Eight men and twelve boys can finish a piece of work in 10 days while six man and eight boys can finish the same work in 14 days. Find the number of days taken by one man alone to complete the work and also one boy alone to complete the work.
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Yes! Here is your answer!
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
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