Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Last updated at July 27, 2020 by Teachoo

Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Transcript

Ex 3.3 ,3 Form the pair of linear equations for the following problems and find their solution by substitution method. (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball. Let cost of one bat = Rs. x & Cost of one ball = Rs. y Given that Coach buys 7 bats and 6 balls for Rs. 3800 So, 7 × (Cost of one bat ) + 6 × (Cost of one ball) = 3800 7x + 6y = 3800 Also, Coach buys 3 bats and 5 balls for Rs 1750 So, 3 × (Cost of one bat ) + 5 × (Cost of one ball) = 1750 3x + 5y = 1750 Thus, our equations are 7x + 6y = 3800 …(1) 3x + 5y = 1750 …(2) From (1) 7x + 6y = 3800 7x = 3800 – 6y x = ((3800−6𝑦)/7) Putting x in (2) 3x + 5y = 1750 3((3800−6𝑦)/7)+5𝑦=1750 Multiplying by 7 both sides 7×(3(3800−6𝑦)/7)+5𝑦×7=1750×7 3( 3800 – 6y ) + 35y = 1750×7 3×3800−6𝑦×3+35𝑦=1750×7 11400 – 18y + 35y = 12250 –18y + 35y = 12250 – 11400 17y = 850 y = 850/17 y = 50 Putting y = 50 in (1) 7x + 6y = 3800 7x + 6(50) = 3800 7x + 300 = 3800 7x = 3800 – 300 7x = 3500 x = 3500/7 x = 500 Thus, x = 500, y = 50 Hence , Cost of one bat = x = Rs. 500 Cost of one ball = y = Rs. 50

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.