Blackbox
Dear
genxer123,
Thank you for responding. My calculations are based on my assumption
Selling Price - Cost Price = Profit. Is this formula incorrect? What I understood from the prompt is that the Gross Profit was affected as per this condition --
"... increased its gross profit on a product from 10 percent of the cost of the product ...". What I
do not understand is why that condition is equated to Selling Price. For e.g., NEW Sell Price = 1.15CP = $92. I am clearly confused

Dear
Blackbox ,
[EDITED TO DELETE DISCUSSION OF "CONDITION." I don't think that is the problem, because your own formula, with the PROFIT recalculated, yielded the answer.]
Of course the formula is correct! I've seen your posts, including the recent one on that horrible problem about a gasoline price increase (and a guy, Ron, who could afford part of the increase, such that we had to calculate by what percentage he must decrease consumption). You nailed it. I solved it. Barely. In FOUR minutes. I got stuck.
So I think you are just a little stuck on this one.
You are not calculating profit correctly in the equation "Selling Price - Cost Price = Profit"! You are forgetting to subtract 100 percent of the cost price from the equation!
I would PM you this whole thing, but math symbols and formulas are not translated any longer on PMs.
Incorrect figures are in red. Strike through will not work on fractions.
Correct figures are in blue.
I beg you: at some point, use multipliers for percentages?
Selling Price -
Cost Price = Profit
Let CP = x
Old Prof =
\(\frac{110}{100}\)(x)
New Prof =
\(\frac{115}{100}\)(x)
Old Prof = \(\frac{110}{100}\)(x) - 1xOld Prof = \(\frac{110}{100}\) (x) - \(\frac{100}{100}\) (x) = \(\frac{10}{100}\)(x)New Prof = \(\frac{115}{100}\)(x) - 1x
New Prof = \(\frac{115}{100}(x) - \frac{100}{100}(x) = \frac{15}{100}(x)\)Old SP = Old Prof + CP [CP is the same in both conditions as given in the prompt]
=> Old SP =
\(\frac{110}{100}\)(x) + x
=> Old SP =
\(\frac{210}{100}\)(x) + x
Old SP = Old Prof + CP [CP is the same in both conditions as given in the prompt]
=> Old SP= \(\frac{10}{100}\)(x) + x
=> Old SP \(\frac{10}{100}\)(x) +\(\frac{100}{100}\)(x)
=> Old SP = \(\frac{110}{100}\)(x)Also, given
New SP = 92
New SP = New Prof + CP [CP is the same in both conditions as given in the prompt]
=>
92 = \(\frac{115}{100}\)(x) + x
=> 92 = \(\frac{15}{100}\)(x) + x
=>92 = \(\frac{115}{100}\)(x)Solving,
REPEAT THE EQUATION FROM ABOVE92 = \(\frac{115}{100}\)(x)x =
\(\frac{92*115}{100}\)x = 92 * \(\frac{100}{115}\)x = 80So, the CP = \(\frac{92*100}{215}\)So, the CP = 80Now, plug this back into the Old SP equation above.
Old SP =
\(\frac{92*100}{215}\) * \(\frac{210}{100}\)Old SP = (92 * \(\frac{100}{115}\) ) * \(\frac{110}{100}\) = $88I so hope that helps. Your formula was fine. (Hard, but fine.) I just redid your math, and voila, the answer!