MHB Problems leading to quadratic equation

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A man spent $78 on cigarettes, and if the price per box had been 50 cents less, he could have purchased one additional box. The original price per box is represented as y, leading to the equations BP = 7800 and (B+1)(P-0.5) = 7800. The correct approach involves using the number of boxes as a variable, simplifying to B^2 + B - 156 = 0, which factors to (B+13)(B-12) = 0. The solution indicates he bought 12 boxes, and a key mistake was using 780 instead of 7800 when calculating in cents. This correction clarifies the problem and confirms the number of boxes purchased.
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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