Question

8 men and 12 women can finish a work in 10 days while 6 men and 8 women can finish in 14 days find time taken by one man and one woman alone to finish the work..

plzzz solve it​

Answer 1

Answer:

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

Answer 2

let number of men be x and number of female be y..

from the first condition,

8x+12y=10------(1)

from the second condition,

6x+8y=14------(2)

multiply equ(1) with 6

::48x+72y=60---(3)

multiply equ(2)with 8

::48x+64y=112---(4)

subtract equation 1 from 2,

::48x+72y=60

48x+64y=112

---------------------------

8y=52

y =14

so women =14 like wise we may get men...

mark as brainliest n follow me