Answer:
(D) 15 cable A, 25 cable B, 15 cable C
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