Ex 3.3, 3
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?
Let Fixed charge = Rs x
& Charge per km = Rs y
Given that Charge paid for 10 km is Rs 105
Fixed charge + 10 × (Charge per km) = Rs 105
x + 10y = 105
Also, Given that
Charge paid for 15 km is ₹ 155
Fixed charge + 15 × (Charge per km) = ₹ 155
x + 15y = 155
Our equations are
x + 10y = 105 …(1)
x + 15y = 155 …(2)
From (1)
x + 10y = 105
x = 105 – 10y
Putting value x in (2)
x + 15y = 155
(105 – 10y) + 15y = 155
105 – 10y + 15y = 155
–10y + 15y = 155 – 105
5y = 50
y = 50/5
y = 10
Putting y = 10 in (1)
x + 10y = 105
x + 10(10) = 105
x + 100 = 105
x = 105 – 100
x = 5
Hence, x = 5, y = 10
Hence,
Fixed charges = x = Rs 5
Change per km = y = Rs 10
We also need to find
How much does a person have to pay for travelling 25 km
Amount paid by person travelling 25 km
= Fixed charges + (Charges per km) × 25
= 5 + 10 × 25
= 5 + 250
= Rs 255

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.