Doubt by Gurpreet
Solution :
Let the fixed charge be ₹ x
and the charge per km be ₹ y
ATQ
x+10y=105 — (1)
x+15y=155 — (2)
Using equation (1)
x+10y=105
x=105-10y — (3)
Putting in equation (2)
[105-10y]+15y=155
105-10y+15y=155
-10y+15y=155-105
5y=50
y=50/5
y=10
Putting in equation (3)
x=105-10(10)
x=105-100
x=5
Fixed charge = ₹ 5
Charge Per Km = ₹10
For a Distance of 25 km
Total Charges
= x+25y
= 5+25(10)
= 5+250
= 255
So, for a distance of 25 km, the person has to pay ₹ 255.
Let the fixed charge be ₹ x
and the charge per km be ₹ y
ATQ
x+10y=105 — (1)
x+15y=155 — (2)
Using equation (1)
x+10y=105
x=105-10y — (3)
Putting in equation (2)
[105-10y]+15y=155
105-10y+15y=155
-10y+15y=155-105
5y=50
y=50/5
y=10
Putting in equation (3)
x=105-10(10)
x=105-100
x=5
Fixed charge = ₹ 5
Charge Per Km = ₹10
For a Distance of 25 km
Total Charges
= x+25y
= 5+25(10)
= 5+250
= 255
So, for a distance of 25 km, the person has to pay ₹ 255.