Intuition behind this algebraic question

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In summary, the problem involves solving two equations with two unknowns (s = students, a = adults) to determine the number of students who attended a concert based on the total number of attendees (950) and the revenue generated ($7550). The equation 2.5s = 1000 arises through algebra and represents the extra revenue that would have been generated had students been charged the same amount as adults. This can be interpreted as the amount of money each student would have paid if they were charged $9, as opposed to the actual price of $6.50.
  • #1
lifelearner
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Hi PF members!

I'd like some guidance with the intuition behind the following problem. I have provided my intuition as far as I can below but if anyone can help clarify it I'd be very thankful.

Question: Tickets to a concert cost $9.00 for adults and $6.50 for students. A total of 950 people paid $7550 to attend. How many students attended the concert?

I. The Algebra
Code:
The answer basically involves solving two equations with two unknowns:

s = students, a = adults

i) a + s = 950 and 9a + 6.5s = 7550

ii) a = 950 - s

iii) sub (ii) into 9a in (i)

iv) 8550 - 7550 = 2.5s

v) s = 400 (?)

II. My intuition (logic) behind the algebra


i) total # of people who came (students and adults) equals 950. Revenue from adults is ($9 per adult)*(total number of adults = a). Revenue from students is ($6.50 per student)*(total number of students = s)

ii) if in total 950 people came and assuming we know how many students came then those that are not students must be adults. In other words, a = 950 - s

iii) 9(950-s) + 6.5(s) = 8550 - 9s + 6.5s = 7550

That is, assume that we charge all 950 people $9. This revenue comes to $8550. Then we take out the students to whom we charged $9, whom are -9s. Then we will be left on one hand with the adults to whom we charged $9 (8550 - 9s). Plus we add the students to whom we charge $6.5 (6.5s). Sum will equal the actual revenue of $7550.


iv) 8550-7550 = 1000 = 2.5s

The revenue that comes from charging students AND adults $9 minus the actual revenue of $7550 from charging ONLY adults $9 gives $1000, which equals the extra $2.5 that all students WOULD'VE paid HAD we charged them $9. How many such students are there? 1000/2.5 = 400.


So, I need your help with iv in part II. How do I interpret equation iv? And that, too, logically?
 
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  • #2
lifelearner said:
Hi PF members!
iv) 8550-7550 = 1000 = 2.5s

The revenue that comes from charging students AND adults $9 minus the actual revenue of $7550 from charging ONLY adults $9 gives $1000, which equals the extra $2.5 that all students WOULD'VE paid HAD we charged them $9. How many such students are there? 1000/2.5 = 400.

It looks like you've got it, what do you need help with?
 
  • #3
How do you interpret in your words 2.5s = 1000?

That is, within the context of this problem, how do you interpret $2.5 per student times some number of students equals $1000?
 
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  • #4
lifelearner,

Am I correct in assuming that you did not find the algebra, but are rather looking at the logic behing the algebra? If so:

Part iv has skipped a couple of steps to get where it is. Basic algebra. Looking at step iii we get:

a = 950 - s
This can be altered to get,
9a = 8550 - 9s
Then substituting into equation i,
8550 - 9s + 6.5s = 7550
Simplify,
8550 - 2.5s = 7550
8550 = 7550 + 2.5s
8550 - 7550 = 2.5s
2.5s = 1000
s = 400

To find a,
a = 950 - s
a = 950 - 400
a = 550

And checking,
9a + 6.5s = 7550
9*550 + 6.5*400 = 7550
4950 + 2600 = 7550


So there you go, the 2.5s merely arises through algebra.

Now, if you did find this already, and you worked out the algebra... I don't know what the problem is :smile:
 
  • #5
I understand completely the algebra. However, the physical meaning of 2.5s is what I don't understand. Somehow I feel Venn diagrams might be useful; i.e: 8550 is revenue you get if you charge all persons $9. When you subtract 9s then you effectively "filter" out the students whom you charge $9, which will leave you with revenue from adults, each of whom paid $9.

I understand 2.5s arises through algebra but can we put some meaning behind it?
 
  • #6
lifelearner said:
I understand completely the algebra. However, the physical meaning of 2.5s is what I don't understand. Somehow I feel Venn diagrams might be useful; i.e: 8550 is revenue you get if you charge all persons $9. When you subtract 9s then you effectively "filter" out the students whom you charge $9, which will leave you with revenue from adults, each of whom paid $9.

I understand 2.5s arises through algebra but can we put some meaning behind it?

I suppose the meaning might be 2.5 times the number of students is 1000?
 
  • #7
What is the 1000 mean in physical meaning? And of what significance is 2.5 times # of students?
 
  • #8
Why should it mean anything at all, other than being provably logically equivalent to a previous statement?

You could multiply that equation with, say, 3.72 and you'd still have a logically equivalent equation, but it wouldn't have much "meaning" in the sense you are seeking.
 
  • #9
arildno said:
Why should it mean anything at all, other than being provably logically equivalent to a previous statement?

You could multiply that equation with, say, 3.72 and you'd still have a logically equivalent equation, but it wouldn't have much "meaning" in the sense you are seeking.

Should it not mean something as we are dealing with a practical question?
 

1. What is the intuition behind algebraic equations?

The main intuition behind algebraic equations is to find the unknown values in a given mathematical problem. It involves using symbols and variables to represent unknown quantities and manipulating them through different operations to arrive at a solution.

2. Why is algebra important in science?

Algebra is important in science because it allows us to describe and solve complex problems using mathematical equations. It also helps us understand patterns and relationships between variables, which is crucial in scientific research and experimentation.

3. How does algebra relate to other branches of mathematics?

Algebra is closely related to other branches of mathematics, such as geometry, calculus, and statistics. It provides a foundation for these branches and is often used to solve problems in these fields. For example, calculus uses algebraic equations to find derivatives and integrals.

4. What are some real-life applications of algebra?

Algebra has many real-life applications, such as calculating distances and speeds in physics, predicting population growth in biology, and designing circuits in electrical engineering. It is also used in finance, computer science, and many other fields.

5. How can I improve my intuition in algebra?

To improve your intuition in algebra, it is important to practice solving various types of equations and problems. You can also try visualizing the equations and understanding the underlying concepts behind them. Additionally, seeking help from a teacher or tutor can also aid in improving your intuition in algebra.

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