Question

150 workers were engaged to finish a piece of work in a certain no. Of days.
Four workers dropped the second day, four more dropped the third day and so on. It takes 8 more days to finish the work now. Find the no. Of days in which the work was completed

Answer fast.​

Answer 1

Answer:

1500 days

Explanation:

150 x 8 = 1500 so there is your answer

Answer 2

||✪✪ QUESTION ✪✪||

150 workers were engaged to finish a piece of work in a certain no. Of days. Four workers dropped the second day, four more dropped the third day and so on. It takes 8 more days to finish the work now. Find the no. Of days in which the work was completed. ?

|| ★★ FORMULA USED ★★ ||

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

_____________________ ||✰✰ ANSWER ✰✰ ||

Let us assume that, The Number of days Required To complete the whole work is n.

Also, Let us assume that, Each worker does 1 unit of work / day.

________

Since , it has been said that, it takes 8 more days to finish the work because of Dropping of 4 workers After Every First day..

with This we can say that, if no worker left, 150 Workers had completed the work in = (n - 8) days.

So,

➳ 1 worker complete work per day = 1 unit.

➳ 150 workers complete work in (n-8) days = 150*(n-8) unit of work = Total work.

----------⟦ Equation ❶ ⟧

__________________________

Now, Lets see first part :-

➪ in 1st day , 150 workers complete = 150*1 = 150 unit work

➪ in 2nd day , 146 workers complete (since 4 dropped the second day) = 146 * 1 = 146 unit of work.

➪ in 3rd day , 142 workers complete (since 4 dropped the Third day) = 142 * 1 = 142 unit of work.

--------------------- So, on ------------------------

So,

☞ Total work = 150 + 146 + 142 _________ upto n terms.

____________

As we can see That, This series is in AP. where,

➻ First term = a = 150

➻ Common Difference = a2 - a1 = 146 - 150 = (-4)

➻ Total terms = n .

Putting all values in above told AP formula we get,

⟼ Total work = (n/2) [ 2*150 + (n -1)(-4) ]

⟼ Total work = (n/2) [ 300 - 4n + 4 ]

⟼ Total work = (n/2) [ 304 - 4n ]

⟼ Total work = (n/2) * 2 [ 152 - 2n ]

⟼ Total work = n * [ 152 - 2n ]

⟼ Total work = (152n - 2n²) ----------

-⟦ Equation ❷ ⟧

___________________________

❁ ✫ Comparing Equation ⓵ and Equation ❷ Now, we get :-

➳ 150(n - 8) = (152n - 2n²)

➳ 150(n - 8) = 2(76n - n²)

Dividing both sides by 2,

➳ 75(n - 8) = (76n - n²)

➳ 75n - 600 = 76n - n²

➳ n² + 75n - 76n - 600 = 0

➳n² - n - 600 = 0

Splitting The Middle - Term Now,

➳ n² - 25n + 24n - 600 = 0

➳ n(n - 25) + 24(n - 25) = 0

➳ (n - 25)(n + 24) = 0

Putting both Equal to Zero now, we get,

➺ n - 25 = 0

➺ n = 25 days.

or,

➺ n + 24 = 0

➺ n = (-24) days.

Since, Number of days are not Possible as Negative value.

∴ The Total no. Of days in which the work was completed is 25 days..