
If the cost of 15 tables is equal to the selling price of 20 tables. Find the loss or gain per cent.
Answer
588.9k+ views
Hint: For answering questions of this type we should assume the cost and selling price of the objects. And compare them and come to a conclusion will it be profit or loss. If cost is greater than selling price then it is loss and otherwise it is profit. Now we will assume the cost of a table will be $x$ and also assume that the selling price of a table will be $y$ in order to solve this question.
Complete step-by-step answer:
As given in the question the cost of 15 tables is equal to the selling price of 20 tables.
So let us assume that the cost of a table is $x$. And also assume that the selling price of a table is $y$.
So, here the cost of 15 tables will be $15x$ and the selling price of 20 tables will be $20y$.
As it is given that we can conclude $15x=20y\Rightarrow 3x=4y$ .
As we have $3x=4y\Rightarrow y=\dfrac{3x}{4}$. Since we know $x>\dfrac{3x}{4}$ .So we can conclude that $x>y$.
That is the cost of a table is greater than the selling price of a table. Hence, there will be loss.
The loss per cent is given by
$\begin{align}
& \dfrac{\text{loss}}{\text{cost}}\times 100 \\
& \Rightarrow \dfrac{\text{cost-selling price}}{\text{cost}}\times 100 \\
\end{align}$
By substituting the values we have $\Rightarrow \dfrac{x-y}{x}\times 100$
Since, we have $3x=4y$ we can say that $y=\dfrac{3x}{4}$.
By substituting the value of $y$, we will have loss per cent equal to
$\begin{align}
& \Rightarrow \dfrac{x-y}{x}\times 100 \\
& \Rightarrow \dfrac{x-\dfrac{3x}{4}}{x}\times 100 \\
& \Rightarrow \dfrac{\dfrac{x}{4}}{x}\times 100 \\
& \Rightarrow \dfrac{1}{4}\times 100 \\
\end{align}$
$\Rightarrow 25\%$
Hence we can conclude that the loss per cent is $25\%$.
Note: The loss percentage is $\dfrac{\text{loss}}{\text{cost}}\times 100\Rightarrow \dfrac{\text{cost-selling price}}{\text{cost}}\times 100$. The profit percentage is $\dfrac{\text{profit}}{\text{cost}}\times 100\Rightarrow \dfrac{\text{selling price-cost}}{\text{cost}}\times 100$ . While answering questions of this type we should be sure whether there will be loss or profit. There is a difference between loss and profit. For profit the selling price will be greater than the cost and for loss it will be reverse.
Complete step-by-step answer:
As given in the question the cost of 15 tables is equal to the selling price of 20 tables.
So let us assume that the cost of a table is $x$. And also assume that the selling price of a table is $y$.
So, here the cost of 15 tables will be $15x$ and the selling price of 20 tables will be $20y$.
As it is given that we can conclude $15x=20y\Rightarrow 3x=4y$ .
As we have $3x=4y\Rightarrow y=\dfrac{3x}{4}$. Since we know $x>\dfrac{3x}{4}$ .So we can conclude that $x>y$.
That is the cost of a table is greater than the selling price of a table. Hence, there will be loss.
The loss per cent is given by
$\begin{align}
& \dfrac{\text{loss}}{\text{cost}}\times 100 \\
& \Rightarrow \dfrac{\text{cost-selling price}}{\text{cost}}\times 100 \\
\end{align}$
By substituting the values we have $\Rightarrow \dfrac{x-y}{x}\times 100$
Since, we have $3x=4y$ we can say that $y=\dfrac{3x}{4}$.
By substituting the value of $y$, we will have loss per cent equal to
$\begin{align}
& \Rightarrow \dfrac{x-y}{x}\times 100 \\
& \Rightarrow \dfrac{x-\dfrac{3x}{4}}{x}\times 100 \\
& \Rightarrow \dfrac{\dfrac{x}{4}}{x}\times 100 \\
& \Rightarrow \dfrac{1}{4}\times 100 \\
\end{align}$
$\Rightarrow 25\%$
Hence we can conclude that the loss per cent is $25\%$.
Note: The loss percentage is $\dfrac{\text{loss}}{\text{cost}}\times 100\Rightarrow \dfrac{\text{cost-selling price}}{\text{cost}}\times 100$. The profit percentage is $\dfrac{\text{profit}}{\text{cost}}\times 100\Rightarrow \dfrac{\text{selling price-cost}}{\text{cost}}\times 100$ . While answering questions of this type we should be sure whether there will be loss or profit. There is a difference between loss and profit. For profit the selling price will be greater than the cost and for loss it will be reverse.
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