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given

Day | MS | ES
1 | x | 0
2 | 2x/3 | x/3
3 | 4x/9 | 1/3 (2x/3) = 2x/9 + x/3 (old batch remains as per question) = 5x/9 (4x/9+5x/9 = X, satistifes)
4 | 2/3 (4x/9) = 8x/27 | 1/3 (4x/9) = 4x/27 + 5x/9 = 19x/27

above i first consider x as total number of workers, let us now consider x workers produce x units

so ES produces in 4 days = 0+x/3 + 5x/9 + 19x/27 = (9x + 15x + 19x)/27 = 43x/27

and Total units produced in 4 days is x+x+x+x = 4x

as number of workers or number of units should be an integer
x should be divisible by 27

trying x = 27 -> ES = 43, which is not there in the options skip
x = 54, ES = 86, looks promising, if x = 54, 4(54) = T =216 good

so are the answers
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Lets assume total workers = x and total units = U

Day 1: Morning (x), Evening (0)
Day 2: Morning (2x/3), Evening (x/3)
Day 3: Morning (4x/9), Evening (5x/9)
Day 4: Morning (8x/27), Evening (19x/27)

Total evening worker shifts = 0 + x/3 + 5x/9 + 19x/27 = 43x/27 = 43WU/27 (Total eve worker units)

Workers and units will have to be integers. Thus, we can assume that worker units is a multiple of 27, lets take WU as 27k.
Thus, eve shift units will now be 43k
Both shift units = 4WU = 4(27k) = 108k (which indicates that both shift units will have to be a multiple of 108)
Now we need to find k

From the given options we can see that 108 is there. In this case k = 1. However we don't have the option of 43 for eve.
We can also see 216 (108x2). Thus, k=2. We can also see that 86 is there (43x2). These both answers match consistently.

Eve units = 86
Total units = 216
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Answer: Both shifts = 216, Evening shift = 86

Let's set up a matrix, w = total number of workers
Morning shift. Evening shift.
D1 (w) 0
D2 2/3(w) 1/3(w)
D3 4/9(w) 1/3(2/3)(w) + 1/3(w) = (5/9)w
D4 8/27(w) 1/3(4/9)(w) + 5/9(w) = 18/27(w)

This means that w must be a multiple of 27.
Let's assume that there are 27 workers.
Counting just the evening workers/shifts, we have 0+9+15+19=43, which means the total number of evening shifts must be a multiple of 43.
The only answer choice that is a multiple of 43 is 86.
Knowing this, we now count the morning workers/shifts; we have 27+18+12+6=63, which means the total number of morning shifts must be a multiple of 63.
The only answer choice that is a multiple of 63 is 126.
Adding both together, we know that the total number of shifts worked = 126 + 86 = 216
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prethinking-worker shift as 0,1/3,5/9,19/27 over days 1-4

evening total-43Nu/27
Both total =4Nu
only 86 fits evening

for both;test the ratio:

86:108 is not equal to 43:108
86:216=43:108
evening =86, both=216
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Let total workers be W and work done by each be u units.

Total work done in both shifts in 4 days = 4 * w * u

Total work done in evening shifts = 0 + 1/3 (w) + 5w/9 + 19w / 27 = 43w/27

Ratio = 86 : 216
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Day1: M= x & E=0
Day2: M=(2/3)x & E=(1/3)x
Day3: M=(2/9)x & E=(5/9)x
Day4: M=(4/27)x & E=(19/27)x
Production(P) = x*(production of whole group in 1 shift)

Evening shift production= 0+(1/3)x+(5/9)x+(19/27)x = (43/27)x
Production of both shifts= 4*x

Test value from choices
If x=27, Both shifts= 4*27=108 & E=43 .....Not in options
If x=54, BS=4*54=216 & E=86.... Both are in choices

Evening shift= 86
Both shifts= 216
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When on day 2, 1/3 of day 1 workers did eve shift,
so for easy calculation, let the totalt workers be 27 (divisible 3)
So on D1:
evening= 0
morning= 27
day 2:
eve= 9
morning= 18
day 3:
eve= 15
mornig= 12
day 4:
eve= 19
morning= 8

total evening = 43
total workers= 108 (27x4)

ratio = 43:108
Just find option from the 2 coloumns in that ratio
that is 86:216
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A plant uses the same group of workers for a 4 day production run.

Each worker works one shift per day, either the morning shift or the evening shift, and each worker produces the same number of units in any shift worked.

On Day 1, all workers work the morning shift. Once a worker is assigned to the evening shift, that worker remains on the evening shift for the rest of the production run.

On Day 2, one third of the workers who worked the morning shift on Day 1 are assigned to the evening shift.

