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 then by one boy alone to finish the work what value are depicted in this question
Answers
Answer:
HI MATE
HERE'S THE 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
HOPE IT WILL HELP YOU
PLEASE MARK AS BRAINLIST
Answer:
Given
8M+12B can finish the work in 10 days
Whether
6M+8B can finish the same work in 14 days
So as we know work in same in both the cases
W1=W2
SIMILARLY
(8M+12B) X 10 = (6M+8B)X14
80M+120B= 84M+112B
120B-112B=84M-80M
8B=4M
M= (8/4)M
1M=2B
it means work done by one man = work done by 2 boys
Total work done by only one boy = (8M + 12B) X 10
8M=16B
(16B+12B)X10
=280days
Total work done by only one man=
(8M+6M)X10
=14MX10
=140days
Step-by-step explanation:
(Men power x time = work done)