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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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