How Many Students Were Initially Planned for the History Class Trip?

In summary, 27 students from a history class were planning a trip that would cost them $189. However, 6 students were unable to participate, leaving the remaining 21 students to cover the cost. To avoid cancellation, each of the 21 students had to pay an additional $2, resulting in a total of $189. This led to a quadratic equation, with the solution being x=27, meaning there were originally 27 students in the class.
  • #1
thua
9
0
Students from a history class prepare a trip that will cost them a total of $189 that will be shared between them. The day before the trip, 6 students discovered that they can't participate in the trip. To avoid cancellation, the rest of the class decided to pay an additional $2 each.
How many students are there in this class?

I have:
(x-6) = the number of students going
189/x = the cost for the whole class before the 6 students couldn't go

2(x-6) + 189/x = 189 <=== not sure if this is even right.
2x - 12 + 189/x = 189
2x^2 - 12x + 189 = 189x
2x^2 - 201x + 189
2(x^2 - 100.5x) + 189
2(x - 50.25)^2 - 5050.125 + 189
2(x - 50.25)^2 - 4861.125

That doesn't seem to make sense. So I hope you can help me.
 
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  • #2
this first needs to be set to a quadradic equation as your title suggests, there should be an x and a y this is because the number of students correlates with the amount each student has to pay (or the amount they would have had to pay before the 6 students left)
 
  • #3
thua said:
Students from a history class prepare a trip that will cost them a total of $189 that will be shared between them. The day before the trip, 6 students discovered that they can't participate in the trip. To avoid cancellation, the rest of the class decided to pay an additional $2 each.
How many students are there in this class?
I have:
(x-6) = the number of students going
189/x = the cost for the whole class before the 6 students couldn't go
2(x-6) + 189/x = 189 <=== not sure if this is even right.
You have every right to be unsure! Think about what each part means. x- 6 is the number of students who are going and I assume the "2" is the $2 extra each is paying so 2(x- 6) is the total extra money paid. But "189/x" is the amount each was paying before. Surely you don't want to add the total extra and the amount each was paying. I think what you intended to say was that, previously each was paying 189/x and now each of those who is going must pay an additional $2- so the amount each of the persons going pays is 2+ 189/x. Now that is the amount paid by x- 6 people: (2+ 189/x)(x-6)= 189.

2x - 12 + 189/x = 189
2x^2 - 12x + 189 = 189x
2x^2 - 201x + 189
2(x^2 - 100.5x) + 189
2(x - 50.25)^2 - 5050.125 + 189
2(x - 50.25)^2 - 4861.125
That doesn't seem to make sense. So I hope you can help me.

Start with (2+ 189/x)(x-6)= 189 and try again.
 
  • #4
i think i got it now... thanks a bunch!
 
Last edited:
  • #5
let x = Initial number of students in the group
189/x = cost per student

when 6 students discovered they cannot participate
x-6 = resulting number of students after the 6 was unable to go with the group
this leads to 189/x-6 being the cost divided among remaining students

To avoid cancellation [(189/x) +2] =amount the rest of the class decided to pay.

In order for the trip to push through:

(189/x-6) = [(189/x) +2]
or
(189/x-6) - 189/x = 2

x^2 -6x-567 = 0

x= [6+/- (sqrt (36+4*567))]/2=27 (discarding the negative root)

therefore x=27 is the original number of students
 

FAQ: How Many Students Were Initially Planned for the History Class Trip?

What is a quadratic equation?

A quadratic equation is a mathematical equation that contains a variable, typically represented as x, raised to the second power (or squared). It also includes a linear term (x), and a constant term (represented by a number). The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants.

How do you solve a quadratic equation?

To solve a quadratic equation, you can use the quadratic formula, which is -b ± √(b^2 - 4ac) / 2a. You can also factor the equation or complete the square to find the roots (or solutions) of the equation.

What is the discriminant of a quadratic equation?

The discriminant of a quadratic equation is the part of the quadratic formula that is under the square root symbol: b^2 - 4ac. It is used to determine the nature of the roots (real or imaginary) and how many solutions the equation has. If the discriminant is positive, there are two real solutions. If it is zero, there is one real solution. And if it is negative, there are two imaginary solutions.

What are the applications of quadratic equations?

Quadratic equations are used in many fields, including physics, engineering, economics, and statistics. They can be used to model real-world situations and make predictions, such as in projectile motion or profit analysis. They are also used in computer graphics and in solving optimization problems.

What are some common mistakes when solving quadratic equations?

Some common mistakes when solving quadratic equations include forgetting to square the linear term when completing the square, using the incorrect values for a, b, and c in the quadratic formula, and not checking for extraneous solutions. It is also important to remember to use the correct order of operations and to double check your work for any errors.

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