Case Study Questions 1
A part of monthly hostel charges in a college is fixed and the remaining depends on the number of days one has taken food in the mess. When a student Anu takes food for 25 days, she has to pay ₹ 4500 as hostel charges, whereas another student Bindu who takes food for 30 days, has to pay ₹ 5200 as hostel charges.
Considering the fixed charges per month by ₹ x and the cost of food per day by ₹ y, then answer the following 5 questions.
1. Represent algebraically the situation faced by both Anu and Bindu.
(a) x + 25y = 4500, x + 30y = 5200
(b) 25x + y = 4500, 30x + y = 5200
(c) x – 25y = 4500, x – 30y = 5200
(d) 25x – y = 4500, 30x – y = 5200
2. The system of linear equations, represented by above situations has.
(a) No solution.
(b) Unique solution.
(c) Infinitely many solutions.
(d) None of these.
3. The cost of food per day is:
(a) ₹ 120 (b) ₹ 130
(c) ₹ 140 (d) ₹ 1300
4. The fixed charges per month for the hostel is:
(a) ₹ 1500 (b) ₹ 1200
(c) ₹ 1000 (d) ₹ 1300
5. If Bindu takes food for 20 days, then what amount she has to pay?
(a) ₹ 4000 (b) ₹ 3500
(c) ₹ 3600 (d) ₹ 3800
Case Study Questions 2
From Bengaluru bus stand, if Riddhima buys 2 tickets to Malleswaram and 3 tickets to Yeswanthpur, then total cost is ₹ 46; but if she buys 3 tickets to Malleswaram and 5 tickets to Yeswanthpur, then total cost is ₹ 74.
Consider the fares from Bengaluru to Malleswaram and that to Yeswanthpur as ₹ x and ₹ y respectively and answer the following questions.
1. 1st situation can be represented algebraically as:
(a) 3x – 5y = 74 (b) 2x + 5y = 74 (d) 2x – 3y = 46 (d) 2x + 3y = 46
2. 2nd situation can be represented algebraically as:
(a) 5x + 3y = 74 (b) 5x – 3y = 74 (d) 3x + 5y = 74 (d) 3x – 5y = 74
3. Fare from Bengaluru to Malleswaram is:
(a) ₹ 6 (b) ₹ 8 (c) ₹ 10 (d) ₹ 2
4. Fare from Bengaluru to Yeswanthpur is:
(a) ₹ 10 (b) ₹ 12 (c) ₹ 14 (d) ₹ 16
5. The system of linear equations represented by both situations has:
(a) Infinitely many solutions. (b) No solution. (c) Unique solution. (d) None of these.
CASE STUDY Solutions
CASE STUDY 1 Solutions:
1. (a) x + 25y = 4500, x + 30y = 5200
Solution:
For student Anu:
Fixed charge + cost of food for 25 days = ₹ 4500
i.e., x + 25y = 4500
For student Bindu:
Fixed charges + cost of food for 30 days = ₹ 5200
i.e., x + 30y = 5200
2. (b) Unique solution.
Solution:
From above, we have a1= 1, b1 = 25
c1 = -4500 and a2 = 1, b2 = 30, c2 = -5200
Thus, system of linear equations has unique solution.
3. (c) ₹ 140
Solution:
We have, x + 25y = 4500
and x + 30y = 5200
Subtracting (i) from (ii), we get
5y = 700 ⇒ y = 140
∴ Cost of food per day is ₹ 140
4. (c) ₹ 1000
Solution:
We have, x + 25y = 4500
⇒ x = 4500 – 25 × 140
⇒ x = 4500 – 3500 = 1000
∴ Fixed charges per month for the hostel is ₹ 100
5. (d) ₹ 3800
Solution:
We have, x = 1000, y = 140 and Bindu takes food for 20 days.
∴ Amount that Bindu has to pay = ₹ (1000 + 20 × 140) = ₹ 3800
CASE STUDY 2 Solutions:
1. (d) 2x + 3y = 46
Solution:
1st situationcanberepresented algebraically as.
2x + 3y = 46
2. (c) 3x + 5y = 74
Solution:
2nd situation can be represented algebraically as:
3x + 5y = 74
3. (b) ₹ 8
Solution:
We have, 2x + 3y = 46………(i)
3x + 5y = 74………(ii)
Multiplying (i) by 5 and (ii) by 3 and then subtracting, we get
10x – 9x = 230 – 222 ⇒ x = 8
∴ Fare from Bengaluru to Malleswaram is ₹ 8.
4. (a) ₹ 10
Solution:
Putting the value of x in equation (i), we g
3y = 46 – 2 × 8 = 30 ⇒ y = 10
∴ Fare from Bengaluru to Yeswanthpur is ₹ 10
5. (c) Unique solution.
Solution:
We have, a1= 2, b1 = 3, c1 = -46 and
a2 = 3, b2 = 5, C2 = -74
Thus system of linear equations has unique solution.