Question

36. The taxi fare in a city is as follows: For the first kilometre the fare is Rs 8 and for the
subsequent distance it is Rs 5 per. Kilometre taking the distance covered as x km and tota
fare is Rs y.write a linear question for this information and draw its graph.
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Answer 1

Step-by-step explanation:

Taxi fare for first kilometer = Rs. 8

Taxi fare for subsequent distance = Rs. 5

Total distance covered =x

Total fare =y

Since the fare for first kilometer = Rs.8

According to problem,

Fare for (x–1) kilometer = 5(x−1)

So, the total fare y=5(x−1)+8

⇒y=5(x−1)+8

⇒y=5x–5+8

⇒y=5x+3

Hence, y=5x+3 is the required linear equation.

Now the equation is

y=5x+3 ...(1)

Now, putting the value x=0 in (1)

y=5×0+3

y=0+3=3 So the solution is (0,3)

Putting the value x=1 in (1)

y=5×1+3

y=5+3=8. So the solution is (1,8)

Putting the value x=2 in (1)

y=5×2+3

y=10+3=13. So the solution is (2,13)

solution

Answer 2

Given :

• A taxi fare in a city charges Rs. 8 for 1st kilometre and for subsequent km it charges Rs. 5 per kilometre .

To Find :

• Linear equation and draw a graph .

Solution :

$\longmapsto\tt{Let\:the\:distance\:covered\:be=x\:km}$

$\longmapsto\tt{Fare\:for\:travelling=Rs.y}$

$\longmapsto\tt{Fare\:for\:1st\:km=Rs.8}$

$\longmapsto\tt{For\:Subsequent\:km=Rs.5}$

A.T.Q :

$\longmapsto\tt{y=8\times{1}+(x-1)\times{(-5)}}$

$\longmapsto\tt{y=8+5x-5}$

$\longmapsto\tt{y=8-5+5x}$

$\longmapsto\tt\bf{y=5x+3}$

So , The Linear Equation is y = 5x+3 ..

Now ,

$\begin{tabular}{|c|c|c|c|c|}\cline{1-5}\sf x & \sf 0 & \sf 1 & \sf 2 & \sf 3\\\cline{1-5}\sf y & \sf 3 &\sf 8&\sf 13&\sf 18\\\cline{1-5}\end{tabular}$

Refer to the Attachment !