Question

A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?

2
3
5
11

Answer 1

Answer:

♠ The correct option is B.

Step-by-step explanation:

Let there were n workers on a job.

Since job was finished when last worker was withdrawn, time taken = n days

♦ Let 1 unit of work is done by one worker in one day.  

Work done on 1st day = n*1 = n

Work done on 2nd day = (n-1)*1 = n-1

n + (n-1) + ... + 2 + 1 = n*2n/3 = 2n2/3

n(n+1)/2 = 2n2/3

3n + 3 = 4n

♦ n = 3

Answer 2

Question :--- we have to find how many workers were there in the group ?

Given :-- on each day one work withdraw , and if no worker withdrawn they did the job in 2/3Rd of the time ....

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:Answer}}}}}}}}}}$

Let initially their were n number of workers in the group ।

..

and let each worker do 1 unit of work daily ..

Than total work = n * 1 = n done in first day ,

similary , on second day = (n-1)*1 unit of work ,

A/q,

n+(n-1)+(n-2)+ _________ +2+1 = n(2n/3) = 2n²/3

we know that now, sum of n terms of a series is =

$\frac{n(n + 1)}{2}$

so,

$\frac{n(n + 1)}{2} = \frac{2 {n}^{2} }{3} \\ \\ \\ 3n + 3 = 4n \\ \\ n = 3$

so, Total workers in the group was = $\large\red{\boxed{\sf </strong><strong>3</strong><strong>\</strong><strong>:</strong><strong>option</strong><strong>(</strong><strong>B</strong><strong>)</strong><strong>}}$