6 men or 10 women can reap a field in 15 days, then the number of days that 12 men and 5 women will take to reap the same field is :

Suppose the man covers first distance in x hrs. and second distance in y hrs.
Then , 4x + 5y = 35 and 5x +4y= 37
Solving these equation , we get
X=5 and y=3
Total time taken = (5+3) hrs= 8hrs

At a speed of 3/4 of the usual speed, the time taken is 4/3 of the usual time
= (4/3 of usual time ) ? (usual time ) = 20min.
? 4x/3 -x = 20
? x/3 =20
? x=60 min

Average speed = 2t₁ x t₂/( t₁ + t₂)
= (2x64x80)/(64+80)km
= (2 x 64 x 80 ) / 144 km/hr.
=71.11km/hr.

6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
? 36+12X2 = 1 day
60 children can do the work in 1 day
Now, 5 men = 10 children
? 10 children can do the work in 6 days.

2 men = 7 boys
? 1 man = 7/2 days

5 women = 7 boys
? 1 women = 7/5 boys

7 men + 5 women + 2 boys = 7 x 7/2 + 5 x 7/5 + 2 = 67/2 boys
Now, B₁ x D₁ = B₂ x D₂
7 X 469 = 67/2 x D₂
D₂ = 98 days

In 1 hour 314 weavers weave = 6594 x 6 shawls
In 1 hour 1 weaver weaves = ( 6594 x 6 ) / 314 shawls = 126 shawls

Ratio of efficiencies of A,B and C = 6 : 5 : 4
? Share of C = [4/(6+5+4)] x 27000 = 7200

Work done by 2 men = 3 woman = 4 boys
? 1 man = 2 boys
1 woman = 4/3 boys
? boys X days = 4 X 52 (boys - days)
Again 1 man + 1 woman + 1 boy = 2 + 4/3 + 1 = 13/3 boys
B₁ x D₁ = B₂ x D₂ B = boys, D = days
4 X 52 = 13/3 x D₂
D₂ = 48 days