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New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

11 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

11 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

11 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

11 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

11 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

11 months ago

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V
Vishal Baghel

Contributor-Level 10

 

Kindly go through the solution

New answer posted

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A
alok kumar singh

Contributor-Level 10

Please consider the following

 

 

New answer posted

11 months ago

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P
Payal Gupta

Contributor-Level 10

106. Let 'x' be the no of days in which 150 workers took to finish the job.

If 150 workers worked for x days then number of workers for x days =150 x.

But given that number of works dropped 4 on 2nd day, then 4 on 3rd day and so on taking 8 more

days to finish the work. i.e., x + 8 days we can express as.

150 x = 150 + (150  4) + (150  4  4)+……+ (x + 8) days.

150 x = 150 + 146 + 142 +……… (x+8) days which

R.H.S. from as A.P. of

a = 150

d = -4 and n = x +8

So, Sn = 150 x

n2 [2*150+ (n1) (4)]=150x

n [ 150 + (n - 1) (-2)] = 150 (n - 8) [ ?  n = x +8 x  8  x]

150n 2n (n - 1) 150n 1200

2n2 + 2n 1200 =0

n2 - n - 6

...more

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