Q. Form the pair of linear equations for the following problem 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 Rs. 105 and for a journey of 15km, 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 25km?
and the charge per km be Rs. y
Sol.
Let the fixed charges be Rs. x
Fixed charges
Rs. 10
Charges per km
Rs. 5
Charges for 1 km =
10
+
5
Charges for 2 km =
10
+
5
Charges for 10 km =
10
+
5
Charges for 15 km =
10
+
5
2 ×
10 ×
15 ×
As per the 1st given condition,
x
+
10
y
= 105
As per the 2nd given condition,
x
+
15
= 155
y
........(ii)
.......(i)
Fixed charges + 2 (charge per km)
Fixed charges + 15 (charge per km)
In all these examples we can see that
The fixed charges and charge per km remains the same
But the no. of km changes
x + y
x + 2y
x + 10y
x + 15y
x + 10y
x + 15 y
Since we don’t know the fixed charges and charge per km
Charges paid for 1 km =
Charges paid for 2 km =
Charges paid for 15 km =
Charges paid for 10 km =
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
Fixed charges + 10 (charge per km)
Fixed charges + charges per km
What we need to find?
In this sum we have considered fixed charges to be ‘x’ and charge per km to ‘y’
Consider eq (i)
x
+
10
y
= 105
x
= 105
-
10y
........(iii)
Substituting (iii) in (ii)
105
-
10
y
+ 15
y
= 155
5
y
= 155
- 105
5
y
= 50
= 10
Substituting y = 10 in (iii)
x
= 105
-
10 (10)
= 105
-
100
= 5
= x
+
25y
= 5
+
25 (10)
= 5
+
250
= 255