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.
Answers
||✪✪ 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)
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|| ✰✰ 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 ⟧
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❁ ✫ 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..
ANSWER:
Given:
a [ first term]= 150
The series according to the question will be 150,146,142 ...
d [common difference]=t(n)- t(n-1)
=t(2) - t(1)
= 146-150
= -4
Total number of workers can be written as,
Total number of days without dropping of workers= (n - 8)
Total number of workers = 150(n - 8)
According to both the situations,
152n-2n² = 150(n - 8)
152n-2n² = 150n - 1200
2n²+150n - 152n - 1200= 0
2n²- 2n- 1200 = 0
2(n²- n -600)=0
n² - n - 600=0
n² - 25n + 24 n -600=0
n (n - 25) + 24 (n - 25)=0
(n - 25) (n + 24)= 0
n - 25=0
n=25
n + 24 =0
n = -24
As the number of workers cannot be negative
The answer is 25.