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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 $300$.
If the school rents 'x' chairs and 'y' tables, answer the following questions :
(i) Write down the pair of linear equations representing the given information.
(ii) (a) Find the number of chairs and number of tables rented by the school.
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.
(iii) What is maximum number of tables that can be rented in ₹30,000 if no chairs are rented?
If the school rents 'x' chairs and 'y' tables, answer the following questions :
(i) Write down the pair of linear equations representing the given information.
(ii) (a) Find the number of chairs and number of tables rented by the school.
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.
(iii) What is maximum number of tables that can be rented in ₹30,000 if no chairs are rented?
Show SolutionHide Solution↓
(i) $x + y = 300$
and $50 x + 200 y = 30000$ or $x + 4y = 600$
(ii) (a) $x + y = 300$ and $x + 4y = 600$
Solving the equations, we get
$x = 200$ and $y = 100$
$\therefore$ Number of chairs and tables rented by the school are $200$ and $100$ respectively.
OR
(b) $x + y = 300$ and $50x + 200y = 27000$ or $x + 4y = 540$
Solving the equations, we get
$x = 220$ and $y = 80$
$\therefore$ Number of chairs and tables rented by the school are $220$ and $80$ respectively.
(iii) Number of tables $= \frac{30000}{200} = 150$
$\therefore$ Maximum number of tables that can be rented is $150$ if no chairs are rented.
and $50 x + 200 y = 30000$ or $x + 4y = 600$
(ii) (a) $x + y = 300$ and $x + 4y = 600$
Solving the equations, we get
$x = 200$ and $y = 100$
$\therefore$ Number of chairs and tables rented by the school are $200$ and $100$ respectively.
OR
(b) $x + y = 300$ and $50x + 200y = 27000$ or $x + 4y = 540$
Solving the equations, we get
$x = 220$ and $y = 80$
$\therefore$ Number of chairs and tables rented by the school are $220$ and $80$ respectively.
(iii) Number of tables $= \frac{30000}{200} = 150$
$\therefore$ Maximum number of tables that can be rented is $150$ if no chairs are rented.