Answer:
Let the fixed charge for 3 days =x and additional charge =y
According to the question
⇒x+4y=27....eq1
⇒x+2y=21....eq2
Subtract eq1 and eq2
⇒(3x+4y=27)−(3x+2y=21)⇒2y=6⇒y=3
put y=3 in eq1
⇒x+4×3=27⇒x=15
Fixed charges = Rs. 15
The charge for each extra day = Rs. 3
Step-by-step explanation:
