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26 Dec 2024, 11:16
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Based on the constraints(Each line must receive more than half its maximum, and the number of hours allocated to each line must be an integer).
A>= 22
B>=11
C>=6
A+B+C = 45
B+C = 45-22 = 23 (assuming A = 22)
Checking the options, B = 16, C = 7, fits the bill
A>= 22
B>=11
C>=6
A+B+C = 45
B+C = 45-22 = 23 (assuming A = 22)
Checking the options, B = 16, C = 7, fits the bill
Bunuel
26 Dec 2024, 11:28
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The answer is B=16 and C=7.
we have A+B+C = 45
Each to have more than its half of maximum,
A = 42/2 = 21 ; B = 20/2 = 10 ; C = 12/2 = 6
So, A>21, B > 10 C> 6
Substitute B = 16, and C = 7 with A = 22
22+16+7 = 45. so we have an answer.
Other values cannot be used.
we have A+B+C = 45
Each to have more than its half of maximum,
A = 42/2 = 21 ; B = 20/2 = 10 ; C = 12/2 = 6
So, A>21, B > 10 C> 6
Substitute B = 16, and C = 7 with A = 22
22+16+7 = 45. so we have an answer.
Other values cannot be used.
26 Dec 2024, 11:29
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To select for B and C
Lets settle the time allotted for A. it will have to be more than the half of the maximum
then the least we can allot for A is (42/2) + 1 = 22
Rest 23 must be allotted for B and C
minimum time to be allotted for B is (20/2) + 1 = 11
minimum time to be allotted for C is (12/2) + 1 = 7
from the options for B the possible time we can allot is 16
from the options for C the possible time we can allot is 7
other options wont satisfy the minimum requirement or the remaining sum of the hours
Hence A = 22, B = 16, C = 7
Lets settle the time allotted for A. it will have to be more than the half of the maximum
then the least we can allot for A is (42/2) + 1 = 22
Rest 23 must be allotted for B and C
minimum time to be allotted for B is (20/2) + 1 = 11
minimum time to be allotted for C is (12/2) + 1 = 7
from the options for B the possible time we can allot is 16
from the options for C the possible time we can allot is 7
other options wont satisfy the minimum requirement or the remaining sum of the hours
Hence A = 22, B = 16, C = 7
Bunuel
