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The first, second- and third-class railway fares in the ratio between two stations are in the ratio as $ 6:4:1 $ , and the number of passengers of the three classes are $ 2:5:50 $ . If the sale of the tickets of three classes amounts to RS 12300, find the total collection from each class.

Answer Verified Verified
Hint: Let us assume that ‘x’ be the railway fares and ‘y’ be the number of passengers. Find the railway fare as well as the number of passengers for each class using the ratios:
Railway fare ration = $ 6:4:1 $
Passengers ratio = $ 2:5:50 $
Sale of the ticket from one class = railway fare $ \times $ number of passengers, i.e. $ x\times y $
Calculate the sale of tickets from each class using $ x\times y $ and add all to get the total collection.
To find the exact collection from each class, equate the total collection in terms of ‘x’ and ‘y’ to Rs. 12300 and get the value of $ x\times y $ , then substitute in the equations of each class ticket sale to get the collection values.

Complete step-by-step answer:
Let us assume that ‘x’ be the railway fares and ‘y’ be the number of passengers.
Since, the first, second- and third-class railway fares in the ratio between two stations are in the ratio as $ 6:4:1 $ .Therefore, we can write railway fare of each class as:
First class = 6x
Second class = 4x
Third class = x
Similarly, we know that the number of passengers of the three classes are in the ratio $ 2:5:50 $ . So, we can write the number of passengers in each class as:
First class = 2y
Second class = 5y
Third class = 50y
Now, sale of tickets from one class = railway fare $ \times $ number of passengers, i.e. $ x\times y $
So, we can compute the sale of ticket from each class as:
First class
 $ \begin{align}
  & =6x\times 2y \\
 & =12xy......(1) \\
\end{align} $
Second class
 $ \begin{align}
  & =4x\times 5y \\
 & =20xy......(2) \\
\end{align} $
Third class
\[\begin{align}
  & =x\times 50y \\
 & =50xy.......(3) \\
\end{align}\]
We have to add equation (1), (2) and (3) to get the total collection of railways,
i.e.
\[\begin{align}
  & =12xy+20xy+50xy \\
 & =82xy.......(4) \\
\end{align}\]
Since, it is given that the total collection = Rs 12300
By substituting the value in equation (4), we get:
 $ \begin{align}
  & 82xy=123000 \\
 & xy=150......(5) \\
\end{align} $
Now, we will put the value of $ x\times y $ from equation (5) in equation (1), (2) and (3) to get the ticket sale from each sale, i.e.,
First class
 $ \begin{align}
  & =12xy \\
 & =12\times 150 \\
 & =1800 \\
\end{align} $
Second class
 $ \begin{align}
  & =20xy \\
 & =20\times 150 \\
 & =3000 \\
\end{align} $
Third class
 $ \begin{align}
  & =50xy \\
 & =50\times 150 \\
 & =7500 \\
\end{align} $
Hence, the total collection from the first class is Rs 1800, the second class is Rs 3000 and the third class is 7500.

Note: Whenever two quantities are given in ratio, it doesn’t mean that the value of the quantity is equal to the given ratio.
For example:
If $ a:b=3:5 $ , it doesn’t mean that a = 3 and b = 5.
Instead, assume that $ a=3x $ and $ b=5x $ where ‘x’ is a constant.