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Question

8 men and 12 boys 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 man alone , and that by one boy alone to finish the work. 


Solution

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


 

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