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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?

Answers (1)

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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.

Posted by

Pankaj Sanodiya

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