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An apparel manufacturer sells premium shirts for $₹1400$ each and regular trousers for $₹900$ each. At the end of a seasonal clearance sale, the store manager realizes that the total revenue generated from selling these two items alone is exactly $₹43,100$. If the store manager also knows that they sold more trousers than premium shirts, how many distinct combinations of shirts and trousers could have been sold? A) 2 B) 3 C) 4 D) 5
Correct Answer: A) 2 ### Step-by-Step Solution #### Step 1: Formulate the Linear Diophantine Equation Let the number of premium shirts sold be $x$ and the number of regular trousers sold be $y$. Since $x$ and $y$ represent quantities of physical items, both must be non-negative integers ($x, y \in \mathbb{Z}_{\ge 0}$). Based on the revenue constraint given in the problem statement: $$1400x + 900y = 43100$$ We can simplify this linear equation by dividing both sides by $100$: $$14x + 9y = 431 \quad \text{--- (Equation 1)}$$ This is a classic Linear Diophantine Equation of the form $ax + by = c$. #### Step 2: Test for the Existence of Integer Solutions A linear Diophantine equation $ax + by = c$ has integer solutions if and only if the Greatest Common Divisor $\text{GCD}(a, b)$ divides $c$. * Here, $a = 14$ and $b = 9$. * $\text{GCD}(14, 9) = 1$ (since 14 and 9 are co-prime). * Since $1$ perfectly divides $431$, integer solutions are guaranteed to exist. #### Step 3: Find an Initial Base Solution $(x_0, y_0)$ We use modular arithmetic on Equation 1 to find our first valid integer coordinate pair. Taking the entire equation modulo $9$ (the smaller coefficient) helps isolate $x$: $$14x + 9y \equiv 431 \pmod 9$$ Simplifying the terms modulo $9$: * $14 \equiv 5 \pmod 9$ * $9y \equiv 0 \pmod 9$ * $431 = (9 \times 47) + 8 \implies 431 \equiv 8 \pmod 9$ Substituting these remainders back into the congruence: $$5x \equiv 8 \pmod 9$$ To solve for $x$, we add multiples of $9$ to the right-hand side until it is perfectly divisible by $5$: $$5x \equiv 8 + 9 \equiv 17 \pmod 9$$ $$5x \equiv 17 + 9 \equiv 26 \pmod 9$$ $$5x \equiv 26 + 9 \equiv 35 \pmod 9$$ Now, dividing both sides by $5$: $$x \equiv 7 \pmod 9$$ Thus, our initial base value for $x$ is $x_0 = 7$. Substitute $x = 7$ back into Equation 1 to solve for the corresponding base value $y_0$: $$14(7) + 9y = 431$$ $$98 + 9y = 431$$ $$9y = 333 \implies y_0 = 37$$ Our baseline valid coordinate solution is $(x_0, y_0) = (7, 37)$. #### Step 4: Write the General Solution Parameter Equation Using the property of Diophantine equations, since $(7, 37)$ is a solution to $14x + 9y = 431$, all other integer solutions will progress along a fixed parametric grid: $$x = 7 + 9t$$ $$y = 37 - 14t$$ where $t$ is any integer ($t \in \mathbb{Z}$). #### Step 5: Apply Non-negative Boundaries and Constraints We must find all values of $t$ that keep both $x$ and $y$ non-negative ($x \ge 0, y \ge 0$): * **For $x \ge 0$:** $$7 + 9t \ge 0 \implies 9t \ge -7 \implies t \ge -0.77$$ * **For $y \ge 0$:** $$37 - 14t \ge 0 \implies 14t \le 37 \implies t \le 2.64$$ Combining these two foundational bounds limits our parameter $t$ to three possible integer values: $$t \in \{0, 1, 2\}$$ Let us compute the complete coordinate pairs for each valid value of $t$ using the layout array below: The raw source code is rendering because the LaTeX syntax requires double backslashes (`\\`) to mark the end of each row inside the array block, but they are being stripped out or misread. Here is the exact corrected block. It uses standard Markdown syntax to ensure it displays perfectly as a clean, structured table without failing: | Parameter Value ($t$) | Shirts Sold ($x = 7 + 9t$) | Trousers Sold ($y = 37 - 14t$) | | --- | --- | --- | | $t = 0$ | $x = 7$ | $y = 37$ | | $t = 1$ | $x = 16$ | $y = 23$ | | $t = 2$ | $x = 25$ | $y = 9$ | #### Step 6: Apply the Final Problem-Specific Condition The problem explicitly states a final critical real-world boundary condition: **the store sold more trousers than premium shirts ($y > x$)**. Let us evaluate our three potential solution cases against this constraint: 1. **Case 1 ($t = 0$):** $x = 7, y = 37 \implies 37 > 7$ (This satisfies $y > x$). 2. **Case 2 ($t = 1$):** $x = 16, y = 23 \implies 23 > 16$ (This satisfies $y > x$). 3. **Case 3 ($t = 2$):** $x = 25, y = 9 \implies 9 \ngtr 25$ (This fails the condition). Only the first two combinations satisfy all algebraic, integer, and contextual problem-specific inequalities. Therefore, there are exactly 2 distinct valid combinations of shirts and trousers that could have been sold.
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