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
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
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 = 12250
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 12 years. He provides courses for Maths and Science at Teachoo.