Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method: (i) 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 Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day. (ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

i) Let the fixed monthly hostel charges be Rs. x and the cost of food per day taken from the mess be y. As per the given condition,
x + 20y = 1000.. (1)
x + 26y = 1180.. (2)
Subtracting eq(1) from eq (2), we get
⇒ 6y = 180
⇒ y = 30
Substitute y = 30 in eq (1), we get
⇒ x + 20(30) = 1000
⇒ x + 600 = 1000
⇒ x = 400
So, the fixed monthly hostel charge is Rs 400 and the cost of food per day is Rs 30.
ii) Let the numerator be x and the denominator be y, so the fraction will be x/y.
(x - 1)/y = 1/3 ⇒ 3x - y = 3 ⇒ y = 3x - 3.. (1)
x/(y + 8) = 1/4 ⇒ 4x = y + 8.. (2)
Substituting for y in (2) we get,
4x = 3x - 3 + 8
x = 5
Putting x = 5 in eq(1), we get
y = 12
∴ fraction = x/y = 5/12