Question

4.
The taxi fare in a city is as follows: For the first kilometre, the fare is 8 and for the
subsequent distance it is 5 per km. Taking the distance covered as x km and total
fare as y, write a linear equation for this information, and draw its graph.​

Answer 1

Answer:

x 0 1 2

y 3 8 13

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)

Answer 2

:

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-- :

x 0  1 2

y 3 8  13

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)