The coach of a cricket team buys 7 bats and 6 balls for 13200. Later he buys 3 bats and 5 balls for 5900. find the cost of each ba

The coach of a cricket team buys 7 bats and 6 balls for 13200. Later he buys 3 bats and 5 balls for 5900. find the cost of each bat and each ball.​

2 thoughts on “The coach of a cricket team buys 7 bats and 6 balls for 13200. Later he buys 3 bats and 5 balls for 5900. find the cost of each ba”

  1. Bat = Rs 1800

    Ball = Rs 100

    Step-by-step explanation:

    Given:

    • Cost of 7 bats and 6 balls is Rs 13200.
    • Cost of 3 bats and 5 balls is Rs 5900.

    To Find:

    • What is the cost of each bat and ball ?

    Solution: Let the cost of each bat and ball be Rs x and y respectively.

    ➙ Cost of 7 bats = Rs 7x

    ➙ Cost of 6 balls = Rs 6y

    [tex]\implies{\rm }[/tex] 7x + 6y = 13200

    • Equation 1

    Now , In second case

    ➙ Cost of 3 bats = Rs 3x

    ➙ Cost of 5 balls = Rs 5y

    [tex]\implies{\rm }[/tex] 3x + 5y = 5900

    • Equation 2

    [ Multiplying 5 in equation 1 and 6 in equation 2 ]

    ➯ (7x + 6y) 5 = 13200 × 5

    • 35x + 30y = 66000ㅤㅤㅤeqⁿ i

    ➯ (3x + 5y) 6 = 5900 × 6

    • 18x + 30y = 35400ㅤㅤeqⁿ ii

    ㅤㅤㅤㅤㅤ35x + 30y = 66000

    ㅤㅤㅤㅤㅤ18x + 30y = 35400

    ㅤㅤㅤㅤㅤ–ㅤㅤ–ㅤㅤ–

    ㅤㅤㅤㅤ━━━━━━━━━━━━━━━

    ㅤㅤㅤㅤㅤ17x = 30600

    ∴ x = 30600/17 = 1800

    Now putting the value of x in equation 1.

    [tex]\implies{\rm }[/tex] 7 × 1800 + 6y = 13200

    [tex]\implies{\rm }[/tex] 6y = 13200 12600

    [tex]\implies{\rm }[/tex] y = 600/6

    [tex]\implies{\rm }[/tex] y = 100

    Hence, the cost of each bat and ball is Rs 1800 & Rs 100 respectively.

  2. Given :-

    The  coach of a cricket team buys 7 bats and 6 balls for 13200. Later he buys 3 bats and 5 balls for 5900

    To Find :-

    Cost of one bat and ball

    Solution :-

    Let

    Price of bat = x

    Price of ball = y

    Now

    7 × x + 6 × y = 13200

    7x + 6y = 13200

    Multiply by 5

    5(7x + 6y) = 5(13200)

    35x + 30y = 66000(1)

    Now

    3 × x + 5 × y = 5900

    3x + 5y = 5900

    Multiply by 6

    6(3x + 5y) = 6(5900)

    18x + 30y = 35400

    Subtracting them

    (35x + 30y) – (18x + 30y) = 66000 – 35400

    35x + 30y – 18x – 30y = 30600

    35x – 18x = 30600

    17x = 30600

    x = 30600/17

    x = 1800

    By putting in 1

    35(1800) + 30y = 66000

    63000 + 30y = 66000

    30y = 66000 – 63000

    30y = 3000

    y = 3000/30

    y = 100

    [tex]\\[/tex]

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