Math, asked by prajjvalh76521, 1 year ago

A and b together can complete a work in 14 2/5 days while b and c together can complete the same work in 10 2/7 days a alone starts work and afyer 8 days b replaced him b did the work for next 12 days and the remaining work is completed by c in next 5 days then find time taken by a b c together to complete that worl if c work with 50% of his usual efficiency

Answers

Answered by amitnrw
3

Given :  A and b together can complete a work in 14 2/5 days while b and c together can complete the same work in 10 2/7 day

To find : time taken by a b c together to complete that work if c work with 50% of his usual efficiency

Solution:

A Alone complete work in A days

B alone complete work in B days

C alone complete work in C days

1/a  + 1/b  = 5/72

1/b + 1/c  = 7/72

8/a  + 12/b  + 5/c  =  1

8 ( 5/72 - 1/b)  + 12/b  +  5(7/72 - 1/b) = 1

=> 40/72 - 8/b + 12/b + 35/72 - 5/b = 1

=> -1/b  = 1 - 75/72

=> -1/b = -3/72

=> b = 24

1/a  + 1/b  = 5/72

=> 1/a + 3/72 = 5/72

=> 1/a = 2/72

=> a = 36

1/b + 1/c  = 7/72

=> 3/72 + 1/c = 7/72

=> 1/c  = 4/72

=> c = 18

1/a  = 1/36

1/b = 1/24

1/c = 1/18

c's efficiency half

=> 1/c = 1/36

1/36 + 1/24 + 1/36

= (2 + 3 + 2)/72

= 72/7

= 10  2/7  Days

time taken by a b c together to complete that work 10  2/7  Days  if  c work with 50% of his usual efficiency

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