Question

A job had to be completed in 12 days. After 10 days only half of the work was
completed. Adding another 16 men to the job allowed it to be completed on time. How
many men were employed at first?

Answer 1

help amigos !

Answer 2

M1* 10*2=(M1+16)*2*2

Solve for M1

Answer 3

4 is the answer?

Answer 4

Okay let me explain this in term of mandays. A "manday" (comes from 'man' and 'day' means work done by one man in one day. So, suppose there are 3 men working for a day, then in one day, 3 mandays worth of work is done.

Let the initial no. of men be x. So, total no. of mandays of work --> 10x
Later on the next two days, (16+x)*2 that is 32+2x mandays of work was used.

Now 32+2x mandays was half the work. Also was the work done in the first 10 days, that is 10x.

So, 10x = 32+2x

Answer 5

yeapp 4 is the answer !! but i couldnt get it at all ! how did u!!

Answer 6

@apoorvk i don't get it =[

Answer 7

Number of men and the work done are inversely proportional.

Answer 8

Okay let me make it simpler.
Let the working power of a man be 'x'. That is he can perform 'x' amount of work in a day.

Let there be 'n' men taken initially.

So n men do n*x work in one day.

In 10 days they did, 10*n*x work - which is half of the total work to be done

Over the next two days, n+16 men were used. They did 2*(n+16)*x amount of work in the those two days, which is the remaining half of the total work to be done.

Since halves are obviously equal,

Work done in the first 10 days = work done in the next 2 days
or, 10*n*x = 2*(n+16)*x
'x' gets cancelled.

Solve and get 'n'.


Understand now? :)

Answer 9

@apoorvk yeah thanks man. i love you.

Answer 10

thank you @FoolForMath too ~ love u too. ;D

Answer 11

You are welcome Sire ;D