Question

The taxi fare in a city is as follows: For the first kilometre, the fare is Rs.30 and for the subsequent distance it is Rs.25 per kilometre. Taking the distance covered as x kilometre and total fare as Rs. Y , write a linear equation for this information and draw its graph​

Answer 1

Answer:

Given,Total distance covered= x km=1+(x-1)km

Fare For first kilometre= ₹8

Fare for subsequent distance= ₹5 per km

 

Fare for next(x-1)km= 5(x-1)

 

A.T.Q

 

Total fare= y

8+5(x-1)=y

8+5x-5=y

5x-y+3=0

Which is the required linear equation.

 

It can also be written as y= 5x+3

When X = 0 ,then Y = 3,When x=1, then y= 5+3=8

When x= 2, then y= 10+3=13

 

[Table & graph are on the attachment]

 

Now plot the points A(0,3), B(1,8), C( 2,13) on the graph paper and join them to form a line BC, which represents the required graph of linear equation.

Step-by-step explanation:

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Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr}}}}}$

Let the distance be x

Fare for 1st km=₹ 30

Fare for subsequent km=₹ 25

ATQ

Distance=x

Taxi fare=₹y

30+25(x-1)=y

=> 30+25x+(-25)=y

=> -25=30+(-25)=y

=>-25=5=y

y+25=5

Therefore the equation is y+25=5