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Let us assume the cost price of the good as x. We can infer that the cost price is increased by 30% initially and then decreased by $15\%$ to get the final selling price.

If a number ‘y’ is increased by k%, it means that ‘y’ is increased by $k\%$ of ‘y’. Numerically, ‘y’ is increased by $k\%$ of ‘y’ can be written as $y+\left( \dfrac{k}{100}\times y \right)$, similarly ‘y’ is decreased by $k\%$ of ‘y’ can be written as $y-\left( \dfrac{k}{100}\times y \right)$.

We should do the problem in two steps.

Step-1 Here is increasing the selling price by $30\%$ of the cost price. From the above relation, substituting y = x and k = 30 in increasing formula, we get

$S.{{P}_{1}}=x+\left( \dfrac{30}{100}\times x \right)=x+0.3x=1.3x$

Step-2 is decreasing the effective selling price after step-1 by $15\%$. From the decreasing formula above, substituting y= $S.{{P}_{1}}$ and k = $15\%$, we get

$\begin{align}

& S.{{P}_{2}}=S.{{P}_{1}}-\left( \dfrac{15}{100}\times S.{{P}_{1}} \right)=1.3x-\left( \dfrac{15}{100}\times 1.3x \right)=1.3x-\left( 0.15\times 1.3 \right)x=1.3x-.195x \\

& S.{{P}_{2}}=1.105x \\

\end{align}$

$S.{{P}_{2}}$ is the final selling price of the product.

Profit or loss is given by the formula Profit or Loss = $\dfrac{S.P-C.P}{C.P}\times 100$.

$S.P=S.{{P}_{2}}=1.105x$

$C.P=x$

Using them in the formula, we get

Profit or loss = $\dfrac{1.105x-x}{x}\times 100=\dfrac{0.105x}{x}\times 100=0.105\times 100=10.5\%$

The answer is positive. So, we can infer that it is a profit or gain of $10.5\%$.