Two friends, Sahil and Meena, travelled by taxi. Let the fixed charge for one taxi trip be ₹x and the additional charge per kilometre be ₹y. Frame a pair of linear equations in x and y to represent the following situation:
- (i) Sahil travelled 6 km and paid a total fare of ₹122.
- (ii) Meena travelled 8 km and paid a total fare of ₹160.
Answer: (i) x + 6y = 122 (ii) x + 8y = 160.
Step-by-step solution
Idea: Fare = fixed charge (paid once per trip) + charge per km × number of km.
(i) Sahil travelled 6 km and paid a total fare of ₹122.
- Fare = x + 6 × y, so x + 6y = 122.1 mark
x + 6y = 122
(ii) Meena travelled 8 km and paid a total fare of ₹160.
- Fare = x + 8 × y, so x + 8y = 160.1 mark
x + 8y = 160
x + 6y = 122 and x + 8y = 160.
Check: Subtracting gives 2y = 38, y = 19 and x = 8: 8 + 114 = 122 ✓, 8 + 152 = 160 ✓.
Answer to write in the exam
(i)
Fare = fixed charge + 6 × charge per km
∴ x + 6y = 122
(ii)
Fare = fixed charge + 8 × charge per km
∴ x + 8y = 160
Common mistakes that cost marks
- Multiplying the fixed charge by the distance (6x + 6y). The fixed charge is paid once per trip.
- Swapping the roles: 6x + y = 122.
How this can come in the exam
Short answer (2 marks)
An auto charges a fixed ₹a plus ₹b per km. A 4 km ride costs ₹70 and a 9 km ride costs ₹120. Form the equations and find a and b.
Show answer
a + 4b = 70, a + 9b = 120 (1 mark). Subtract: 5b = 50 ⇒ b = 10, a = 30 (1 mark).Try one yourself
A gym charges a joining fee ₹x plus ₹y per month. 3 months cost ₹2300 and 5 months cost ₹3500. Form the equations.
Show answer
x + 3y = 2300 and x + 5y = 3500 (giving y = 600, x = 500).
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