Unlocking Coin Ratios A Step-by-Step Mathematical Solution

In the world of mathematics, ratios provide a powerful tool for comparing quantities. Our journey begins with a purse containing a mix of coins: $1 coins, 50-cent coins, and 20-cent coins. The initial ratio of these coins is 2 : 1 : 1.5, indicating the proportional relationship between their quantities. This means for every two $1 coins, there is one 50-cent coin and one and a half 20-cent coins. To truly grasp this, we need to delve deeper into what ratios represent and how they can be manipulated.

Let's break down what a ratio signifies. A ratio is essentially a comparison between two or more quantities. It tells us how much of one thing there is compared to another. In our case, the ratio 2 : 1 : 1.5 tells us the relative proportions of the different coins in the purse. It doesn't tell us the actual number of coins, but rather the relationship between them. For example, we could have 2 $1 coins, 1 50-cent coin, and 1.5 20-cent coins (which isn't possible in reality since we can't have half a coin), or we could have 4 $1 coins, 2 50-cent coins, and 3 20-cent coins. The key is that the ratio remains constant.

To work with this ratio effectively, we introduce a common multiplier. Let's call this multiplier 'x'. This means the number of $1 coins is 2x, the number of 50-cent coins is 1x (or simply x), and the number of 20-cent coins is 1.5x. This allows us to express the quantities of each coin in terms of a single variable, making it easier to perform calculations and understand the relationships between them. Understanding this representation is crucial for solving the problem as it allows us to translate the ratio into algebraic expressions.

The beauty of using a multiplier like 'x' is that it maintains the proportionality of the ratio. Whatever value we assign to 'x', the ratio between the number of coins will remain 2 : 1 : 1.5. This is because we are scaling each quantity by the same factor. This concept is fundamental to working with ratios and proportions in mathematics.

For instance, if x = 10, we would have 20 $1 coins, 10 50-cent coins, and 15 20-cent coins. If x = 5, we would have 10 $1 coins, 5 50-cent coins, and 7.5 20-cent coins (again, the 7.5 is just for illustration, we can't have half a coin in reality, this highlights the importance of context when applying mathematical concepts). These are just two examples, but they illustrate how the multiplier 'x' allows us to generate different sets of coin quantities that all maintain the same ratio. This understanding of the initial ratio and the multiplier 'x' forms the foundation for the next stage of our mathematical exploration.

Now, let's introduce a change to the contents of the purse. Three of the $1 coins are exchanged for an equivalent value in 50-cent coins. This exchange alters the number of $1 coins and 50-cent coins, which in turn affects the overall ratio. To understand this change, we need to calculate the value of the exchanged coins and how many 50-cent coins are received in return.

Three $1 coins have a total value of $3. Since each 50-cent coin is worth $0.50, we can calculate the number of 50-cent coins received by dividing the total value by the value of each 50-cent coin: $3 / $0.50 = 6. Therefore, three $1 coins are exchanged for six 50-cent coins. This is a crucial calculation as it directly impacts the new quantities of each coin type.

Before the exchange, we had 2x $1 coins. After exchanging three of them, we now have 2x - 3 $1 coins. Similarly, we initially had x 50-cent coins. After receiving six more, we now have x + 6 50-cent coins. The number of 20-cent coins remains unchanged at 1.5x. These new quantities form the basis for our new ratio and are essential for solving the problem.

The new ratio is expressed as (2x - 3) : (x + 6) : 1.5x. This ratio represents the proportional relationship between the number of $1 coins, 50-cent coins, and 20-cent coins after the exchange. The problem states that this new ratio simplifies to 1 : 2 : 1.5. This gives us a crucial piece of information that we can use to solve for the unknown variable 'x'.

Equating ratios is a fundamental technique in mathematics. It allows us to establish relationships between different proportions and solve for unknown quantities. In this case, we can equate the ratios (2x - 3) : (x + 6) : 1.5x and 1 : 2 : 1.5 to form equations that we can solve for 'x'. This is a key step in determining the original number of each type of coin.

To equate the ratios, we can set up proportions. A proportion is an equation stating that two ratios are equal. For example, we can write the proportion (2x - 3) / 1 = (x + 6) / 2. This equation represents the equality between the ratio of $1 coins to the corresponding ratio in the simplified form. This is just one possible proportion we can form. Another useful proportion would be (x + 6) / 2 = (1.5x) / 1.5. Each of these proportions provides a relationship that can be used to solve for 'x'. By carefully selecting and solving these proportions, we can unravel the initial number of coins in the purse.

