Two men and 7 children complete a certain piece of work in 4 days while 4 men and 4 children

1 day work of 1men and 7children=1/4 ----1
1 day work of 4men and 4children=1/3 ----2
solving 1 and 2 we get 1men work =15 days
10 years agoHelpfull: Yes(9) No(3)
(2m+7c)4=(4m+4c)3
on solving, 1m=4c which means M:C::4:1
substitute in Lhs or Rhs u will get total work done as 60.
60/4=15
10 years agoHelpfull: Yes(6) No(0)
2x+7y=1/4
4x+4y=1/3

8x+28y=1...........(1)
28x=28y=7/3.........(2)

by solving 1 & 2
x=1/15

so 15days
9 years agoHelpfull: Yes(5) No(0)
given,
2m+7c=1/4 ; 4m+4c=1/3
by solving above two eqns , we get m=1/15 and c=1/60
ans: 1 man can complete same work in 15 days.
10 years agoHelpfull: Yes(3) No(0)
(2/m)+(7/c)=1/4 ie. 8a+28b=1
and (4/m)+(4/c)=1/3 ie. 12a+12b=1
96a-336a=-16
240a=16
a=15=1/m
no of days 1 men can complete = 15
10 years agoHelpfull: Yes(1) No(0)
15 days
2m +7c=1/4
4m+4c=1/3
after solving
m=15
8 years agoHelpfull: Yes(0) No(0)

Two men and 7 children complete a certain piece of work in 4 days while 4 men and 4 children complete the same work in only 3 days. The number of days required by 1 man to complete the work is

60 days
15 days
6 days
51 days