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605-655 Level|   Math Related|         
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 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

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A plant must distribute 45 hours weekly among three production lines: A, B, and C.

  • Line A: Maximum 42 hours
  • Line B: Maximum 20 hours
  • Line C: Maximum 12 hours

Each line must receive more than half its maximum, all 45 hours must be used, and the number of hours allocated to each line must be an integer.

Select for Line B the hours that could have been allocated to Line B, and select for Line C the hours that could have been allocated to Line C that would be jointly consistent with the given information. Make only two selections, one in each column.
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 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.
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 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
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A plant must distribute 45 hours weekly among three production lines: A, B, and C.

  • Line A: Maximum 42 hours
  • Line B: Maximum 20 hours
  • Line C: Maximum 12 hours

Each line must receive more than half its maximum, all 45 hours must be used, and the number of hours allocated to each line must be an integer.

Select for Line B the hours that could have been allocated to Line B, and select for Line C the hours that could have been allocated to Line C that would be jointly consistent with the given information. Make only two selections, one in each column.
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 11:31
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We need to allocate 45 hours among the three production lines (A, B, and C) such that all conditions are satisfied
Constraints:

- Line A: Maximum 42 hours, Minimum 42/2 = 21 hours. Hence, Line A must get at least 22 hours
- Line B: Maximum 20 hours, Minimum 20/2 =10 hours. Hence, Line B must get at least 11 hours
- Line C: Maximum 12 hours, Minimum 12/2 =6 hours. Hence, Line C must get at least 7 hours


Let x,y and z be hours for Line A, B and C.

Therefore,\( x+y+z = 45 , x>=22 , y>=11, z >=7\) .......(1)

We test with the values,
\(x=22,y=11 and z=7\), we get 40 which is not enough.

Trying with \(x=22,y=16 and z=7,\) we get 45 which is a valid solution.

Therefore, \(y=16 \) and \( z=7\)
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 11:33
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Total 45 hours,
1.Each line must receive more than half its maximum
2.all 45 hours must be used,
3.and the number of hours allocated to each line must be an integer.

A=42---- half is 21
B=20---- half is 10
C=12---- half is 6

Lets use the options now... we cannot take 6 for C due to condition 1. Lets take next option C=7 and let A=22(more than half) this will give B=16, which is available.

For confirmation check with C=8, and let A=22(more than half), B=15, but 15 is not available in options. Other options will decrease B to less than 11- which breaks the constraint number 1.

Therefore answer is B=16 and C=7
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 11:53
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Total number of hours = 45
Line A: Maximum 42 hours, Minimum 22 hours
Line B: Maximum 20 hours, Minimum 11 hours
Line C: Maximum 12 hours, Minimum 7 hours

Considering the minimum number of hours for each,
22 + 11 + 7 = 40

Number of hours left = 5

Looking at the given options, 5 hours need to be added to B

Line B = 16 hours
Line C = 7 hours
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 12:05
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minimum for A =22
minimum for B=11
minimum for C=7

As per options, if B=16, & A=22..C=7

Answer 16 & 7
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 12:47
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Bunuel
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A plant must distribute 45 hours weekly among three production lines: A, B, and C.

  • Line A: Maximum 42 hours
  • Line B: Maximum 20 hours
  • Line C: Maximum 12 hours

Each line must receive more than half its maximum, all 45 hours must be used, and the number of hours allocated to each line must be an integer.

Select for Line B the hours that could have been allocated to Line B, and select for Line C the hours that could have been allocated to Line C that would be jointly consistent with the given information. Make only two selections, one in each column.
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Line A : Min = 21

Line B : Min = 10

Line C : Min = 6

Hence, from the give options line B can be 16 or 17

If Line B = 16, suppose we give Line A = 22 , Line C = 45 - 38 = 7

We can have this value.

Line B = 16
Line C = 7
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 12:58
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Line A: hours >=22
Line B: hours >=11
Line C: hours >=7

Total minimum=22+11+7=40

5 hours to be allocated among a,b and c.

Checking the values: b=16>=11 and c=7>=7

b=16 and c=7
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 13:48
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Hi everyone :)

This a logic question involving some constraints.

Constraints:
1. Each line must receive more than half its maximum.
2. All 45 hours must be used.
3. The number of hours allocated to each line must be an integer.

Line A=42 => 42/2 = 21 Thus 22 is the minimum.
Line B=20 => 20/2 = 10 Thus 11 is the minimum.
Line A=12 => 12/2 = 6 Thus 7 is the minimum.
22+11+7=40 => Thus 5 hours is the only possible addition to the lines.

We have 2 numbers above 11 that relate to line B:
16/17 - if we use 17, we add 6 to 11 and we said we have only 5. so 17 can be eliminated.
16 is our Line B

Line C: 22+16+C=45
C=45-38 = 7


B=16 C=7
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 15:02
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Line B = 16, Line C = 7

Line A > 21
Line B > 10
Line C > 6

We can eliminate 6 from B and C because under minimum.
We can eliminate 7 and 8 from B because under minimum.
We can eliminate 16 and 17 from C because exceeds maximum.
If C = 7, then A + B = 38. If A = 22, then B = 16. A cannot be 21, so B cannot be 17. C = 7 and B = 16.
If C = 8, then A + B = 37. If A = 22, then B = 15, which is not a choice.
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 16:03
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Line A: Minimum 22 hours.
Line B: Minimum 11 hours
Line C: Minimum 7 hours.

