Math Problem Statement
Solution
The problem reads as follows:
Q.1 The total cost of 3 mobiles and 2 TVs is ₹35,000. The shopkeeper wants a profit of ₹1,000 per mobile and ₹5,000 per TV. He sells 2 mobiles and 1 TV and gets total revenue of ₹21,500. Find the selling price of a mobile and a TV.
Let's solve this step by step:
Let:
- The cost price of one mobile be .
- The cost price of one TV be .
From the given information:
-
Cost equation:
The total cost of 3 mobiles and 2 TVs is ₹35,000: -
Profit conditions:
The shopkeeper wants a profit of ₹1,000 per mobile and ₹5,000 per TV. Therefore:- The selling price of one mobile is .
- The selling price of one TV is .
-
Revenue equation:
He sells 2 mobiles and 1 TV, and the total revenue from the sales is ₹21,500:
We now have the following system of equations:
Let me solve these equations.The solution indicates that the cost price of one mobile is ₹-6,000, which does not make practical sense, and the cost price of one TV is ₹26,500.
Since negative cost prices are not logical in this context, there seems to be an issue with the provided conditions in the problem. We should revisit the assumptions or check for errors in the problem's setup.
Would you like to reconsider the conditions or clarify anything further? I can also help troubleshoot alternative approaches.
Here are 5 related questions to expand your understanding:
- How could the profit margin affect pricing models in a retail business?
- Can you think of real-world scenarios where negative values might emerge in mathematical models?
- How would the solution change if the profit on each item was calculated as a percentage instead of a fixed amount?
- What other costs (like taxes or overhead) might complicate the basic cost/profit calculations in real life?
- How would we approach the problem differently if we were missing information?
Tip: When encountering illogical results in a problem, rechecking assumptions and conditions can often lead to clearer insights.
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Math Problem Analysis
Mathematical Concepts
Algebra
Simultaneous Equations
Linear Equations
Formulas
Cost equation: 3x + 2y = 35,000
Revenue equation: 2(x + 1,000) + (y + 5,000) = 21,500
Theorems
-
Suitable Grade Level
Grades 8-10
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