The taxi fare in a city is as follows: For the first kilometre, the fare is Rs. $8$ and the subsequent distance it is Rs. $5$ per km. Taking the distance covered as x km and fare as Rs y, write a linear equation.

y = 4 + 6 x

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y = 3 + 5 x

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y = 4 + 5 x

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y = 3 + 6 x

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Solution

The correct option is D $(y = 3 + 5 x)$
Since, for the first $1, k m$ fare $= 8$
and further subsequent distance it is Rs. $5$ per km.
To find fare $(y)$ for $x k m$ : first $1$ km, we have to subtract from $x$ ,
So $(x - 1) k m$ , it cost Rs. $5$ per each km.
So, Fare $= (x - 1) 5$
Now we can add $1$ km fare separately as follow:
Thus, total fare, $y = 5 (x - 1) + 8 = 5 x - 5 + 8 = 3 + 5 x$

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