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# Form the pair of linear equations for the following problems and find their solution by substitution method.  (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge

Q3.    Form the pair of linear equations for the following problems and find their solution by substitution method.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

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Let the fixed charge is x and the per km charge is y.

Now According to the question

$x+10y=105.......(1)$

And

$x+15y=155.......(2)$

Now, From (1) we have,

$x=105-10y........(3)$

Substituting this value of x in (2), we have

$105-10y+15y=155$

$\Rightarrow 5y=155-105$

$\Rightarrow 5y=50$

$\Rightarrow y=10$

Now, Substituting this value in (3)

$x=105-10y=105-10(10)=105-100=5$

Hence, the fixed charge is 5 Rs and the per km charge is 10 Rs.

Now, Fair For 25 km :

$\Rightarrow x+25y=5+25(10)=5+250=255$

Hence fair for 25km is 255 Rs.

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