Question

A 20% hike in bus fare results in a 10% fall in passenger traffic still the daily collection at the bus depot is increased by Rs. 150. The daily collection at the depot after the fare hike is
(A). 1600
(B). 1750
(C). 2025
(D). 1875

Answer 1

Hint: To solve this question, we will find the daily collection at the bus depot on the day before the hike and after calculating that, we will find the daily collection at the bus depot after the hike. We have to use as few variables as possible.

Complete step-by-step solution -
Now, we will use the information given in the question, to make the equations. Let the fair before the hike is ‘x’ and the total passengers coming before the hike is ‘y’. The total collection before the hike is denoted by ‘C’. Now the total collection can be obtained by multiplying the fair of one passenger to the total number of passengers. Thus, we will get the following equation:
$C = x \times y$
$ \Rightarrow C = xy$ …………….(i)
Now, it is given that the fair is increased by 20%.
So the new fair = (original fair)+(20% of original fair)
New fare $ = x + \left( {20\% \,of\,x} \right)$
New fare $ = x + \dfrac{{20}}{{100}} \times x$
New fare $ = x + \dfrac{{20x}}{{100}}$
New fare $ = \dfrac{{120x}}{{100}}$
Also, it is given that, when the fair is increased, the number of passengers gets reduced by 10%. So the new number of passengers = (original number)-(10% of original number)
So the new number of passengers $ = y - \dfrac{{10}}{{100}} \times y$
So the new number of passengers $ = \dfrac{{90y}}{{100}}$
The total collection in this case is denoted by C’. Thus, we will get the following equation:
$C' = \left( {\dfrac{{120x}}{{100}}} \right)\left( {\dfrac{{90y}}{{100}}} \right)$
$C' = \dfrac{{108xy}}{{100}}$ …………….(ii)
Now ,we will divide the equation (i) by (ii). After doing this, we will get:
$\dfrac{C}{{C'}} = \dfrac{{xy}}{{\left( {\dfrac{{108xy}}{{100}}} \right)}}$
$\dfrac{C}{{C'}} = \dfrac{{xy \times 100}}{{108xy}}$
$ \Rightarrow \dfrac{C}{{C'}} = \dfrac{{100}}{{108}}$
$ \Rightarrow C' = \dfrac{{108C}}{{100}}$ ………….(iii)
Another information given in question is that, $new\,\,fair - old\,\,fair = \,Rs.\,150$.Thus, we get the following equation:
$C' - C = Rs.\,150$ …………….(iv)
Now, we can see that (iii) and (iv) are linear equations in two variables. So we will solve them by substitution method. So from (iii) and (iv), we have:
$ \Rightarrow \dfrac{{108C}}{{100}} - C = 150$
$ \Rightarrow \dfrac{{108C - 100C}}{{100}} = 150$
$ \Rightarrow \dfrac{{8C}}{{100}} = 150$
$ \Rightarrow C = \dfrac{{150 \times 100}}{8}$
$ \Rightarrow C = 1875$
Now, we will put the value of C in (iv). After doing this, we will get:
$ \Rightarrow C' - 1875 = 150$
$ \Rightarrow C' = 1875 + 150$
$ \Rightarrow C' = Rs.\,2025$
So the new daily collection = Rs.2025
Hence, the option (c) is correct.

Note: We can also solve the linear equations (iii) and (iv) by other methods such as elimination methods, graphical method, or cross multiplication method. The answer would still be the same.