Math, asked by llegshshll8, 5 months ago

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-?​

Answers

Answered by anilahirwar0002
1

Let consider chair as x and table as y

According to the question

2x + 3y = ₹ 1300…..eq 1

3x +2y = ₹ 1200……eq 2

Multiply 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 +4×300 =₹ 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 Ans…

Answered by MissZiddi
1

Answer:

Let consider chair as x and table as y

Let consider chair as x and table as yAccording to the question

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 1

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 2

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹3900

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 2

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 26x +4×300 =₹ 2400

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 26x +4×300 =₹ 24006x +1200 =₹ 2400

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 26x +4×300 =₹ 24006x +1200 =₹ 24006x = ₹ (2400–1200)

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 26x +4×300 =₹ 24006x +1200 =₹ 24006x = ₹ (2400–1200)6x = ₹1200

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 26x +4×300 =₹ 24006x +1200 =₹ 24006x = ₹ (2400–1200)6x = ₹1200X = ₹ 200

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 26x +4×300 =₹ 24006x +1200 =₹ 24006x = ₹ (2400–1200)6x = ₹1200X = ₹ 200Chair price = x = ₹200

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 26x +4×300 =₹ 24006x +1200 =₹ 24006x = ₹ (2400–1200)6x = ₹1200X = ₹ 200Chair price = x = ₹200Table price = y = ₹300

Let consider chair as x and table as yAccording to the question2x + 3y = ₹ 1300…..eq 13x +2y = ₹ 1200……eq 2Multiply eq 1 by 3 and eq 2 by 26x + 9y = ₹39006x + 4y = ₹2400Now subtract eq 2 from eq 1, we get5y = ₹1500Y = ₹300Now put the value in eq 26x +4×300 =₹ 24006x +1200 =₹ 24006x = ₹ (2400–1200)6x = ₹1200X = ₹ 200Chair price = x = ₹200Table price = y = ₹300Cost of each table is more than that of chair = ₹(300–200) = ₹100

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