The amount of work in a leather factory is increased by 50%. By what percentage is it necessary to increase the number of workers to complete the new amount of work in previously planned time, if the productivity of the new labour is 25% more?
A. 60%
B. $66\dfrac{2}{3}\% $
C. 40%
D. $33\dfrac{1}{3}\% $
Answer
Verified
116.4k+ views
Hint: Let us consider that initially there were 100 men in the factory and the planned time is 1 unit. Find the amount of work by the formula ${\text{men}} \times {\text{time = work}}$. Find the required number of men for the new work according to the given relation.
Complete step by step answer:
Let us consider that initially there were 100 men in the factory and the planned time is 1 unit.
We know that the ${\text{men}} \times {\text{time = work}}$
Substitute the value 100 for men and 1 for time , we get work as
$100 \times 1 = 100$
These parameters represent the initial value of the factory.
According to the question, the work in the factory has increased by 50%.
Thus the new work will be initial work + 50% of initial work
The value for initial work was calculated as 100.
Therefore the new work is $100 + 50 = 150$.
We have to complete the work in the previous planned time, that is 1 unit. Therefore the number of men required can be calculated from the formula ${\text{men}} \times {\text{time = work}}$.
${\text{men}} \times {\text{1 = 150}}$
Total men required = 150
However 100 men are already present in the factory. Thus the number of new men required is $150 - 100 = 50$ .
We are given that the efficiency of new men is 25% more than the already present men. Thus, the new men should be hired 25% less.
That is \[50 - 0.25 \times 50 = 40\].
Thus the actual requirement for new men is 40.
Percentage requirement of new men is \[\dfrac{{40}}{{100}} \times 100 = 40\% \], as the initial number of labour was 100.
Thus option C is the correct answer.
Note: The number of times required and the number of men is directly proportional to the amount of work done, that is, ${\text{men}} \times {\text{time = work}}$. Percentage requirement is calculated as 40% in this question, which means the factory requires 40% more men of the total men in the factory.
Complete step by step answer:
Let us consider that initially there were 100 men in the factory and the planned time is 1 unit.
We know that the ${\text{men}} \times {\text{time = work}}$
Substitute the value 100 for men and 1 for time , we get work as
$100 \times 1 = 100$
These parameters represent the initial value of the factory.
According to the question, the work in the factory has increased by 50%.
Thus the new work will be initial work + 50% of initial work
The value for initial work was calculated as 100.
Therefore the new work is $100 + 50 = 150$.
We have to complete the work in the previous planned time, that is 1 unit. Therefore the number of men required can be calculated from the formula ${\text{men}} \times {\text{time = work}}$.
${\text{men}} \times {\text{1 = 150}}$
Total men required = 150
However 100 men are already present in the factory. Thus the number of new men required is $150 - 100 = 50$ .
We are given that the efficiency of new men is 25% more than the already present men. Thus, the new men should be hired 25% less.
That is \[50 - 0.25 \times 50 = 40\].
Thus the actual requirement for new men is 40.
Percentage requirement of new men is \[\dfrac{{40}}{{100}} \times 100 = 40\% \], as the initial number of labour was 100.
Thus option C is the correct answer.
Note: The number of times required and the number of men is directly proportional to the amount of work done, that is, ${\text{men}} \times {\text{time = work}}$. Percentage requirement is calculated as 40% in this question, which means the factory requires 40% more men of the total men in the factory.
Recently Updated Pages
JEE Main Cut Off 2025: Category-wise Details and Admission Criteria
JEE Main 2025 Question Paper Pattern (Revised) and Marking Scheme
JEE Main 2025 Exam Date: Check Important Dates and Schedule
JEE Main Chemistry Question Answers- Exam Preparation Tips
JEE Main Question Answers 2023- Exam Preparation Tips
JEE Main Mock Test For 2025 - FREE Test Series
Trending doubts
JEE Main 2025 Maths Online - FREE Mock Test Series
JEE Main 2024 Physics Question Paper with Solutions 27 January Shift 1
JEE Main Results 2025: Updates, Toppers, Scorecard, and Cut-Offs
JEE Main Maths Class 12 Mock Test
Purely Resistive Inductive and Capacitive Circuits for JEE Main 2025 Physics
Charging and Discharging of Capacitor