Question

8meb and 12boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days .find the time taken by one men alone and that by one alone to finish the work.

Answer 1

Let the time taken by man be x days

time taken by boys be y days

work done by 1 man in one day=1/x

work done by 1 boy in one day=1/y

8/x+12/y = 1/10............(i)

6/x+ 8/y = 1/14..........(ii)

Let u=1/x & v=1/y

8u+ 12y=1/10.........(iii)

6u+8v= 1/14...........(iv)

Multiplying (iii) by 2 & (iv) by 3

16u+ 24y=2/10.........(v)
18u+24v= 3/14.........(vi)

subtract (v) & (vi)
-2u= -3/14+2/10
-2u = (-3×10 +2×14)/140

-2u=( -30+28)/140

-2u=-2/140

u=1/140

put this value in (v)

16u+ 24y=2/10

16×1/140 +24y =1/5

4/35+24y= 1/5

24y= 1/5-4/35

24y= (1×7 -4)/35

24y = (7-4)/35

24y= 3/35

y= 3/35×24

y= 1/280

u=1/x

1/140 = 1/x

x=140

v=1/y

1/280 = 1/y
y=280

A man complete the work in 140 days
A boy can complete the work in 280 days