Tabulate the data:
\[ \begin{array}{|c|c|c|c|c|} \hline \textbf{Year} & \textbf{Profit (y)} & t & t^2 & t \cdot y \\ \hline 2004 & 114 & -3 & 9 & -342 \\ 2005 & 130 & -2 & 4 & -260 \\ 2006 & 126 & -1 & 1 & -126 \\ 2007 & 144 & 0 & 0 & 0 \\ 2008 & 138 & 1 & 1 & 138 \\ 2009 & 156 & 2 & 4 & 312 \\ 2010 & 164 & 3 & 9 & 492 \\ \hline \textbf{Total} & 972 & 0 & 28 & 214 \\ \hline \end{array} \]Step 3: Use the least squares formula:
\[ y = a + bt, \]where:
\[ a = \frac{\sum y}{N}, \quad b = \frac{\sum (t \cdot y)}{\sum t^2}. \]Substituting the values:
\[ a = \frac{972}{7} = 138.86, \quad b = \frac{214}{28} = 7.64. \]Step 4: Write the equation:
\[ y = 138.86 + 7.64t. \]A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
Rupal, Shanu and Trisha were partners in a firm sharing profits and losses in the ratio of 4:3:1. Their Balance Sheet as at 31st March, 2024 was as follows: 
(i) Trisha's share of profit was entirely taken by Shanu.
(ii) Fixed assets were found to be undervalued by Rs 2,40,000.
(iii) Stock was revalued at Rs 2,00,000.
(iv) Goodwill of the firm was valued at Rs 8,00,000 on Trisha's retirement.
(v) The total capital of the new firm was fixed at Rs 16,00,000 which was adjusted according to the new profit sharing ratio of the partners. For this necessary cash was paid off or brought in by the partners as the case may be.
Prepare Revaluation Account and Partners' Capital Accounts.