Maximizing Purchases: S/.40 & S/.100 Items Within S/.1000 Budget

Hey guys! Ever wondered how to stretch your budget to get the most out of your money? Let's dive into a real-world problem where Juan needs to figure out how to buy the maximum number of items with a limited budget. This is a classic optimization problem that we can solve with a bit of math and logical thinking. This guide will walk you through a step-by-step solution, ensuring you understand every part of the process. So, buckle up and let's get started!

Understanding the Problem

In this scenario, Juan has a budget of S/.1000 and wants to purchase items that come in two different price ranges: S/.40 and S/.100 per item. Our main goal is to figure out the maximum number of items Juan can buy without exceeding his budget. This problem involves a bit of number juggling, but don't worry, we'll break it down so it's super easy to follow. The key here is to understand the constraints: Juan has a fixed budget, and we need to find the optimal combination of items to maximize the quantity he buys. Think of it like a puzzle where each piece (the number of items at each price) needs to fit perfectly to reach the maximum number of items within the budget. It’s not just about buying as many cheap items as possible or a few expensive ones; it's about finding the right balance. This requires a strategic approach, considering the trade-offs between buying more of the cheaper items versus fewer of the expensive ones. Let's put on our thinking caps and explore how we can help Juan make the smartest purchasing decision possible!

Setting Up the Equation

To solve this problem effectively, we need to translate the word problem into a mathematical equation. This helps us represent the situation clearly and makes it easier to find a solution. Let's define our variables:

• Let x be the number of items Juan buys at S/.40.
• Let y be the number of items Juan buys at S/.100.

Now, we can express Juan's total spending as an equation. The total cost of the items should be exactly S/.1000. So, the equation is:

40x + 100y = 1000

This equation is the heart of our solution. It tells us that the sum of the money spent on the S/.40 items and the S/.100 items must equal Juan's budget. This equation sets the stage for us to explore different combinations of x and y that satisfy this condition. Think of it as a balancing act: we need to find the right combination of the number of cheaper items (x) and the number of pricier items (y) so that the scale tips perfectly at S/.1000. By setting up this equation, we've transformed a real-world problem into a mathematical one, making it much easier to tackle. Now, we can use mathematical techniques to find the values of x and y that give us the maximum number of items.

Simplifying the Equation

Before we start plugging in numbers, let's make our equation a little easier to work with. We can simplify the equation 40x + 100y = 1000 by dividing all terms by their greatest common divisor, which is 20. This will give us smaller coefficients, making the calculations simpler.

Dividing each term by 20, we get:

(40x / 20) + (100y / 20) = 1000 / 20

Which simplifies to:

2x + 5y = 50

This simplified equation is much more manageable. It represents the same relationship between x and y as the original equation but with smaller numbers. This simplification is a crucial step because it reduces the chances of making errors in our calculations and makes it easier to spot potential solutions. Think of it as cleaning up a cluttered workspace before starting a project; it helps you focus on the task at hand. With this simplified equation, we're now ready to find the integer solutions that maximize the total number of items Juan can buy. We've essentially streamlined the problem, setting ourselves up for a more efficient solution-finding process. So, let's move on to finding those solutions!

Finding Possible Solutions

Now that we have our simplified equation, 2x + 5y = 50, we need to find integer solutions for x and y. Remember, x and y represent the number of items, so they must be whole numbers (you can't buy a fraction of an item!). To find the maximum number of items, we’ll explore different values for y (the number of S/.100 items) and then solve for x (the number of S/.40 items).

Here's how we can approach this:

1. Start with possible values for y: Since y represents the number of S/.100 items, and Juan has a budget of S/.1000, the maximum value for y is 10 (1000 / 100 = 10). We’ll start with y = 0 and go up to y = 10.
2. Solve for x: For each value of y, we’ll plug it into the equation and solve for x. Remember, x must be a non-negative integer.
3. Check for Integer Solutions: If x turns out to be a whole number, we have a valid solution. If not, we discard that value of y.
4. Calculate Total Items: For each valid solution, we’ll calculate the total number of items (x + y).

Let's walk through some examples:

• If y = 0:
  • 2x + 5(0) = 50
  • 2x = 50
  • x = 25
  • Total items: 25 + 0 = 25
• If y = 1:
  • 2x + 5(1) = 50
  • 2x = 45
  • x = 22.5 (Not an integer, so we discard this solution)
• If y = 2:
  • 2x + 5(2) = 50
  • 2x = 40
  • x = 20
  • Total items: 20 + 2 = 22

We’ll continue this process for all possible values of y (up to 10) and keep track of the solutions that give us integer values for x. This methodical approach ensures we don't miss any potential solutions and helps us identify the one that maximizes the total number of items Juan can buy. It's like carefully checking all the boxes to make sure we've explored every possibility. So, let's keep crunching those numbers and see what we find!

Identifying the Maximum

After systematically testing different values for y and solving for x, we need to compile our results and identify the solution that gives Juan the maximum number of items. Remember, we're looking for integer solutions (whole numbers) for both x and y, as they represent the number of items.

Here’s a summary of the valid solutions we found:

• y = 0, x = 25: Total items = 25
• y = 2, x = 20: Total items = 22
• y = 4, x = 15: Total items = 19
• y = 6, x = 10: Total items = 16
• y = 8, x = 5: Total items = 13
• y = 10, x = 0: Total items = 10

Now, let's analyze these solutions. We can see that when Juan buys 25 items at S/.40 each (x = 25) and no items at S/.100 (y = 0), he maximizes the total number of items he can purchase within his S/.1000 budget.

Therefore, the maximum number of items Juan can buy is 25.

This step is crucial because it's where we put all our calculations into perspective and draw a conclusion. It's not just about finding a solution; it's about finding the best solution. By comparing the total number of items for each valid combination of x and y, we can confidently say that Juan can get the most items by sticking to the S/.40 ones. This part of the process highlights the importance of a thorough analysis to ensure we've truly optimized Juan's spending. So, with our solution in hand, let’s move on to the final conclusion!

Conclusion

Alright guys, we've made it to the end! After carefully analyzing the problem and crunching the numbers, we've found the solution for Juan. To recap, Juan wants to buy items that cost S/.40 and S/.100, and he has a budget of S/.1000. Our goal was to determine the maximum number of items he can purchase.

By setting up an equation, simplifying it, and exploring different possibilities, we discovered that Juan can buy a maximum of 25 items. This is achieved when he buys 25 items at S/.40 each and no items at S/.100.

So, the final answer is:

Juan can buy a maximum of 25 items.

This exercise demonstrates how we can use math to solve real-world problems and make smart purchasing decisions. Understanding constraints and optimizing resources is a valuable skill in many areas of life. This problem, while seemingly simple, touches on important concepts in optimization and resource management. It shows us that sometimes, the best solution isn't always the most obvious one, and it's worth taking the time to work through the numbers to find the optimal outcome. Plus, it’s pretty cool to help Juan get the most bang for his buck! I hope you found this breakdown helpful and maybe even a little fun. Until next time, keep those problem-solving skills sharp!