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)