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

(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