Question

The cost of 2 chairs and 3 tables is Rs.1300. The cost of 3 chairs and 2 tables is Rs.1200. The cost of each table is more than that of each chair by-?​

Answer 1

Given:-

The cost of 2 chairs and 3 tables = Rs.1300

The cost of 3 chairs and 2 tables = Rs.1200

To Find:-

By how much is the cost of table more than the cost of chair.

Solution:-

Let the cost of each chair be x

And cost of each table be y

Case 1:-

Cost of 2 chairs + 3 tables = 1300

$\sf \implies 2x + 3y = 1300.....(i)$

Case 2:-

Cost of 3 chairs + 2 tables = 1200

$\sf \implies 3x + 2y = 1200.....(ii)$

Multiply equation (i) with 3 and (ii) with 2

$\sf \implies 3(2x + 3y = 1300)$

$\sf \implies 6x + 9y = 3900......(iii)$

$\sf \implies 2(3x + 2y = 1200)$

$\sf \implies 6x + 4y = 2400......(iv)$

Equation (iii) - (iv)

$\sf \implies 6x + 9y - (6x + 4y)= 3900 - 2400$

$\sf \implies 6x + 9y - 6x - 4y= 1500$

$\sf \implies 5y = 1500$

$\sf \implies 5y = \dfrac{1500}{5} = 300$

Substitute y = 300 in equation (i)

$\sf \implies 2x + 3y = 1300$

$\sf \implies 2x + 3(300) = 1300$

$\sf \implies 2x = 1300 - 900$

$\sf \implies 2x = 400$

$\sf \implies x = 200$

Therefore:-

Cost of a chair = Rs.200

Cost of a table = Rs.300

Hence,

The cost of each table is more than that of each chair by Rs.100

Answer 2

Answer:-

✏ Let consider chair as x and table as y

According to the question

2x + 3y = ₹ 1300....eq 1

3x + 2y = ₹ 1200....eq 2

Mutiply eq 1 by 3 and eq 2 by 2

6x + 9y = ₹ 3900

6x + 4y = ₹ 2400

Now subtract eq 2 from eq 1, we get

5y = ₹ 1500

y = ₹ 300

Now put the value in eq 2

6x + 4 × 3000 = ₹ 2400

6x + 1200 = ₹ 2400

6x = ₹ (2400 - 1200)

6x = ₹ 1200

x = ₹ 200

Chair price x = ₹ 200

Table price y = ₹ 300

Cost of each table is more than that of chair = ₹ (300 - 200) = ₹ 100

Result:-

✒ ₹ 100 is your answer...✔