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16 January, 11:27

A company makes 3 types of cable. Cable A requires 3 black, 3 white, and 2 red wires. B requires 1 black, 2 white, and 1 red. C requires 2 black, 1 white, and 2 red. They used 100 black, 110 white and 85 red wires. How many of each cable were made? A) 15 cable A, 98 cable B, 15 cable CB) 15 cable A, 25 cable B, 88 cable CC) 15 cable A, 15 cable B, 25 cable CD) 15 cable A, 25 cable B, 15 cable C

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  1. 16 January, 15:06
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    (D) 15 cable A, 25 cable B, 15 cable C

    Step-by-step explanation

    Total number of black wires = 100

    Total number of white wires = 110

    Total number red wires = 85

    For black wire

    3A + 1B + 2C = 100 ... (1)

    For white wire

    3A + 2B + 1C = 110 ... (2)

    For red wire

    2A + 1B + 2C = 85 ... (3)

    Using elimination method

    Considering equation 1 and 2, multiply (1) by 2 and (2) by 1. We have

    6A + 2B + 4C = 200 ... (4)

    3A + 2B + 1C = 110 ... (5)

    Subtract (5) from (4). So we have

    3A + 3C = 90 ... (6)

    Considering equation 2band 3, multiply (2) by 1 and (3) by 2. We have

    3A + 2B + 1C = 110 ... (7)

    4A + 2B + 4C = 170 ... (8)

    Subtract (8) from (7). So we have

    -A - 3C = - 60

    = A + 3C = 60 ... (9)

    From equation (9)

    A = 60 - 3C

    Put A = 60 - 3C in equation (6)

    3 (60 - 3C) + 3C = 90

    180 - 9C + 3C = 90

    180 - 6C = 90

    -6C = 90 - 180

    -6C = - 90

    C = - 90/-6

    C = 15

    Put C = 15 in equation (9)

    A + 3 (15) = 60

    A + 45 = 60

    A = 60 - 45

    A = 15

    Put A = 15 and C = 15 into equation (1)

    3 (15) + 1B + 2 (15) = 100

    45 + B + 30 = 100

    B + 75 = 100

    B = 100 - 75

    B = 25

    A = 15 cables

    B = 25 cables

    C = 15 cables
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