[CBSE 2025] A school is organizing a grand cultural event to show . . .

Case Study Based Question on Chapter - 3 
(Pair of Linear Equations in Two Variables)

A school is organizing a grand cultural event to show the talent of its students. To accommodate the guests, the school plans to rent chairs and tables from a local supplier. It finds that rent for each chair is ₹ 50 and for each table is ₹ 200. The school spends ₹ 30,000 for renting the chairs and tables. Also, the total number of items (chairs and tables) rented are 300.

(i) Write down the pair of linear equations representing the given information. [1 Marks]

(ii) (a) Find the number of chairs and number of tables rented by the school. [2 Marks]
OR

 (b) If the school wants to spend a maximum of ₹ 27,000 on 300 items (tables and chairs), then find the number of chairs and tables it can rent. [2 Marks]

(iii) What is maximum number of tables that can be rented in ₹ 30,000 if no chairs are rented ? [1 Marks]

CBSE 2025 

Detailed Solution : 

(i) Write down the pair of linear equations representing the given information. [1 Marks]

Ans : 
Let number of chairs =  x

and number of tables = y 

ATQ

50x+200y=30000

50[x+4y]=30000

x+4y=30000/50

x+4y=600 — (1) 

Also 

x+y=300 — (2) 

(ii) (a) Find the number of chairs and number of tables rented by the school. [2 Marks]

Ans : 
Solving equation (1) and (2) 

x + 4y=600

x +   y=300
-  -        -
----------------
0+3y=300
----------------

3y=300

y=300/3

y=100

putting in equation (2) 

x+y=300
x+100=300

x=300-100

x=200 

Number of chairs = x = 200

Number of tables = y = 100

OR

(b) If the school wants to spend a maximum of ₹ 27,000 on 300 items (tables and chairs), then find the number of chairs and tables it can rent. [2 Marks]

Ans : 

50x+200y=27000

50[x+4y]=27000

x+4y=540 

x+y=30 

Solving the above equations 

x+4y=540

x+  y=300
- -      -
---------------
0+3y=240
---------------

y=240/3

y=80

putting in equation (2) 

x+80=300

x=300-80

x=220

Number of chairs = 220

Number of tables = 80

(iii) What is maximum number of tables that can be rented in ₹ 30,000 if no chairs are rented ? [1 Marks]

Ans : 

Using Equation (1) 

x + 4y=600

x=0
y=?

0+4y=600

4y=600

y=600/4
y=150

Maximum number of tables which can be rented in 30,000 is 150, if no chairs are rented.