Question

4 chairs and 3 tables cost is 2100 and 5 chairs and 2 tables cost is 1750 and find cost of one chair and one table in substitution method answer please


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Answer 1

ANSWER:-

Given:

4 chairs & 3 tables cost is Rs.2100 & 5 chairs & 2 tables cost is Rs.1750.

To find:

The cost of 1 chair & 1 table.

Solution:

Let the cost of 1 table be Rs. R &

Let the cost of 1 chair be Rs. M

According to the question:

3R + 4M= 2100.............(1)

2R + 5M = 1750............(2)

From equation (1), we get;

=) 3R= 2100 - 4M

=) R= 2100 - 4M/3...........(3)

Putting the value of R in equation (2), we get;

$= > 2( \frac{2100 - 4M}{3} ) + 5M = 1750 \\ \\ = > \frac{4200 - 8M}{3} + 5M = 1750 \\ \\ = > 4200 - 8M + 15M = 5250 \\ \\ = > 4200 + 7M = 5250 \\ \\ = > 7M = 5250 - 4200 \\ \\ = > 7M = 1050 \\ \\ = > M = \frac{1050}{7} \\ \\ = > M = 150$

So,

Putting the value of M in equation (3), we get;

$r = \frac{2100 - 4M}{3} \\ \\ = > R= \frac{2100 - 4(150)}{3} \\ \\ = > R = \frac{2100 - 600}{3} \\ \\ = > R= \frac{1500}{3} \\ \\ = > R= 500$

Thus,

The cost of 1 chair & 1 table is Rs.150 &

Rs.500 respectively.

Hope it helps ☺️

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]