Math, asked by shrutigupta1909, 10 months ago

A person C can complete 21% of work in 10 days while working with 233 1/3% of his efficiency. B is 11 1/9% more efficient than C. A while working with his half efficiency can complete the work in half time as compared to the taken by B. Find the time taken by A & B together to complete the 50% of whole work ?

Answers

Answered by amitnrw
5

Given :  person C can complete 21% of work in 10 days while working with 233 1/3% of his efficiency.

To find :  time taken by A & B together to complete the 50% of whole work

Solution:

person C can complete 21% of work in 10 days   while working with 700/3 % efficiency

=> person C can complete 100% of work with 7/3 efficiency in 10 * 100/ 21 days  = 1000/21 days

person C can complete 100% of work with  normal Efficiency  = (1000/21) * 7/3

= 1000/9 days

=> C 's 1 day work = 9/1000

B is 11 1/9% more efficient than C =>   100/9 % efficient = 1/9 more efficient

B's 1 Days work  = 9/1000( 1 + 1/9) = (9/1000)(10/9)  = 1/100

=> B can complete work in 100 Days

A with half efficiency can complete  in 50 Days

A with full efficiency will take  25 Days to complete

time taken by A & B together to complete the 50% of whole work  in d days

d(1/100 + 1/25 )  = 1/2

=> d (5/100) = 1/2

=> d = 10

10 Days  time taken by A & B together to complete the 50% of whole work

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Answered by wajahatkincsem
1

The person B can do the whole job is 16 days.

Step-by-step explanation:

A can do the job in 24 days.

The job A can do in a day =1/24

B is 50% more efficient than A.

The job B can do in a day = (1 / 24) x 150 / 100

The job B can do in a day = 150 / 2400 = 1 / 16

So, B can do the whole job in 1 / (1 / 16) = 16 days

Thus B can do the whole job is 16 days.

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