Two mixers and one TV cost Rs. 500, while two TVs and one mixer cost Rs. 700. The cost of one TV is:
Rs. 300
Understanding the Mixer and TV Cost Problem
This problem involves finding the cost of a single television (TV) and a single mixer using the information given about the combined costs of different quantities of these items. We can solve this by setting up a system of linear equations based on the given data.
Setting Up the Cost Equations
Let's assign variables to represent the unknown costs:
- Let \(M\) be the cost of one mixer in Rupees.
- Let \(T\) be the cost of one TV in Rupees.
Based on the problem statement, we can form two equations:
- "Two mixers and one TV cost Rs. 500" can be written as: \[2M + T = 500 \quad \text{(Equation 1)}\]
- "Two TVs and one mixer cost Rs. 700" can be written as: \[M + 2T = 700 \quad \text{(Equation 2)}\]
We now have a system of two linear equations with two variables, \(M\) and \(T\). Our goal is to find the value of \(T\), the cost of one TV.
Solving the System of Equations for TV Cost
We can solve this system using methods like substitution or elimination. Let's use the elimination method to eliminate \(M\) and solve for \(T\).
We can multiply Equation 2 by 2 so that the coefficient of \(M\) becomes the same as in Equation 1:
\[2 \times (M + 2T) = 2 \times 700\] \[2M + 4T = 1400 \quad \text{(Equation 3)}\]Now we have two equations (Equation 1 and Equation 3) with the same coefficient for \(M\):
\[2M + T = 500 \quad \text{(Equation 1)}\] \[2M + 4T = 1400 \quad \text{(Equation 3)}\]Subtract Equation 1 from Equation 3:
\[(2M + 4T) - (2M + T) = 1400 - 500\] \[2M + 4T - 2M - T = 900\]Combine like terms:
\[(2M - 2M) + (4T - T) = 900\] \[0M + 3T = 900\] \[3T = 900\]Now, solve for \(T\) by dividing both sides by 3:
\[T = \frac{900}{3}\] \[T = 300\]So, the cost of one TV (\(T\)) is Rs. 300.
Verification (Optional but Recommended)
We can also find the cost of one mixer (\(M\)) by substituting the value of \(T = 300\) into either original equation. Using Equation 1:
\[2M + T = 500\] \[2M + 300 = 500\]Subtract 300 from both sides:
\[2M = 500 - 300\] \[2M = 200\]Divide by 2:
\[M = \frac{200}{2}\] \[M = 100\]The cost of one mixer is Rs. 100. Let's check if these values satisfy Equation 2:
\[M + 2T = 700\] \[100 + 2(300) = 100 + 600 = 700\]The values \(M=100\) and \(T=300\) satisfy both equations, confirming our solution is correct.
Conclusion on the Cost
Based on our calculations, the cost of one TV is Rs. 300.
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Variable | A symbol (like \(M\) or \(T\)) used to represent an unknown quantity. | Representing the cost of mixers and TVs. |
| Linear Equation | An equation where the highest power of the variable(s) is 1. | Each statement about total cost translates into a linear equation. |
| System of Linear Equations | A set of two or more linear equations involving the same variables. | We had two equations with two variables (\(M\) and \(T\)). |
| Elimination Method | A technique to solve systems of equations by eliminating one variable. | Used here to eliminate \(M\) and solve for \(T\). |
| Substitution Method | Another technique to solve systems by substituting an expression from one equation into another. | Could also be used to solve this problem. |
Additional Information on Word Problems and Algebra
Word problems like this help us translate real-world scenarios into mathematical models. The key steps usually involve:
- Reading Carefully: Understand what quantities are involved and what is being asked.
- Defining Variables: Assign letters to the unknown quantities.
- Formulating Equations: Translate the given information into mathematical equations based on the relationships described.
- Solving the Equations: Use algebraic techniques (like substitution, elimination, or matrix methods for larger systems) to find the values of the variables.
- Checking the Solution: Substitute the calculated values back into the original word problem or equations to ensure they make sense and satisfy all conditions.
This specific problem is a classic example of a problem solvable with a system of two linear equations in two variables. Mastering these types of problems is fundamental in algebra.
Important Questions from Linear Equation in 2 or more Variables
If \(3x+6y+9z = \dfrac{20}{3}, 6x+9y + 3z = \dfrac{17}{3}\) and \(18x+ 27y - z = \dfrac{113}{9}\) , then what is the value of \(75x+113y \ ?\)
If the system of equations 2x - 3y - 3 and -4x + qy - p/2 is inconsistent which of the following cannot be the value of p ?
If \(4x + \dfrac{6}{y}= 15 \) and \(6x - \dfrac{8}{y} = 14\) , then the value of p in y = px - 2 is :
Simplify: 7x + 3x(x – 4) = ?
A. 10x + 12
B. 10x – 12
C. 3x2 + 5x
D. 3x2 – 5xIf 5x + y = 17 and xy = 6, then what is the value of 125x3 + y3 ?
