Math Problems Averages, Gifts, And Parking Costs

Let's dive into the first mathematical puzzle that involves the concept of averages. Average weight can often be a tricky concept to grasp, but it becomes clearer when we break it down step by step. The question poses a scenario where we have a group of men, and we know the average weight of a subset of them, and the overall average weight of the entire group. Our task is to determine the weight of the missing member, in this case, the fourth man.

To solve this, we need to first understand the fundamental definition of average. The average weight is calculated by summing the weights of all the individuals and then dividing by the total number of individuals. Mathematically, this can be expressed as:

Average = (Sum of Weights) / (Number of Individuals)

In our scenario, we are given that the average weight of 4 men is 79 kg. This means that if we were to add up the weights of all 4 men and divide by 4, the result would be 79 kg. We can represent this mathematically as:

79 kg = (Weight of Man 1 + Weight of Man 2 + Weight of Man 3 + Weight of Man 4) / 4

Similarly, we are also given that the average weight of 3 men is 82 kg. This can be expressed as:

82 kg = (Weight of Man 1 + Weight of Man 2 + Weight of Man 3) / 3

Now, our goal is to find the weight of the 4th man. To do this, we can use the information we have gathered from the two average calculations. First, let's calculate the total weight of the 4 men. We can do this by multiplying the average weight of the 4 men by the number of men:

Total weight of 4 men = 79 kg * 4 = 316 kg

Similarly, let's calculate the total weight of the 3 men:

Total weight of 3 men = 82 kg * 3 = 246 kg

Now, we have the total weight of the 4 men and the total weight of the 3 men. The difference between these two values will give us the weight of the 4th man. This is because the total weight of the 4 men includes the weight of the 4th man, while the total weight of the 3 men does not.

Weight of 4th man = Total weight of 4 men - Total weight of 3 men

Weight of 4th man = 316 kg - 246 kg = 70 kg

Therefore, the weight of the 4th man is 70 kg. This problem demonstrates how understanding the concept of averages and using basic algebraic manipulations can help us solve real-world problems. Remember, the average weight is just one application of the concept of averages, which can be used in various fields, from statistics to finance. The key is to break down the problem into smaller, manageable steps and apply the relevant formulas and concepts.

Moving on to our second intriguing mathematical challenge, we encounter a scenario involving a gift exchange at a party. This problem delves into the realm of combinatorics, a branch of mathematics concerned with counting and arranging objects. The problem states that in a party, 7 people each gave a gift to every other person. The question we need to answer is: How many gifts were given in total?

This may seem like a simple counting problem at first glance, but it's important to approach it systematically to avoid errors. A common mistake is to simply multiply the number of people by the number of gifts each person gave. However, this would lead to overcounting, as it would count each gift twice (once for the giver and once for the receiver). To accurately solve this problem, we need to carefully consider how each gift exchange contributes to the total count.

Let's break down the problem. We have 7 people at the party. Each person gives a gift to every other person, which means each person gives gifts to 6 other people (since they don't give a gift to themselves). If we were to simply multiply 7 people by 6 gifts each, we would get 42. However, as mentioned earlier, this overcounts the gifts.

To understand why, consider two people, person A and person B. Person A gives a gift to person B, and person B gives a gift to person A. Our initial calculation would count this as two gifts, but in reality, it's just one exchange of gifts. To correct for this overcounting, we can use the concept of combinations.

In combinatorics, a combination is a way of selecting items from a set where the order of selection doesn't matter. In our case, we are selecting pairs of people who exchange gifts. The number of ways to choose 2 people from a group of 7 is given by the combination formula:

nCr = n! / (r! * (n-r)!)

Where:

• n is the total number of items (in our case, 7 people)
• r is the number of items being chosen (in our case, 2 people)
• ! denotes the factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1)

Plugging in our values, we get:

7C2 = 7! / (2! * 5!)

7C2 = (7 * 6 * 5 * 4 * 3 * 2 * 1) / (2 * 1 * 5 * 4 * 3 * 2 * 1)

7C2 = (7 * 6) / (2 * 1)

7C2 = 42 / 2

7C2 = 21

Therefore, there are 21 ways to choose 2 people from a group of 7. This means that there are 21 gift exchanges in total. Each gift exchange involves two gifts being given (one in each direction), so the total number of gifts given is 21.

This problem highlights the importance of careful counting and understanding combinatorial principles. By using the combination formula, we were able to accurately determine the total number of gifts exchanged without overcounting. This type of problem demonstrates how mathematical concepts can be applied to everyday scenarios, making problem-solving more engaging and intuitive.

Our final mathematical challenge takes us to the realm of parking costs. This problem introduces a real-world scenario where the cost of parking varies depending on the duration. We are given a tiered pricing structure: a fixed rate for the initial hours and an additional hourly rate for the subsequent hours. The challenge is to calculate the total parking cost based on the given rates and the total parking time.

The problem states that parking costs P15.00 for the first 4 hours and P8.00 for each additional hour or part thereof. To effectively solve this problem, we need to break it down into two distinct parts: the cost for the initial 4 hours and the cost for the additional hours. This approach allows us to apply the correct pricing structure to each segment of the parking time.

First, let's consider the cost for the initial 4 hours. The problem explicitly states that the cost for the first 4 hours is P15.00. This is a fixed cost, regardless of whether the car is parked for the full 4 hours or less. Therefore, if a car is parked for 4 hours or less, the parking cost will be P15.00.

Now, let's consider the cost for the additional hours. For every hour or part thereof exceeding the initial 4 hours, an additional charge of P8.00 is incurred. This means that if a car is parked for, say, 5 hours, there is an additional hour beyond the initial 4 hours, and an additional charge of P8.00 will be added to the base cost of P15.00.

To calculate the total parking cost, we need to determine the number of additional hours and multiply it by the additional hourly rate. If the total parking time is less than or equal to 4 hours, the number of additional hours is zero. If the total parking time exceeds 4 hours, we subtract 4 from the total parking time to get the number of additional hours.

For example, if a car is parked for 6 hours, the number of additional hours is 6 - 4 = 2 hours. The cost for these additional 2 hours would be 2 hours * P8.00/hour = P16.00. The total parking cost would then be the base cost of P15.00 plus the additional cost of P16.00, which equals P31.00.

Let's generalize this process into a formula. Let T be the total parking time in hours, and let C be the total parking cost. Then, the parking cost can be calculated as follows:

If T <= 4, then C = P15.00

If T > 4, then C = P15.00 + (T - 4) * P8.00

This formula encapsulates the tiered pricing structure of the parking fees. It accurately calculates the total cost based on the total parking time, taking into account the fixed rate for the initial hours and the additional hourly rate for subsequent hours.

This problem demonstrates how mathematical concepts, such as tiered pricing structures and conditional calculations, are used in everyday life. By breaking down the problem into smaller parts and applying the appropriate formulas, we can easily calculate the total parking cost. This type of problem highlights the practical application of mathematics in managing finances and making informed decisions.

In conclusion, we have explored three diverse mathematical problems ranging from averages and combinatorics to tiered pricing structures. Each problem presented a unique challenge and required a different set of mathematical tools to solve. By understanding the underlying concepts and applying logical reasoning, we were able to successfully unravel each puzzle. These examples showcase the versatility and relevance of mathematics in everyday life, emphasizing the importance of developing strong problem-solving skills.