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A group of workers with same efficiency can finish a job in 24 hours working together from start to end. Instead, they start after equal intervals one by one and continue working till the end. Wages being proportional to the time for which the work is done. The first worker (who starts the work) gets 11 times as much wages as the last worker. In how many hours can the work be finished now?
- a)44
- b)64
- c)33
- d)78
Correct answer is option 'A'. Can you explain this answer?
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A group of workers with same efficiency can finish a job in 24 hours w...
►Let there be x workers and efficiency of each of them be 1 unit/hour.
►The work is worth 24x man-hours. Next, if the last person is working for y hours, the 1st person, who starts the work, works for 11y hours. No. of workers is still x.
►Since they have to do the same work 24x = 11y + ...... + 2y + y (x terms).
►24x = x/2 [11y + y] i.e. y = 4 i.e. 11y = 44.
► ∴ The work is finished in 44 hours.
Most Upvoted Answer
A group of workers with same efficiency can finish a job in 24 hours w...
Given:
A group of workers can finish a job in 24 hours working together.
The first worker gets 11 times as much wages as the last worker.
To find:
The time required to finish the job when the workers start after equal intervals.
Solution:
Let the number of workers be n.
Let the total wages paid be W.
From the given information, we can write two equations:
Equation 1: W = (t1 + t2 + … + tn)*k
where ti is the time taken by the ith worker to finish the job and k is the proportionality constant for wages.
Equation 2: t1 = 11tn
where t1 is the time taken by the first worker and tn is the time taken by the last worker.
When the workers work together, the work done by each worker is equal. So, we can write:
Equation 3: (1/t1 + 1/t2 + … + 1/tn) = 1/24
This is because the total work done by all the workers in 24 hours is equal to the work done by one worker in t1 + t2 + … + tn hours.
Using Equation 2, we can write:
Equation 4: (1/t1 + 1/(t1/11) + … + 1/tn) = 1/24
Simplifying Equation 4, we get:
(11/12)*(1/t1 + 1/t2 + … + 1/tn) = 1/24
Using Equation 3, we can substitute 1/24 as (1/t1 + 1/t2 + … + 1/tn).
So, (11/12)*(1/t1 + 1/t2 + … + 1/tn) = (1/24)
Simplifying further, we get:
(1/t1 + 1/t2 + … + 1/tn) = (12/11)*(1/24)
(1/t1 + 1/t2 + … + 1/tn) = (1/22)
Using Equation 1, we can write:
W = (t1 + t2 + … + tn)*k
Substituting t1 = 11tn, we get:
W = (11tn + t2 + … + tn)*k
W = ((11+n-1)*tn/2)*k
W = (10.5n*tn)*k
W = 10.5ntn*k
Using Equation 2, we can write:
tn = t1/11
Substituting this in the above equation, we get:
W = (10.5n*t1^2*k)/121
We know that the total work done is the same in both cases (when working together and when working one by one). So, we can write:
Equation 5: W = (t1 + t2 + … + tn)*k
where t1 + t2 + … + tn is the time taken to finish the job when the workers start after equal intervals.
Substituting the value of W from Equation 4 in Equation 5, we get:
(10.5n*t1^2*k)/121 = (t1 + t2 + … + tn)*k
Simplifying, we get:
(10.5
A group of workers can finish a job in 24 hours working together.
The first worker gets 11 times as much wages as the last worker.
To find:
The time required to finish the job when the workers start after equal intervals.
Solution:
Let the number of workers be n.
Let the total wages paid be W.
From the given information, we can write two equations:
Equation 1: W = (t1 + t2 + … + tn)*k
where ti is the time taken by the ith worker to finish the job and k is the proportionality constant for wages.
Equation 2: t1 = 11tn
where t1 is the time taken by the first worker and tn is the time taken by the last worker.
When the workers work together, the work done by each worker is equal. So, we can write:
Equation 3: (1/t1 + 1/t2 + … + 1/tn) = 1/24
This is because the total work done by all the workers in 24 hours is equal to the work done by one worker in t1 + t2 + … + tn hours.
Using Equation 2, we can write:
Equation 4: (1/t1 + 1/(t1/11) + … + 1/tn) = 1/24
Simplifying Equation 4, we get:
(11/12)*(1/t1 + 1/t2 + … + 1/tn) = 1/24
Using Equation 3, we can substitute 1/24 as (1/t1 + 1/t2 + … + 1/tn).
So, (11/12)*(1/t1 + 1/t2 + … + 1/tn) = (1/24)
Simplifying further, we get:
(1/t1 + 1/t2 + … + 1/tn) = (12/11)*(1/24)
(1/t1 + 1/t2 + … + 1/tn) = (1/22)
Using Equation 1, we can write:
W = (t1 + t2 + … + tn)*k
Substituting t1 = 11tn, we get:
W = (11tn + t2 + … + tn)*k
W = ((11+n-1)*tn/2)*k
W = (10.5n*tn)*k
W = 10.5ntn*k
Using Equation 2, we can write:
tn = t1/11
Substituting this in the above equation, we get:
W = (10.5n*t1^2*k)/121
We know that the total work done is the same in both cases (when working together and when working one by one). So, we can write:
Equation 5: W = (t1 + t2 + … + tn)*k
where t1 + t2 + … + tn is the time taken to finish the job when the workers start after equal intervals.
Substituting the value of W from Equation 4 in Equation 5, we get:
(10.5n*t1^2*k)/121 = (t1 + t2 + … + tn)*k
Simplifying, we get:
(10.5
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A group of workers with same efficiency can finish a job in 24 hours working together from start to end. Instead, they start after equal intervals one by one and continue working till the end. Wages being proportional to the time for which the work is done. The first worker (who starts the work) gets 11 times as much wages as the last worker. In how many hours can the work be finished now?a)44b)64c)33d)78Correct answer is option 'A'. Can you explain this answer?
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