If 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. Then, the total no of days by one man alone and that by one boy alone to finish the work is -
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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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