Find the Price of a Shoe on the Second Day | Simple Word Problem Solution

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The problem involves calculating the price of shoes on the second day after a reduction. Initially, shoes were priced at $30, and sales increased by 50% while revenue rose by 25%. An algebraic approach was attempted, but the user struggled to formulate a second equation. It was clarified that one variable would drop out, simplifying the solution. The final price of the shoes sold on the second day is determined to be $25.
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Homework Statement


On the first day, shoes in a shop were sold for $30 each. On the second day, the price of a shoe was reduced and the number of shoes sold on the second day increased by 50% while the money collected increased by 25%. Find the price of a shoe sold on the second day.

The Attempt at a Solution



well, I tried using an algebraic solution but got stuck. I used two unknowns, where

x=number of shoes sold on the first day
y=amount reduced for the price of a shoe sold on the second day.

I only managed to yield one equation from the problem which was not enough obviously,

amount collected on saturday:
1.5x(30-y)=30x\times1.25

please help, I must be missing another equation somewhere. Also, the final answer should be $25.
 
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Oerg said:
amount collected on saturday:
1.5x(30-y)=30x\times1.25

please help, I must be missing another equation somewhere.
Looks fine to me. (Note that one of your variables drops out.) Be sure to express the answer as the new price of the shoes, not the reduction in price.
 
thanks doc ai I didn't realize one of my variables would drop out.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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