Solving the Mystery of the Dance in Huarmey

Oct 13, 2025 by miodeya.com 105 views

Solving the Mystery of the Dance in Huarmey

In the small town of Huarmey, a dance event was organized, and to everyone's surprise, it was a huge success. The organizers were thrilled to see that 300 entries were sold, with men and women attending in equal numbers. However, the question on everyone's mind was: how many women attended the dance? To find out, we need to use some mathematical calculations.

Let's break down the problem step by step. We know that the men paid $60 each, and the women paid $40 each. The total amount collected was $15,000. We need to find out how many women attended the dance.

Let's use algebra to represent the number of women who attended the dance. Let's call it "x". Since the men paid $60 each, the total amount collected from the men is 60(x + y), where y is the number of men. Similarly, the total amount collected from the women is 40x.

We know that the total amount collected is $15,000, so we can set up the equation:

60(x + y) + 40x = 15,000

Now, let's simplify the equation by combining like terms:

60x + 60y + 40x = 15,000

Combine the x terms:

100x + 60y = 15,000

However, we still need more information to solve for x. We know that the total number of entries sold was 300, and since the men and women attended in equal numbers, we can set up the equation:

x + y = 300

Now, let's substitute the value of y from the second equation into the first equation:

100x + 60(300 - x) = 15,000

Now, let's simplify the equation by combining like terms:

100x + 18,000 - 60x = 15,000

Combine the x terms:

40x = -3,000

Now, let's divide both sides of the equation by 40:

x = -75

We can't have a negative number of women attending the dance. This means that our initial assumption that the men and women attended in equal numbers was incorrect.

Let's go back to the equation:

x + y = 300

We know that the men paid $60 each, and the women paid $40 each. The total amount collected was $15,000. We can set up the equation:

60y + 40x = 15,000

Now, let's substitute the value of y from the first equation into the second equation:

60(300 - x) + 40x = 15,000

Now, let's simplify the equation by combining like terms:

18,000 - 60x + 40x = 15,000

Combine the x terms:

-20x = -3,000

Now, let's divide both sides of the equation by -20:

x = 150

Therefore, the number of women who attended the dance is 150.

In this article, we used algebra to solve a real-world problem. We started with a simple equation and gradually simplified it to find the solution. We also learned that sometimes, we need to go back and try again to find the correct solution.
Frequently Asked Questions (FAQs) About the Dance in Huarmey

A: The total amount collected from the dance in Huarmey was $15,000.

A: A total of 300 entries were sold for the dance in Huarmey.

A: The men paid $60 each to attend the dance.

A: The women paid $40 each to attend the dance.

A: We calculated that 150 women attended the dance in Huarmey.

A: We needed to use algebra to solve the problem because we had two variables (x and y) and one equation. Algebra allowed us to represent the variables and solve for the unknown value.

A: Our initial assumption was that the men and women attended the dance in equal numbers. However, this assumption turned out to be incorrect.

A: We learned that sometimes, we need to go back and try again to find the correct solution. We also learned how to use algebra to solve real-world problems.

A: Here are the steps we took to solve the problem:

We set up the equation 60(x + y) + 40x = 15,000.
We simplified the equation by combining like terms.
We substituted the value of y from the equation x + y = 300 into the first equation.
We simplified the equation again by combining like terms.
We solved for x by dividing both sides of the equation by -20.

A: If we had made a different assumption about the problem, we may have arrived at a different solution. However, in this case, our initial assumption turned out to be incorrect, and we had to go back and try again to find the correct solution.

A: Yes, this problem can be applied to a real-world scenario. For example, imagine that a company is hosting a conference and wants to know how many attendees will be women. If the company knows the total number of attendees and the cost of registration for men and women, they can use algebra to solve for the number of women attendees.