MHB How Do You Calculate Loss Percentage in a Quadratic Equation Problem?

NotaMathPerson
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A person, selling a horse for $72, finds that his loss per cent is one-eight of the number of dollars that he paid for the horse; what was the cost price?

Can anybody explain the part " loss per cent" and how do I express that algebraically. Thanks!
 
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NotaMathPerson said:
A person, selling a horse for $72, finds that his loss per cent is one-eight of the number of dollars that he paid for the horse; what was the cost price?

Can anybody explain the part " loss per cent" and how do I express that algebraically. Thanks!

Hey NotaMathPerson! ;)

That is quite ambiguous.

Suppose the cost price is $C$, then his loss is $C - 72$.

It could mean:
1. His loss per cent of the cost price (which would be my expectation). That would mean that we have $\frac{C-72}{100C} = \frac C 8$.
2. His loss per cent of the selling price. That would mean that we have $\frac{C-72}{7200} = \frac C 8$.
3. If there is a typo, his loss percentage of the cost price, meaning $\frac{C-72}{C} = \frac C 8$.
4. If there is a typo, his loss percentage of the selling price, meaning $\frac{C-72}{72} = \frac C 8$.

Options 1 and 3 do not have a solution, so for now I'm inclined to assume we're talking about option 2, but seeing the result I wouldn't be surprised if option 4 was intended.
Anyway, how about solving it for option 2? (Wondering)

EDIT: I have just noticed that your title mentions solving a quadratic equation.
That suggests that option 1 is intended after all...
 
Looks to me like your teacher is getting ready to show the class
that a percentage is really a fraction; like 20% = 20/100 = 1/5.

With your problem: 1/8 = .125, or 12.5%.
82.28 - .125*82.22 = 82.28 - 10.28 = 72.
 
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