În Doi Saci Sunt 140 Kg De Făină. Dacă Din Primul Sac Se Scot 2000 Dag Și Se Pun În Al Doilea, Atunci Cantitățile Din Cei Doi Saci Devin Egale. Câte Kg De Făină Erau Inițial În Fiecare Sac?
The Problem Unveiled
We are presented with a classic mathematical problem that involves the distribution of a quantity between two containers, followed by a transfer that equalizes the contents. In this case, we have two sacks containing a total of 140 kg of flour. The challenge lies in determining the initial amount of flour in each sack, given that if 2000 dag (decagrams) are transferred from the first sack to the second, the quantity of flour in both sacks becomes equal. This problem requires careful conversion of units and the application of algebraic principles to solve for the unknowns. Understanding the relationship between the initial amounts, the transfer, and the final equal quantities is crucial for arriving at the correct solution. The problem not only tests mathematical skills but also logical reasoning and the ability to translate real-world scenarios into mathematical equations. To effectively solve this problem, we need to break it down into smaller, manageable steps, starting with converting the units and then setting up the equations that represent the given information. The use of variables to represent the unknown quantities will allow us to formulate the problem in a clear and concise manner, paving the way for a systematic solution. The elegance of this problem lies in its simplicity, yet it requires a thorough understanding of fundamental mathematical concepts to solve it accurately.
Decoding the Decagram: Unit Conversion is Key
Before diving into the algebraic manipulation, a crucial step is to ensure consistency in units. The problem mixes kilograms (kg) and decagrams (dag), necessitating a conversion. Since the final answer is desired in kilograms, converting decagrams to kilograms is the logical choice. We know that 1 kg is equal to 1000 grams, and 1 dag is equal to 10 grams. Therefore, 1 kg is equal to 100 dag. Consequently, 2000 dag is equivalent to 2000 / 100 = 20 kg. This conversion is not merely a technicality; it's a fundamental aspect of problem-solving. Ignoring unit conversions can lead to significant errors and incorrect conclusions. In this particular problem, the transfer of 2000 dag, if not properly converted, would be misinterpreted, leading to a flawed equation and an inaccurate solution. The conversion step highlights the importance of attention to detail and a methodical approach to mathematical problems. It also underscores the interconnectedness of mathematical concepts, where unit conversions serve as a bridge between different measurement systems, enabling us to work with them seamlessly. By converting the decagrams to kilograms upfront, we ensure that all subsequent calculations are performed using a consistent unit, simplifying the process and reducing the risk of errors. The clarity gained from this step sets the stage for a more confident and accurate resolution of the problem.
Setting Up the Equations: The Language of Mathematics
With the units aligned, we can now translate the word problem into the language of mathematics: equations. Let's denote the initial amount of flour in the first sack as 'x' kg and the initial amount in the second sack as 'y' kg. The first piece of information, that the total amount of flour is 140 kg, can be expressed as a simple equation: x + y = 140. This equation captures the fundamental relationship between the two unknowns, providing a crucial starting point for solving the problem. The second piece of information involves the transfer of 20 kg of flour. After the transfer, the first sack will contain x - 20 kg, and the second sack will contain y + 20 kg. The problem states that after this transfer, the quantities are equal. This leads to our second equation: x - 20 = y + 20. These two equations form a system of linear equations, a powerful tool for solving problems with multiple unknowns. The system of equations encapsulates all the information provided in the problem, allowing us to manipulate these equations to isolate the unknowns and find their values. The process of setting up the equations is not just about translating words into symbols; it's about capturing the underlying relationships and constraints inherent in the problem. A well-formulated system of equations is the foundation for a successful solution, providing a clear and structured pathway to the answer. The equations represent a concise and unambiguous representation of the problem, allowing us to apply mathematical techniques to solve it effectively.
Solving the System: Unveiling the Unknowns
We now have a system of two equations with two unknowns: 1) x + y = 140 and 2) x - 20 = y + 20. There are several methods to solve such a system, including substitution and elimination. Let's use the substitution method. From the first equation, we can express y in terms of x: y = 140 - x. Now, substitute this expression for y into the second equation: x - 20 = (140 - x) + 20. Simplify the equation: x - 20 = 160 - x. Add x to both sides: 2x - 20 = 160. Add 20 to both sides: 2x = 180. Divide both sides by 2: x = 90. Now that we have the value of x, we can substitute it back into the equation y = 140 - x to find y: y = 140 - 90 = 50. Therefore, the initial amount of flour in the first sack (x) is 90 kg, and the initial amount in the second sack (y) is 50 kg. This solution represents the unique set of values that satisfies both equations simultaneously, providing the answer to the problem. The process of solving the system of equations demonstrates the power of algebraic manipulation in unraveling complex relationships. By systematically isolating the unknowns, we were able to determine their values, providing a concrete solution to the problem. The solution not only answers the specific question but also validates the initial setup of the equations, confirming that our mathematical model accurately reflects the given information.
The Solution: Flour Distribution Revealed
Having solved the system of equations, we arrive at the solution: Initially, the first sack contained 90 kg of flour, and the second sack contained 50 kg of flour. This distribution satisfies the conditions of the problem. The total amount of flour is 90 kg + 50 kg = 140 kg, as given. If 20 kg are transferred from the first sack to the second, the first sack will have 90 kg - 20 kg = 70 kg, and the second sack will have 50 kg + 20 kg = 70 kg. This confirms that after the transfer, the amount of flour in both sacks is equal, as stated in the problem. The solution not only provides the numerical answers but also offers a sense of closure, validating the entire problem-solving process. The initial translation of the word problem into mathematical equations, the careful manipulation of those equations, and the final interpretation of the results all contribute to a comprehensive understanding of the problem. This mathematical journey highlights the power of logical reasoning and algebraic techniques in solving real-world problems. The final answer is not just a set of numbers; it's a testament to the effectiveness of a systematic and methodical approach to problem-solving. The solution provides a clear and concise answer to the question, leaving no room for ambiguity and demonstrating the precision of mathematical analysis.
Problem-Solving Strategies: A Retrospective
This problem, while seemingly simple, encapsulates several important problem-solving strategies applicable across various mathematical domains. First and foremost, understanding the problem is paramount. This involves carefully reading the problem statement, identifying the unknowns, and recognizing the given conditions. In this case, we needed to understand the relationship between the initial amounts of flour, the transfer, and the final equal quantities. Secondly, unit conversion is crucial for accuracy. The problem's use of both kilograms and decagrams highlighted the importance of ensuring consistent units before performing calculations. Neglecting this step could lead to significant errors. Thirdly, translating the problem into mathematical language is essential. This involves representing the unknowns with variables and expressing the given conditions as equations. The ability to formulate a system of equations is a powerful problem-solving tool. Fourthly, choosing an appropriate method to solve the equations is key. In this case, we used the substitution method, but other methods, such as elimination, could also have been employed. Finally, verifying the solution is a critical step. We checked that our solution satisfied the conditions of the problem, ensuring that it was both mathematically correct and logically consistent. These strategies, applied systematically, can empower individuals to tackle a wide range of mathematical challenges with confidence and competence. The problem-solving process is not just about arriving at the answer; it's about developing a robust and adaptable approach to mathematical thinking.