Get live Maths 1-on-1 Classs - Class 6 to 12

Ex 3.5

Ex 3.5, 1 (i)
Deleted for CBSE Board 2023 Exams

Ex 3.5, 1 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 3.5, 1 (iii) Deleted for CBSE Board 2023 Exams

Ex 3.5, 1 (iv) Important Deleted for CBSE Board 2023 Exams

Ex 3.5, 2 (i) Important

Ex 3.5, 2 (ii)

Ex 3.5, 3

Ex 3.5, 4 (i) You are here

Ex 3.5, 4 (ii)

Ex 3.5, 4 (iii)

Ex 3.5, 4 (iv) Important

Ex 3.5, 4 (v) Important

Last updated at March 29, 2023 by Teachoo

Ex 3.5, 4 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. Let Fixed charge = Rs x & Charge per day taken food = Rs y per day Given that charge paid by student A for 20 days is Rs 1000 Fixed charge + 20 × (Charge per day) = 1000 x + 20y = 1000 Also, Charge paid by student B for 26 days is Rs 1180 Fixed charge + 26 × (Charge per day) = Rs 1180 x + 26y = 1180 Hence, our equations are x + 20y = 1000 …(1) x + 26y = 1180 …(2) From (1) x + 20y = 10000 x = 1000 – 20y Putting value of x in (2) x + 26y = 1180 (1000 – 20y) + 26y = 1180 – 20y + 26y = 1180 – 1000 6y = 180 y = 180/6 y = 30 Putting y = 30 in equation (1) x + 20y = 1000 x + 20(30) = 1000 x + 600 = 1000 x = 1000 – 600 x = 400 Hence, x = 400, y = 30 is the solution of the equations So, Fixed charges = x = Rs 400 Charge per km = y = Rs 30 per day