| Market | Products | ||
| I | 10000 | 2000 | 18000 |
| II | 6000 | 20000 | 8000 |
(a) The unit sale prices of \(x, y\), and \(z\) are respectively given as Rs 2.50, Rs 1.50, and Rs 1.00. Consequently, the total revenue in market I can be represented in the form of a matrix as:
\(\begin{bmatrix}10000& 2000& 18000\end{bmatrix}\begin{bmatrix}2.50\\ 1.50\\ 1.00\end{bmatrix}\)
\(=10000\times2.50+2000\times1.50+18000\times1.00]\)
\(=25000+3000+18000\)
\(=46000\)
The total revenue in market II can be represented in the form of a matrix as:
\(\begin{bmatrix}6000& 20000& 8000\end{bmatrix}\begin{bmatrix}2.50\\ 1.50\\ 1.00\end{bmatrix}\)
\(=6000\times2.50+20000\times1.50+8000\times1.00\)
\(=15000+30000+8000\)
\(=53000\)
Therefore, the total revenue in market I is Rs 46000 and the same in market II is Rs 53000.
(b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50 paise.
Consequently, the total cost prices of all the products in market I can be represented in the form of a matrix as:
\(\begin{bmatrix}10000& 2000& 18000\end{bmatrix}\begin{bmatrix}2.00\\ 1.00\\ 0.50\end{bmatrix}\)
\(=10000\times2.00+2000\times1.00+18000\times0.50\)
\(=20000+2000+9000\)
\(=31000\)
Since the total revenue in market I is Rs 46000, the gross profit in this market is (Rs 46000−Rs 31000) Rs 15000.
The total cost prices of all the products in market II can be represented in the form of a matrix as:
\(\begin{bmatrix}6000& 20000& 8000\end{bmatrix}\begin{bmatrix}2.00\\ 1.00\\ 0.50\end{bmatrix}\)
\(=6000\times2.00+20000\times1.00+8000\times0.50\)
\(=12000+20000+4000\)
\(=Rs 36000\)
Since the total revenue in market II isRs 53000, the gross profit in this market is (Rs53000 − Rs 36000) Rs 17000.
Let \( A = \begin{bmatrix} \alpha & -1 \\ 6 & \beta \end{bmatrix} , \ \alpha > 0 \), such that \( \det(A) = 0 \) and \( \alpha + \beta = 1. \) If \( I \) denotes the \( 2 \times 2 \) identity matrix, then the matrix \( (I + A)^8 \) is:
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]
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.