On Day 3, one third of the workers who worked the morning shift on Day 2 are assigned to the evening shift.

On Day 4, one third of the workers who worked the morning shift on Day 3 are assigned to the evening shift.

Let the number of workers on Day 1 (all morning shifts) be x and number of units produced per shift be y.

Day 1: Workers = morning shifts = x; Unit produced = xy; Units produced in evening shift = 0
Day 2: Workers = x; Morning shifts = 2x/3; Evening shifts = x/3; Units produced = xy; Units produced in evening shift = xy/3
Day 3: Workers = x; Morning shifts = (2/3)(2x/3) = 4x/9; Evening shifts = 5x/9; Units produced = xy; Units produced in evening shift = 5xy/9
Day 4: Workers = x; Morning shifts = (2/3)(4x/9) = 8x/27; Evening shifts = 19x/27; Units produced = xy; Units produced in evening shift = 19xy/27

Total number of units produced by the evening shift over the four days (E) = 0 + xy/3 + 5xy/9 + 19xy/27 = (9+15+19)xy/27 = 43xy/27
Total number of units produced by both shifts over the four days (B) = 4xy

E/B = (43xy/27)/(4xy) = 43/108 = 86/216

Evening shiftBoth shifts
86216
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A plant uses the same group of workers for a 4 day production run. Each worker works one shift per day, either the morning shift or the evening shift, and each worker produces the same number of units in any shift worked.

On Day 1, all workers work the morning shift. Once a worker is assigned to the evening shift, that worker remains on the evening shift for the rest of the production run. On Day 2, one third of the workers who worked the morning shift on Day 1 are assigned to the evening shift. On Day 3, one third of the workers who worked the morning shift on Day 2 are assigned to the evening shift. On Day 4, one third of the workers who worked the morning shift on Day 3 are assigned to the evening shift.


given that there are

two shifts in a day and ; group of workers are same for a 4 day production run...

1/3 workers shift each day to evening or morning shift
number of workers has to be x = 27*n

evening shift workers :

day 1 all workers are there = 27n ( morning shift) only ; so evening shift is 0
day 2 27n/3; 9n ; morning will be 18n
day 3 18n/3 6n + 9n ; 15n ; morning 12n
day 4 12n/3 + 15n ; 19n ; morning 8n

total of evening shift = 0 + 9n+15n+19n ; 43n

both shifts will be 27*4*n = 108n


upon checking value of n as integer 1,2 we get


Evening Shift the possible total number of units produced by the evening shift over the four days,
43n ; 43*2 ; 86

Both Shifts the possible total number of units produced by both shifts together over the four days,
108n ; 108 *2 ; 216

86; 216 are correct options
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Bunuel
A plant uses the same group of workers for a 4 day production run. Each worker works one shift per day, either the morning shift or the evening shift, and each worker produces the same number of units in any shift worked.

On Day 1, all workers work the morning shift. Once a worker is assigned to the evening shift, that worker remains on the evening shift for the rest of the production run. On Day 2, one third of the workers who worked the morning shift on Day 1 are assigned to the evening shift. On Day 3, one third of the workers who worked the morning shift on Day 2 are assigned to the evening shift. On Day 4, one third of the workers who worked the morning shift on Day 3 are assigned to the evening shift.

Select for Evening Shift the possible total number of units produced by the evening shift over the four days, and select for Both Shifts the possible total number of units produced by both shifts together over the four days, that would be jointly consistent with the given information. Make only two selections, one in each column.

Total Worker = 27x

DaysMorningEvening
D127x
D218x9x
D312x15x
D48x19x

Total Evening = 9x + 15x + 19x = 43x

So the option should be a multiple of 43. From the given options 86 is the multiple of 43.

43x = 86

x = 2

Total production across all the days = 27x * 4 = 27*2*4 = 27*8 = 216

Evening Shift = 86
Both Shifts = 216
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If total workers = x

Day 1- MS - x
Day 2-
MS- 2x/3
ES- x/3
Day 3- ES- x/3+2x/9 = 5x/9
MS- x-5x/9 = 4x/9

Day 4- ES- 5x/9 + 4x/27 = 19x/27
MS= 8x/27

If each worker produced a units, then total units by ES workers = a (x/3 + 5x/9 + 19x/27) = 43xa/27

Now total units of evening shift should be a multiple of 43, out of all the choices it could only be 86.

