
A workforce of 420 men with a contractor can finish a certain piece of work in 9 months. How many extra men must be employed to complete the job in 7 months?
Answer
153k+ views
Hint: We start solving the problem by assigning the variables for total work that need to be done and total men required to complete the job in 7 months. We then find the total work by multiplying 420 months with 9 months. We then equate this with the multiplication of 7 months with the total people required. Once we find the total no. of people required we subtract it with 420 to get the number of extra men that need to be employed.
Complete step-by-step answer:
According to the problem, we have given that a certain piece of work can be completed in 9 months by a workforce of 420 men. We need to find the number of extra men to be employed in order to complete the job in 7 months.
Let us assume the total work to be done as x and the total men required to complete the job in 7 months be y. So, 420 men takes 9 months to complete work x.
We get the total work done by all 420 men in 9 months.
So, we get $x=420\times 9$.
$\Rightarrow x=3780$ ---(1).
Now, the amount of work obtained in equation (1) has to be completed in 7 months by y men.
So, the total amount of work done by y men in 7 months is $7\times y$ which is equal to x.
So, we get $7y=x$.
From equation (1) we get,
$\Rightarrow 7y=3780$.
$\Rightarrow y=\dfrac{3780}{7}$.
$\Rightarrow y=540$.
We have found a total of 540 men required to complete the job in 7 months. But we already have 420 men working for us. In order to find the extra men employed, we subtract 420 from the 540 people.
The extra men employed is $\left( 540-420 \right)=120$.
We have found the extra men that need to be employed in order to complete the work in 7 months as 120.
∴ The extra men that need to be employed in order to complete the work in 7 months is 120 people.
Note: We can see that there is an inverse relation between the total number of men employed and time required to complete the work done. We can also find the work done by each man in 9 months by dividing x with the total man power. Similarly, we can expect problems to find the work done by each man in a day or in a month using the obtained values.
Complete step-by-step answer:
According to the problem, we have given that a certain piece of work can be completed in 9 months by a workforce of 420 men. We need to find the number of extra men to be employed in order to complete the job in 7 months.
Let us assume the total work to be done as x and the total men required to complete the job in 7 months be y. So, 420 men takes 9 months to complete work x.
We get the total work done by all 420 men in 9 months.
So, we get $x=420\times 9$.
$\Rightarrow x=3780$ ---(1).
Now, the amount of work obtained in equation (1) has to be completed in 7 months by y men.
So, the total amount of work done by y men in 7 months is $7\times y$ which is equal to x.
So, we get $7y=x$.
From equation (1) we get,
$\Rightarrow 7y=3780$.
$\Rightarrow y=\dfrac{3780}{7}$.
$\Rightarrow y=540$.
We have found a total of 540 men required to complete the job in 7 months. But we already have 420 men working for us. In order to find the extra men employed, we subtract 420 from the 540 people.
The extra men employed is $\left( 540-420 \right)=120$.
We have found the extra men that need to be employed in order to complete the work in 7 months as 120.
∴ The extra men that need to be employed in order to complete the work in 7 months is 120 people.
Note: We can see that there is an inverse relation between the total number of men employed and time required to complete the work done. We can also find the work done by each man in 9 months by dividing x with the total man power. Similarly, we can expect problems to find the work done by each man in a day or in a month using the obtained values.
Recently Updated Pages
JEE Main 2022 (June 29th Shift 2) Maths Question Paper with Answer Key

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 29th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 26th Shift 2) Chemistry Question Paper with Answer Key

JEE Main 2022 (June 26th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (June 29th Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Electrical Field of Charged Spherical Shell - JEE

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
