Question

8mens and 12boys can finish a piece of work in 10 days and 6men and 8boys can finish the work in 14 days. find the time taken by one man alone and that of one boy

Answer 1

Let the time taken by a man be x days

time taken by a boy 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............1

6/x+ 8/y = 1/14..........2

Let a=1/x & b=1/y

8a+ 12b=1/10.........3

6a+8b= 1/14...........4

Multiplying eqn 3 by 2 & eqn 4 by 3

16a+ 24b=2/10.........5
18a+24b= 3/14.........6

subtract eqn 5 & 6
-2a= -3/14+2/10
-2a = (-3×10 +2×14)/140

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

-2a=-2/140

a=1/140

put this value in eqn 5

16a+ 24b=2/10

16×1/140 +24b =1/5

4/35+24b= 1/5

24b= 1/5-4/35

24b= (1×7 -4)/35

24b = (7-4)/35

24b= 3/35

b= 3/35×24

b= 1/280

a=1/x

1/140 = 1/x

x=140

b=1/y

1/280 = 1/y
y=280

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

Answer 2

Hey

Here is your answer,

In attachment

Hope it helps you!