This also gives us xa = 54

Total units produced by all workers- After summing the morning + evening units, it gives 108xa/27 = 4xa

If xa = 54, total units = 216

Correct answers- 86 and 216 respectively
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I made the following table below to solve this question with supposing total worker working = 27x

Below shift timeDAY 1 DAY 2 DAY 3 DAY 4
Morning27x18x12x8x
Evening9x15x19x

As all worker had same working rate so we could assume it as 'r' units produced/worker

Total number of units in EVENING SHIFT = (9x+15x+19x)r = 43xr = 86
the number would be multiple of 43 and 86 is only there so we must multiply 43xr by 2 and if we multiply it by 2 then we must multiply other by 2 as well as it might be rate '2r' for all

Total number of units in all shifts = (27x+18x+12x+8x+43x)*(2r) = 108x*2r = 216r
Only 216 is the multiple of 216r hence it would be the answer

evening shift 86
both shifts 216
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Bunuel
A plant uses the same group of workers for a 4 day production run. Each worker works one shift per day, either the morning shift or the evening shift, and each worker produces the same number of units in any shift worked.

On Day 1, all workers work the morning shift. Once a worker is assigned to the evening shift, that worker remains on the evening shift for the rest of the production run. On Day 2, one third of the workers who worked the morning shift on Day 1 are assigned to the evening shift. On Day 3, one third of the workers who worked the morning shift on Day 2 are assigned to the evening shift. On Day 4, one third of the workers who worked the morning shift on Day 3 are assigned to the evening shift.

Select for Evening Shift the possible total number of units produced by the evening shift over the four days, and select for Both Shifts the possible total number of units produced by both shifts together over the four days, that would be jointly consistent with the given information. Make only two selections, one in each column.


Let the number of workers be N and each worker make r units per shift.

Evening workers each day:

Day 1 = 0
Day 2 = N/3
Day 3 = N/3 + 2N/9 = 5N/9
Day 4 = 5N/9 + 4N/27 = 19N/27

Total evening worker-shifts:

0 + N/3 + 5N/9 + 19N/27

= 9N/27 + 15N/27 + 19N/27

= 43N/27

So: Evening units = (43N/27) * r

Total units from both shifts: 4 * N * r

Therefore:

Evening : Both Shifts = (43N/27)r : 4Nr

= 43 : 108

Now check the choices.

86 : 216 = 43 : 108

Matches exactly.

So the selections are:

Evening Shift = 86

Both Shifts = 216.
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Day M E
D1 n 0
D2 2n/3 n/3
D3 4n/9 5n/9
D4 8n/27 19n/27
T = n*4(total worker = n each day) = 4n
T(E) = 43n/27
so, for evening, it should be a multiple of 43. There 86(43*2); n = 54
T = 4*54 = 216
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Let's assume number of workers per day is x
Day 1: Morning shift workers = x, Evening shift workers = 0
Day 2: Morning shift workers =2x/3, Evening shift workers = x/3
Day 3: Morning shift workers =4x/9, Evening shift workers = 5x/9
Day 4: Morning shift workers =8x/27, Evening shift workers = 19x/27

Total evening shift workers = 43x/27. Total workers both shift = 4x
For this to be an integer, minimum value of x=27. That would give us evening shift workers = 43. But it's not there in the options.
Let's try x=27*2=54, total evening shift workers will be = 86 and Total workers both shift = 216. That is our answer
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Take the number of workers as 270. Presuming each worker produces 1 unit per day
Total units will be 270 x 4 for the 4 days.

Day 1
Morning = 270

Day 2
Morning 180
Evening 90

Day 3
Morning 120
Evening 150

Day 4
Morning 80
Evening 190

Total number of evening units = 90 + 150 + 190 = 430
Hence total and evening will be in the ratio 270 *4 : 430
i.e 108:86

Answer choice 216 and 86 fits




Bunuel
A plant uses the same group of workers for a 4 day production run. Each worker works one shift per day, either the morning shift or the evening shift, and each worker produces the same number of units in any shift worked.

On Day 1, all workers work the morning shift. Once a worker is assigned to the evening shift, that worker remains on the evening shift for the rest of the production run. On Day 2, one third of the workers who worked the morning shift on Day 1 are assigned to the evening shift. On Day 3, one third of the workers who worked the morning shift on Day 2 are assigned to the evening shift. On Day 4, one third of the workers who worked the morning shift on Day 3 are assigned to the evening shift.

Select for Evening Shift the possible total number of units produced by the evening shift over the four days, and select for Both Shifts the possible total number of units produced by both shifts together over the four days, that would be jointly consistent with the given information. Make only two selections, one in each column.
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morning shift evening shift
Day 1 1 0
Day 2 2/3 1/3
day 3 1/3 2/3
day 4 0 1

morning = evening shift
1:1
one shift is half of the total and only 108 is half of 206 ANS (for first and second part respectively)
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