Blackbox wrote:
I am not sure I am following this question (or rather the solution that
Bunuel responded with). Here is my thought:
Selling Price - Cost Price = Profit
Let CP = x
Old Prof = \(\frac{110}{100}\)(x)
New Prof =\(\frac{115}{100}\)(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)
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
Solving,
x = \(\frac{92*100}{215}\)
So, the CP = \(\frac{92*100}{215}\)
Now, plug this back into the Old SP equation above.
Old SP = \(\frac{92*100}{215}\) * \(\frac{210}{100}\)
This yields a value of 89.86. Where am I going wrong? This, of course, is a loooong-winded math.
Blackbox, I think you added in CP twice. See, e.g., partial quote of your post below.
You define CP as x, but then also define old SP, it seems, as
\(\frac{110}{100}\)(x) + x =
1.10
x +
xYour original "100" base of the percentage of CP gets used twice, once invisibly in the second term, x. Given your direction, I
think your equation should have been:
\(\frac{10}{100}\)(x) + \(\frac{100}{100}\)(x) =
\(\frac{110}{100}\)(x) = Old SP
Old SP is not \(\frac{210}{100}\)(x), or 2.10x
Old SP is 1.10x. See my solution above.
Similarly, you're right on track when you define new SP as \(\frac{115}{100}\)(x). So you could just write
\(\frac{115}{100}\)(x) = 92, then solve. You would get cost price of 80.
I didn't run your math without the "extra" CP. But that's at least one issue. Easy mistake to make. Hope that helps.
Quote:
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)
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