Total minimum hours = 22 + 11 + 7 = 40 hours.

This leaves a surplus of 5 hours to be distributed.
  • Line B can take a maximum of 16 hours (since B needs at least 11, and only 5 hours remain to distribute).
  • Line B = 16 hours, then Line C = 7 hours

Answer: Line B - 16 ; Line C - 7
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 19:20
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A plant must distribute 45 hours weekly among three production lines: A, B, and C.

Max Max/2 >Max/2
Line A: 42 hours 21 22
Line B: 20 hours 10 11
Line C: 12 hours 6 7
Sum: 40

Out of 45, at least 40 are needed to fulfill the minimum requirements, and we must use the remaining 5 hours.

So, the options 16 - 7 are perfectly aligned with the information given.
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 22:29
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1. Analyze given info:
A + B + C = 45
42>=A> 42/2 ----> 42>= A > 21 => A>=22
20>=B >20/2 ----> 20>=B > 10 => B>=11
12>=C> 12/2 ----> 12>= C > 6 => C>= 7

B+ C = 45- A = 45-22 = 23
=> only possible pairs that meets all requirements is B = 16 & C =7
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 26 Dec 2024, 23:59
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Given, Each line must receive more than half its maximum, so-

Line A: Maximum 42 hours/2= 21
Line B: Maximum 20 hours/2= 10
Line C: Maximum 12 hours/2= 6

a+b+c= 45

LineA: 22<= a<=42
LineB: 11<=b<=20
LineC: 7<=c<=12

from the given opt.-
Line B= 16/17
and Line C= 7/8

Doing hit and trial- The only opt. that satisfies the condition is-
B= 16, C= 7 and A= 22

Ans. B= 16, C= 7
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 27 Dec 2024, 00:41
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Line A: Needs at least 22 hours.

Line B: Needs at least 11 hours.

Line C: Needs at least 7 hours.

Total: 45 hours.

If Line B gets 16 hours:
Line A gets 22 hours, Line C gets 7 hours.
Total = 45 hours, which works.

If Line B gets 17 hours:
Line C would get 6 hours, violating the minimum.

Final Answer:
Line B: 16 hours
Line C: 7 hours
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 27 Dec 2024, 01:37
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Please Find the attached Solution
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 27 Dec 2024, 02:21
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Maximum hours for each line :
A = 42; B = 20 & C = 12
Each line must receive :
A = > 21; B = > 10 & C = >6 & A + B + C = 45

Using the constraints & the available options, we get :
B = 16 & C = 7
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A plant must distribute 45 hours weekly among three production lines: A, B, and C.

  • Line A: Maximum 42 hours
  • Line B: Maximum 20 hours
  • Line C: Maximum 12 hours

Each line must receive more than half its maximum, all 45 hours must be used, and the number of hours allocated to each line must be an integer.

Select for Line B the hours that could have been allocated to Line B, and select for Line C the hours that could have been allocated to Line C that would be jointly consistent with the given information. Make only two selections, one in each column.
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 27 Dec 2024, 02:28
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Bunuel
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A plant must distribute 45 hours weekly among three production lines: A, B, and C.

  • Line A: Maximum 42 hours
  • Line B: Maximum 20 hours
  • Line C: Maximum 12 hours

Each line must receive more than half its maximum, all 45 hours must be used, and the number of hours allocated to each line must be an integer.

Select for Line B the hours that could have been allocated to Line B, and select for Line C the hours that could have been allocated to Line C that would be jointly consistent with the given information. Make only two selections, one in each column.
Show more
Determine the minimum hours for each line:
  • Line A: Minimum hours = > 42 / 2 = >21 hours
  • Line B: Minimum hours = > 20 / 2 = >10 hours
  • Line C: Minimum hours = > 12 / 2 = >6 hours

Find possible combinations:
We need to find integer values for Line B and Line C that satisfy the following conditions:
  • Line B: Between 10 and 20 hours
  • Line C: Between 6 and 12 hours
  • Total hours: Line A + Line B + Line C = 45

Check possible combinations:
We know that Line B hours will be greater than 10, therefore, possible values for B and corresponding values of C
  • Line B = 16 hours, Line C = 7 hours:
    • Line A = 45 - 16 - 7 = 22 hours (within limits)
  • Line B = 17 hours, Line C = 6 hours:
    • Line A = 45 - 17 - 6 = 22 hours (within limits), however C>6, therefore not possible.

Therefore, one possible solution is:
  • Line B = 16 hours
  • Line C = 7 hours
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12 Days of Christmas GMAT Competition - Day 9: A plant must distribute 27 Dec 2024, 02:35
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Ans: B = 16 and C = 7
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A plant must distribute 45 hours weekly among three production lines: A, B, and C.

  • Line A: Maximum 42 hours
  • Line B: Maximum 20 hours
  • Line C: Maximum 12 hours

Each line must receive more than half its maximum, all 45 hours must be used, and the number of hours allocated to each line must be an integer.

Select for Line B the hours that could have been allocated to Line B, and select for Line C the hours that could have been allocated to Line C that would be jointly consistent with the given information. Make only two selections, one in each column.
Show more
Given = a+b+c =45
a > 21, b > 10 , and c > 6

this means from options b can only take 16 and 17 and c can take values from 7 and higher
start here b = 16 and c = 7 -> b+c = 23 so a = 22 (fits)
we know other trial will increase the value of either b or c or both and that will decrease the value of a from 22, we also know that a>21 so this is the only available solution.
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