Question

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 Rs.y, write a linear equation for this information and draw its graph.


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Answer 1

Question :-

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 Rs.y, write a linear equation for this information and draw its graph.

Given :-

• ↬ fare for 1st kilometre = rs.8
• ↬ fare for the rest of the distance = rs.5

To find :-

• ↬ total distance covered

Solution :-

fare for the rest of the distance = 5 × remaining distance

= rs 5 (X - 1)

therefore the total fare = fare for the 1st km + fare for rest (X - 1)km

now according to question :-

$\rm :\longmapsto\: y = 8 + (x - 1)5 \\ \\\rm :\longmapsto\: y = 8 + 5x - 5 \\ \\ \rm :\longmapsto\: y = 5x + 3$

now to draw the graph we need at least two solution of the equation :-

❍ putting X = 0 we get,

$\rm :\longmapsto\: y = 5(0) + 3 \\ \\ \rm :\longmapsto\: y = 0 + 3 \\ \\ \rm :\longmapsto\: y = 3 \\ \\ \longrightarrow \sf\bold{\red{0 \: and \: 3}}$

so (0,3) is a solution of the equation 1st

❍ putting X = 1 we get,

$\rm :\longmapsto\: y = 5(1) + 3 \\ \\ \rm :\longmapsto\: y = 5 + 3 \\ \\ \rm :\longmapsto\: y = 8 \\ \\ \longrightarrow \sf\bold{\red{1 \: and \: 8}}$

so (1,8) is a solution of the equation 2nd

now let's point this 1st and 2nd equation on graph.

Note :-

• graph is in the attachment.

Answer 2

fare for 1st kilometre = rs.8

↬ fare for the rest of the distance = rs.5

↬ total distance covered=?

Solution :-

fare for the rest of the distance = 5 × remaining distance

= rs 5 (X - 1)

total fare = fare for the 1st km + fare for rest (X - 1)km

now according to question :-

\begin{gathered}\rm :\longmapsto\: y = 8 + (x - 1)5 \\ \\\rm :\longmapsto\: y = 8 + 5x - 5 \\ \\ \rm :\longmapsto\: y = 5x + 3 \\ \\ \end{gathered}

y=8+(x−1)5

y=8+5x−5

y=5x+3

❍ putting X = 0 we get,

\begin{gathered}\rm :\longmapsto\: y = 5(0) + 3 \\ \\ \rm :\longmapsto\: y = 0 + 3 \\ \\ \rm :\longmapsto\: y = 3 \\ \\ \longrightarrow \sf\bold{\red{0 \: and \: 3}}\end{gathered}

y=5(0)+3

y=0+3

y=3

0and3

so (0,3) is a solution of the equation 1st

❍ putting X = 1 we get,

\begin{gathered}\rm :\longmapsto\: y = 5(1) + 3 \\ \\ \rm :\longmapsto\: y = 5 + 3 \\ \\ \rm :\longmapsto\: y = 8 \\ \\ \longrightarrow \sf\bold{\red{1 \: and \: 8}}\end{gathered}

y=5(1)+3

y=5+3

y=8

1and8

so (1,8) is a solution of the equation(2)

Now draw a graph and get the lines..