Question

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

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 ₹ 105 and for a journey of 15 km, the charge paid is ₹ 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?

Answer 1

Let the fixed fare of taxi be x rupees.

And the fare for each additional kilometre is Rs.

Situation I

x + 10y = 105       ...(i)

Situation II

x + 15y = 155       …(ii)

From equation (i)

x + 10y = 105

⇒ x = 105 – 10y

Now on putting the value of x in equation (ii)

x + 15y = 155

⇒ (105 – 10y) + 15y = 155

⇒ 105 + 5y = 155

⇒ 5y = 155 - 105

⇒ 5y = 50

⇒ y = `50/5`

⇒ y = 10

Now putting y = 10 in equation (i)

⇒ x = 105 – 10y

⇒ x = 105 – 10(10)

⇒ x = 105 – 100

⇒ x = 5

Hence, the fixed fare is Rs 5 and the additional fare is Rs 10.

Fare for 25 km = x + 25y

= 5 + 25(10)

= 5 + 250

= Rs 255