The taxi fare in a city is as follows. For the first km the fare is $R s .10$ and subsequent distance is $R s .6 / k m .$ Taking the distance covered as $x k m$ and fare as $R s y$ ,write a linear equation.

y = 4 + 6 x

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y = 4 + 5 x

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y = 3 + 6 x

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y = 3 + 5 x

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Solution

The correct option is A $y = 4 + 6 x$
First $k m$ fare = $R s .10$
Subsequent distance fare= $R s 6 / k m$

Then fare $x k m$ of distance
$y = (x - 1) \times 6 + 10$
$y = 6 x - 6 + 10$

$y = 6 x + 4$

$A$ is correct

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Similar questions

Q. The taxi fare in a city is as follows: For the first kilometre, the fare is Rs.

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per km. Taking the distance covered as x km and fare as Rs y, write a linear equation.

Q. The taxi fare in a city is as follows. For the first km the fare is

R s 10

and subsequent distance is

R s 6 / k m

. Taking the distance covered as

x k m

and fare as

R s y

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The taxi fare in a city is as follows: For the first kilometre, the fares is Rs 8 and for the subsequent distance it is Rs 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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The taxi fare in a city is as follows. For the first km the fare is Rs.10 and subsequent distance is Rs.6/km. Taking the distance covered as x km and fare as Rs y ,write a linear equation.

The correct option is A y=4+6xFirst km fare =Rs.10Subsequent distance fare=Rs 6/kmThen fare x km of distance y=(x−1)×6+10y=6x−6+10y=6x+4A is correct

The taxi fare in a city is as follows. For the first km the fare is $R s .10$ and subsequent distance is $R s .6 / k m .$ Taking the distance covered as $x k m$ and fare as $R s y$ ,write a linear equation.

The correct option is A $y = 4 + 6 x$
First $k m$ fare = $R s .10$
Subsequent distance fare= $R s 6 / k m$

Then fare $x k m$ of distance
$y = (x - 1) \times 6 + 10$
$y = 6 x - 6 + 10$

$y = 6 x + 4$

$A$ is correct