Problems leading to quadratic equation

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Discussion Overview

The discussion revolves around a problem that leads to a quadratic equation related to the purchase of cigarettes. Participants explore different approaches to formulating the problem mathematically, focusing on the relationship between the price per box and the number of boxes bought, with an emphasis on identifying errors in the initial setup.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant proposes using $y$ as the original price per box and $y-50$ as the new price, leading to a quadratic equation $y^2-50y-39000=0$.
  • Another participant suggests letting $B$ be the number of boxes and $P$ the price per box, leading to the equations $BP=78$ and $(B+1)(P-\frac{1}{2})=78$.
  • A later reply points out that the first participant should use 7800 instead of 780, indicating a potential error in the unit conversion from dollars to cents.
  • There is a suggestion that formulating the problem in terms of the number of boxes rather than the price may provide a more straightforward approach.

Areas of Agreement / Disagreement

Participants express differing approaches to the problem without reaching a consensus on the best method. There is acknowledgment of mistakes in the initial setup, but no agreement on a single correct solution.

Contextual Notes

There are unresolved issues regarding the correct formulation of the equations and the appropriate units to use, which may affect the outcome of the calculations.

NotaMathPerson
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a man spent 78 dollars for cigarettes. has the price per box been .50 cents less, he could have had one more box. How many boxes did he buy?

Heres what I tried

let $y=$ original price per box
$y-50=$ new price per box

Now,

$\frac{780}{y}=$ original number of boxes bought
$\frac{780}{y-50}=$ new number of boxes bought$\frac{780}{y-50}=\frac{780}{y}+1=$

$y^2-50y-39000=0$

Solving for y I get decimal number.

Can you tell me where my mistake is?

Thanks!

All variables are y. I edited it.
 
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I would let $B$ be the number of boxes he bought...and $P$ be the price (in dollars) per box, such that we may state:

$$BP=78$$

$$(B+1)\left(P-\frac{1}{2}\right)=78$$

I am assuming you meant "had the price per box been 50 cents less."

Solving the first equation for $P$ and substituting into the second, we obtain:

$$(B+1)\left(\frac{78}{B}-\frac{1}{2}\right)=78$$

Multiply through by $2B\ne0$:

$$(B+1)(156-B)=156B$$

Expand and write in standard form:

$$B^2+B-156=0$$

Factor:

$$(B+13)(B-12)=0$$

Discard the negative root, and we have:

$$B=12$$
 
Hello MarkFl!

Can you tell what I did wrong in my attempt above?
 
NotaMathPerson said:
Hello MarkFl!

Can you tell what I did wrong in my attempt above?

Well, you are being asked for the number of boxes, so you want to get an equation using a variable that represents the number of boxes. That's more direct than solving for the price...but you can answer the question this way.

In your equations you should be using 7800 instead of 780, since you are using cents instead of dollars. See if that fixes things...:D
 
MarkFL said:
Well, you are being asked for the number of boxes, so you want to get an equation using a variable that represents the number of boxes. That's more direct than solving for the price...but you can answer the question this way.

In your equations you should be using 7800 instead of 780, since you are using cents instead of dollars. See if that fixes things...:D

Oh yes! That careless mistake. Thank you very much!
 

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