
If the cost of 15 tables is equal to the selling price of 20 tables. Find the loss or gain per cent.
Answer
579.6k+ 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.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

A moving boat is observed from the top of a 150 m high class 10 maths CBSE

