Question

Eight men and 12 boys can finish a piece of work in 10 days while six men and eight boys can finish the work in 14 days find the time taken by one man alone and that by one boy alone to finish the work

Answer 1

Step-by-step explanation:

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............eq 1

6/x+ 8/y = 1/14..........eq2

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

8u+ 12y=1/10.........eq3

6u+8v= 1/14...........eq4

Multiplying eq3 by 2 & eq4 by 3

16u+ 24y=2/10.........eq5

18u+24v= 3/14.........eq6

subtract eq 5 & 6

-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 eq 5

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