A scholarship is a sum of money provided to a student to help him or her pay for education. Some students are granted scholarships based on their academic achievements, while others are rewarded based on their financial needs.
Every year a school offers scholarships to girl children and meritorious achievers based on certain criteria. In the session 2022–23, the school offered monthly scholarships of ₹3,000 each to some girl students and ₹4,000 each to meritorious achievers in academics as well as sports.
In all, 50 students were given the scholarships, and the monthly expenditure incurred by the school on scholarships was ₹1,80,000.
Based on the above information, answer the following questions:
Express the given information algebraically using matrices.
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Solution and Explanation
Let: \[ x = \text{number of girl child scholarships}, \] \[ y = \text{number of meritorious achievers}. \] Step 2: Formulate the equations
We are given: \[ x + y = 50 \quad \text{(1)}, \] \[ 3000x + 4000y = 180000 \quad \text{(2)}. \] Divide equation (2) by 1000: \[ 3x + 4y = 180 \quad \text{(3)}. \] Step 3: Represent the equations in matrix form
The equations can be expressed in matrix form as: \[ \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 50 \\ 180 \end{bmatrix}. \]
Check whether the system of matrix equations so obtained is consistent or not.
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Solution and Explanation
The coefficient matrix is: \[ A = \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix}. \] Step 2: Compute the determinant of \(A\)
The determinant is: \[ \det(A) = (1)(4) - (1)(3) = 4 - 3 = 1. \] Step 3: Check consistency
Since \( \det(A) \neq 0 \), the system of equations is consistent, meaning it has a unique solution.
Find the number of scholarships of each kind given by the school using matrices.
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Solution and Explanation
The matrix equation is: \[ \begin{bmatrix} 1 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 50 \\ 180 \end{bmatrix}. \] Step 2: Find the inverse of \(A\)
The inverse of \(A\) is: \[ A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} 4 & -1 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} 4 & -1 \\ -3 & 1 \end{bmatrix}. \] Step 3: Solve for \(X\)
Using \(X = A^{-1}B\): \[ X = \begin{bmatrix} 4 & -1 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 50 \\ 180 \end{bmatrix}. \] Multiply the matrices: \[ X = \begin{bmatrix} 4(50) - 1(180) \\ -3(50) + 1(180) \end{bmatrix} = \begin{bmatrix} 200 - 180 \\ -150 + 180 \end{bmatrix} = \begin{bmatrix} 20 \\ 30 \end{bmatrix}. \] Thus, \( x = 20 \) (girl students) and \( y = 30 \) (meritorious achievers).
Had the amount of scholarship given to each girl child and meritorious student been interchanged, what would be the monthly expenditure incurred by the school?
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Solution and Explanation
Recalculate the expenditure:
If the scholarship amounts are interchanged, girl students receive ₹4,000, and meritorious achievers receive ₹3,000.
The expenditure becomes: \[ {Expenditure} = 30(3000) + 20(4000) = 90000 + 80000 = 170000. \]
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