Question

A shopkeeper marks his goods $30\%$ above the cost price and gives a discount of $15\%$ on the market price. What gain or loss did he make?

Answer 1

Hint: To solve this question, we have to assume the cost price as x. 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 can infer that the cost price is increased by $30\%$ initially and then decreased by $15\%$ to get the final selling price. Using the above relation by taking y = x and k = 30 we get the increased cost price of the good. After that, by taking the new cost price and k = 15 in the decreased formula, we get the final selling price. By using the relation Profit or Loss = $\dfrac{S.P-C.P}{C.P}\times 100$ we get the answer. If it is positive, it is profit or gains and if it is negative, it is a loss.

Complete step-by-step solution:
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 $10.5\%$.
$\therefore $ The shopkeeper will get a gain of $10.5\%$.

Note: Students can make a common mistake by just subtracting $15\%$ from $30\%$ and give an answer as $15\%$. This is done by not taking into account that the later $15\%$ discount should be applied to the increased selling price but not on the initial cost price. A systematic approach like in the above procedure will help the student is not making any mistake like this one.