Question

Ajay, Sameer and Meenal have Rs. 20/- each and some footballs, basketballs and volleyballs in their shops. In a week, Ajay sold 3 footballs and a volleyball but he bought 2 basketballs for his shop and he has Rs. 35/- now. In same duration, Sameer sold 2 basketballs and 2 volleyballs but he bought a football for his shop and he has Rs. 95/- now. Similarly, Meenal sold 2 footballs and a basketball but she bought 3 volleyballs for her shop and she has Rs. 15/- now. Find the cost of a football, a basketball and a volleyball by the help of matrices.

Hint:

In multiple-choice matrix problems, instead of fully solving the system, you can plug the options into the equations to see which one satisfies all three. For instance, \( 3(15) - 2(25) + 20 = 45 - 50 + 20 = 15 \), which confirms Option A.

Answer 1

Step 1: Understanding the Concept:
The problem involves translating a word problem into a system of linear equations and solving them using matrix algebra.
Net Change in cash = (Income from Sales) - (Expenditure on Purchases).
Step 2: Key Formula or Approach:
Let \( x, y, z \) be the cost of a football, a basketball, and a volleyball respectively.
The net change in Ajay's cash: \( 35 - 20 = 15 \).
The net change in Sameer's cash: \( 95 - 20 = 75 \).
The net change in Meenal's cash: \( 15 - 20 = -5 \).
Step 3: Detailed Explanation:
From the given data, we form the following equations:
For Ajay: \( 3x - 2y + z = 15 \) ... (1)
For Sameer: \( -x + 2y + 2z = 75 \) ... (2)
For Meenal: \( 2x + y - 3z = -5 \) ... (3)
This can be written in matrix form \( AX = B \):
\[ \begin{bmatrix} 3 & -2 & 1 \\ -1 & 2 & 2 \\ 2 & 1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 15 \\ 75 \\ -5 \end{bmatrix} \] Solving by substitution (or matrix inversion):
Adding (1) and (2): \( 2x + 3z = 90 \implies x = \frac{90 - 3z}{2} \).
Multiplying (3) by 2 and adding to (1):
\( (3x - 2y + z) + 2(2x + y - 3z) = 15 + 2(-5) \)
\( 3x - 2y + z + 4x + 2y - 6z = 5 \)
\( 7x - 5z = 5 \).
Substitute \( x = \frac{90 - 3z}{2} \) into \( 7x - 5z = 5 \):
\[ 7\left(\frac{90 - 3z}{2}\right) - 5z = 5 \implies \frac{630 - 21z - 10z}{2} = 5 \] \[ 630 - 31z = 10 \implies 31z = 620 \implies z = 20 \].
Find \( x \): \( x = \frac{90 - 3(20)}{2} = \frac{30}{2} = 15 \).
Find \( y \) using (3): \( 2(15) + y - 3(20) = -5 \implies 30 + y - 60 = -5 \implies y = 25 \).
Step 4: Final Answer:
The cost of a football is Rs. 15, a basketball is Rs. 25, and a volleyball is Rs. 20.