Question

if 3 men or 6 boys can do a piece of work in 10 days, working 7 hours a day; how man
days will it take to compete a piece of work twice as large with 6 men and 2
working together for 8 hours a day ?​

Answer 1

Answer:

Step-by-step explanation:

Let man is doing a part of work in 1 hour and boy is doing b part of work in 1 hour

total work = 10(7a×3+7b×6)=70(3a+6b)

work to be done 2×70(3a+6b)

let double work completed in d days

so 2×70(3a+6b)=d(6a×8+2b×8)

140×3(a+2b)=d×16(3a+b)

d=420(a+2b)/16(3a+b)=105(a+2b)/4(3a+b)

note :- the result needs a relation between per day work of man and boy.

for an example if man performs double work than a boy in one work , then a=2b

d=105(2b+2b)/4(6b+b)=105×4b/4×7b= 15 days

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Answer 2

Answer:

15days

Step-by-step explanation:

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