3 men does same work as 4 women in 43 days.

multiply 3 and 4 and 43 for a supposed amount of total work (516).

Now calculate how much work is done by 3 men in one day by dividing 516 by 43 (12). Then calculate one man work in one day by dividing 12 by 3 (4).

Same (12) amount of work is done by 4 women in one day. so one women does 12/4 (3) in one day.

now see there are 7 men and 5 women in the final team.
total amount of work in one day will be

7(4)+5(3) = 43

Now divide total work (516) by total work in one day (43) to get the answer (12)

Time is
directly proportional to number of sofas
inversely proportional to number of people No. of days
= (6/8)*(8/10)*2
= 1.2 days

10 workers make 10 tables in 10 days.

1 worker makes 1 table in 10 days. Similarly,

5 workers can make 5 tables in 10 days.

For the work to be completed in 9 days,
6 men are required. Time and men are inversely proportional. For the work to be completed in 3 days,
6*9/3 = 18 men are required

Assume that cost of keeping a cow for 1 day = c ,

cost of keeping a goat for 1 day = g
Cost of keeping 20 cows and 40 goats for 10 days = 460

Cost of keeping 20 cows and 40 goats for 1 day = 460/10 = 46

=> 20c + 40g = 46

=> 10c + 20g = 23 —(1)

Given that 5g = c
Hence equation (1) can be written as 10c + 4c = 23 => 14c =23

=> c=23/14
cost of keeping 50 cows and 30 goats for 1 day

= 50c + 30g

= 50c + 6c (substituted 5g = c)

= 56 c = 56×23/14

= 92
Cost of keeping 50 cows and 30 goats for 12 days = 12×92 = 1104