Question

2 men and 5 boys together can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by 1 man alone to finish the work and than taken by 1 boy alone.

Answer 1

let the time taken by 1 men alone to finish the work be x days
and the time taken by 1 boy alone to finish the work be y days then
1 men's 1 day's work= 1/x
1 boy's 1 day's work=1/y
therefore 2men's 1 days work=2/x
and 5 boys 1 days work= 5/y
since 2 men and 5 boys can together finish one work in 4 days
therefore
4(2/x+5/y)=1
2/x+5/y=1/4
again,3 men's 1 days work=3/x
and
6 boys's 1 day's work=6/y
since
3 men's and 6 boys can together finish one work in 3 days
therefore
3(3/x+6/y)=1
3/x+6/y=1/3....(1)
put 1/x=u.....(2)
and 1/y=v....(3)
then equation (1) and (2) can be rewritten as
2u+5v=1/4...(5)
3u+6v=1/3...(6)
multiplying equation (5) by 3 and equation (6) by 2 we get
6u+15v=3/4...(7)
6u+12v=2/3...(8)
subtracting, equation (8) from equation (7) we get
3v=1/12
v=1/36...(9)
substituting this value of v in equation (5) we get
u=1/18
from equation (3) and equation (10) we get
x=18
and y=36
hence the time taken by 1 men alone to finish the work is 18days and the time taken by 1 boy alone to finish the work is 36 days



I hope this help you

Answer 2

Step-by-step explanation:

Let the amount of work done by a Man alone in a day be M and by a Boy alone be B

Given, 2 men and 5 boys together can finish a piece of work in 4 days

Amount of work they can do in one day 2M+5B =

4

1

-- (1)

Similarly, 3M+6B=

3

1

--- (2)

Solving equations (1) and (2), we get B=

36

1

Substituting B in equation (2), we get M=

18

1

Hence, one man alone will take 18 days to do the work and one boy will take 36 days.