Question

4 CHAIRS & 3 TABLES COST RS.2100. 5 CHAIRS AND 2 TABLES COST RS. 1750. FIND THE COST OF CHAIR AND TABLE SEPERATLY
using substitution method

Answer 1

Answer:

Let the cost of one table be x Rs. and the cost of one chair be y Rs.

Then, ATQ,

3x + 4y = 2100                    (1)

and, 2x + 5y = 1750            (2)

Now, from (1),

x = (2100 - 4y)÷3                 (3)

Using (3) in (2),

We get:

y = 150                               (4)

Using y = 150 in (1), we get,

x = 500

Hence, the cost of a table is 500 Rs. and the cost of a chair is 150 Rs.

Answer 2

Given data : 4 chairs and 3 tables cost Rs 2100 and 5 chairs and 2 tables cost Rs 1750.

Solution : Let, the cost of one chair be x and the cost of one table be y.

Now, according to the given data;

• 4x + 3y = 2100 ----{1}

• 5x + 2y = 1750 ----{2}

Now, multiply eq. {1} by 2

• 8x + 6y = 4200 ----{3} and similarly,

Multiply eq. {2} by 3

• 15x + 6y = 5250 ----{4}

Now, subtract eq. {3} from eq. {4}

15x + 6y = 5250

- (8 + 6y) = 4200

______________

7x = 1050

x = 1050/7

x = 150

Now, put value of x in eq. {1}

➜ 4x + 3y = 2100

➜ 4*(150) + 3y = 2100

➜ 600 + 3y = 2100

➜ 3y = 2100 - 600

➜ 3y = 1500

➜ y = 1500/3

➜ y = 500

Answer : Hence, the cost of one chair is Rs 150 and the cost of one table is Rs 500.

[Verification : Put vale of x and y in eq. {1}

➜ 4x + 3y = 2100

➜ 4*(150) + 3*(500) = 2100

➜ 600 + 1500 = 2100

➜ 2100 = 2100

Hence, it's verified]