Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method.
$A$ part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student $A$ takes food for $20$ days she has to pay ₹ $1000$ as hostel charges,whereas a student $B$,who takes food for $26$ days,pays ₹ $1180$ as hostel charges. Find the fixed charges and the cost of food per day.

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(N/A) Let $x$ be the fixed monthly hostel charge and $y$ be the cost of food per day.
According to the given information:
For student $A$: $x + 20y = 1000$ $...(1)$
For student $B$: $x + 26y = 1180$ $...(2)$
Subtracting equation $(1)$ from equation $(2)$,we get:
$(x + 26y) - (x + 20y) = 1180 - 1000$
$6y = 180$
$y = 30$
Substituting $y = 30$ in equation $(1)$:
$x + 20(30) = 1000$
$x + 600 = 1000$
$x = 400$
Thus,the fixed monthly charge is ₹ $400$ and the cost of food per day is ₹ $30$.

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