A group of workers was put on a job. From second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?

Answer:- 1

Explanation:-

Solution:

Let initially X number of workers be there.
Now, Using work equivalence method,
X + (X - 1) + (X - 2) + . . . . . + 1 = X × 55% of X

\frac{X \times (X + 1)}{2} = \frac{5 X}{100}

$\frac{{{\text{X}} \times \left( {{\text{X}} + 1} \right)}}{2} = \frac{{5{\text{X}}}}{{100}}$
[series is in AP. Sum of AP = {No. of terms (first term + last term)/2}]
X = 10 workers.

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[series is in AP. Sum of AP = {No. of terms (first term + last term)/2}]
X = 10 workers. ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

Let initially X number of workers be there.
Now, Using work equivalence method,
X + (X - 1) + (X - 2) + . . . . . + 1 = X × 55% of X

\frac{X \times (X + 1)}{2} = \frac{5 X}{100}

$\frac{{{\text{X}} \times \left( {{\text{X}} + 1} \right)}}{2} = \frac{{5{\text{X}}}}{{100}}$
[series is in AP. Sum of AP = {No. of terms (first term + last term)/2}]
X = 10 workers. ", "dateCreated": "7/24/2019 10:09:12 AM" } }