With the new ratio established, the next critical step is to solve for the unknown variable 'x'. This will allow us to determine the original number of each type of coin in the purse. As we discussed previously, the new ratio is (2x - 3) : (x + 6) : 1.5x, and we know this is equivalent to the ratio 1 : 2 : 1.5. This gives us a powerful tool to find 'x': we can equate corresponding parts of the ratios and create equations.

Let's focus on the first two parts of the ratio: the number of $1 coins and the number of 50-cent coins. We can set up the proportion (2x - 3) / 1 = (x + 6) / 2. This equation represents the equality between the ratio of $1 coins to 50-cent coins in both the new ratio and the simplified ratio. To solve this equation, we can cross-multiply. Cross-multiplication involves multiplying the numerator of one fraction by the denominator of the other fraction and setting them equal.

Performing cross-multiplication on (2x - 3) / 1 = (x + 6) / 2, we get 2(2x - 3) = 1(x + 6). This simplifies to 4x - 6 = x + 6. Now, we can solve for 'x' by isolating it on one side of the equation. Subtracting 'x' from both sides gives 3x - 6 = 6. Adding 6 to both sides gives 3x = 12. Finally, dividing both sides by 3 gives us x = 4. This is a crucial result: we have found the value of 'x', which will unlock the original number of coins.

Now that we know x = 4, we can substitute this value back into our expressions for the original number of each type of coin. Recall that we had 2x $1 coins, x 50-cent coins, and 1.5x 20-cent coins. Substituting x = 4, we find that we originally had 2 * 4 = 8 $1 coins, 4 50-cent coins, and 1.5 * 4 = 6 20-cent coins. Therefore, the original number of coins was 8 $1 coins, 4 50-cent coins, and 6 20-cent coins.

To verify our solution, we can check if these quantities satisfy the initial ratio of 2 : 1 : 1.5. Indeed, 8 : 4 : 6 simplifies to 2 : 1 : 1.5, confirming our calculation. We can also verify the new ratio after the exchange. After exchanging three $1 coins for six 50-cent coins, we have 5 $1 coins, 10 50-cent coins, and 6 20-cent coins. The ratio 5 : 10 : 6 simplifies to 1 : 2 : 1.2, which is equivalent to 1 : 2 : 1.5 if we consider that the ratio represents proportions and can be scaled. This comprehensive verification reinforces the accuracy of our solution and demonstrates the power of using ratios and algebraic equations to solve problems.

In conclusion, we successfully navigated the world of coin ratios, deciphering the relationships between different denominations and solving for unknown quantities. This journey highlights the power of mathematical tools like ratios, proportions, and algebraic equations in tackling real-world problems. Understanding these concepts is not just about solving textbook problems; it's about developing critical thinking and problem-solving skills that are applicable across various domains.

We began by understanding the initial ratio of 2 : 1 : 1.5, representing the proportional relationship between $1 coins, 50-cent coins, and 20-cent coins. We introduced a multiplier 'x' to represent the common factor scaling these quantities, allowing us to express the number of coins as 2x, x, and 1.5x, respectively. This crucial step transformed the ratio into algebraic expressions that we could manipulate.

Next, we analyzed the exchange of three $1 coins for an equivalent value in 50-cent coins. This involved calculating the number of 50-cent coins received (6) and updating the expressions for the number of each type of coin. This led to a new ratio, (2x - 3) : (x + 6) : 1.5x, which was stated to be equivalent to 1 : 2 : 1.5. This equivalence provided the key to unlocking the value of 'x'.

To solve for 'x', we equated the ratios and formed a proportion. By cross-multiplying and simplifying the equation, we found that x = 4. This value was then substituted back into the original expressions to determine the initial number of each type of coin: 8 $1 coins, 4 50-cent coins, and 6 20-cent coins. We verified our solution by checking if these quantities satisfied both the initial ratio and the new ratio after the exchange.

This problem demonstrates the importance of breaking down complex scenarios into smaller, manageable steps. By carefully analyzing the given information, translating it into mathematical expressions, and applying appropriate techniques, we can arrive at a solution. The use of ratios and proportions is a fundamental skill in mathematics, with applications ranging from everyday calculations to advanced scientific modeling.

Furthermore, this problem underscores the importance of verification. Checking our solution against the original conditions ensures that our calculations are accurate and our answer is valid. This is a crucial step in any problem-solving process, as it helps us identify potential errors and build confidence in our results.

In conclusion, exploring coin ratios is more than just a mathematical exercise; it's a journey into the world of problem-solving, critical thinking, and the power of mathematical tools. By mastering these concepts, we can tackle a wide range of challenges and gain a deeper appreciation for the elegance and applicability of mathematics. The ability to work with ratios, proportions, and equations is a valuable asset in various fields, from finance and engineering to science and everyday decision-making. This problem serves as a testament to the importance of mathematical literacy and its role in empowering us to understand and navigate the world around us.