Question

A taxi ride in a city costs one rupee for the first km and sixty paise for each additional km. The cost for each km is incurred at the beginning of the km so that the rider pays for the whole km. What is the average cost for $ 2\dfrac{3}{4} $ km?

Answer 1

Hint: Let y be the total fare and x be the total distance. Fare for the first km is Rs. 1 and after the 1st km, the fare for each subsequent km is Rs. 0.60 (60 paise). The total fare y for x km is $ Rs.1 + \left( {x - 1} \right)Rs.0.60 $ . And take the ceiling function for the no. of kilometres, because the rider pays for the whole km even though he did not travel.

Complete step-by-step answer:
We are given that a taxi ride costs Rs. 1 for the first km and sixty paise for each additional km.
We have to find the average cost for $ 2\dfrac{3}{4} $ km if the rider pays for the whole km even though not travelling the whole km.
Let the total fare be y for x km.
Fare for first km is Rs. 1 and fare for the remaining
$ \left( {x - 1} \right) $ km is $ \left( {x - 1} \right)Rs.0.60 $
Therefore, the total fare y is
$\Rightarrow y = Rs.1 + \left( {x - 1} \right)Rs.0.60 $
We are given that the distance is $ 2\dfrac{3}{4} $ km, which is also equal to
$ \dfrac{{\left( {4 \times 2} \right) + 3}}{4} = \dfrac{{11}}{4} $ km.
Therefore the ceiling function of
$ \dfrac{{11}}{4} $ is $ \dfrac{{11}}{4} = 2.75 = 3 $ this means the rider pays for 3 km even though he travels 2.75 km.
Therefore, x is 3km.
Y will be
$\Rightarrow y = Rs.1 + \left( {3 - 1} \right)Rs.0.60 = Rs.1 + \left( {2 \times Rs.0.60} \right) = Rs.1 + Rs.1.20 = Rs.2.20 $
The total fare for $ 2\dfrac{3}{4} $ km is Rs. 2.20.
The average cost for $ 2\dfrac{3}{4} $ km is $ \dfrac{{2.20}}{{\left( {\dfrac{{11}}{4}} \right)}} = \dfrac{{2.20 \times 4}}{{11}} = 0.2 \times 4 = Rs.0.8 $
So, the correct answer is “RS 0.8”.

Note: The fare for 1st km is given in rupees and the fare for additional km is given in paise. So we had to convert the both fares into a similar type of units (rupee or paise). If the number is a decimal number, then the ceiling function of that number gives the next integer like for 3.4 we get 4. Whereas in floor function we get the previous integer like for 3.4 the result is 3.