6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
∴ 36+12X2 = 1 day
60 children can do the work in 1 day

Correct Option: A

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.

Work done by 6 men = work done by 10 women.
Work done by 1 man = work done by 10 / 6 = 5/3 women
∴ 12 men + 5 women = 12 x ( 5 / 3) + 5 = 25 women
∴ W₁ x D ₁ = W₂ x

Correct Option: B

Work done by 6 men = work done by 10 women.
Work done by 1 man = work done by 10 / 6 = 5/3 women
∴ 12 men + 5 women = 12 x ( 5 / 3) + 5 = 25 women
∴ W₁ x D ₁ = W₂ x D₂ W = women, D=days
10 X 15 = 25 x D₂
D₂ = 6

Suppose the man covers first distance in x hrs. and second distance in y hrs.

Correct Option: D

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

Correct Option: B

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

Let C alone can finish the job in T hours.
According to the question,
Work done by A, B and C in 1h = 1/2
⇒ 1/6 +1/5 + 1/T = 1/2

Correct Option: B

Let C alone can finish the job in T hours.
According to the question,
Work done by A, B and C in 1h = 1/2
⇒ 1/6 +1/5 + 1/T = 1/2
⇒ 1/T = 1/2 - 1/6- 1/5 = (15 - 5- 6)/30
⇒ 1/T = 4/30 = 2/15
∴ T = 7¹/₂ h