A can do a piece of work in 4 hours . A and C together can do it in just 2 hours, while B and C together need 3 hours to finish the same work. In how many hours B can complete the work ?

Answer: Option C

Explanation:

Work done by A in l0 days = (1/25) *10 = 2/5
Remaining work = 1 - (2/5) = 3/5
(A+B)s 1 days work = (1/25) + (1/20) = 9/100

9/100 work is done by them in 1 day.
hence 3/5 work will be done by them in (3/5)*(100/9)
= 20/3days.

Total time taken = (10 + 20/3) = 16*(2/3) days

Answer: Option B

Explanation:

Let 1 man's 1 day work = x
and 1 woman's 1 days work = y.
Then, 4x + 6y = 1/8
and 3x+7y = 1/10
solving, we get y = 1/400 [means work done by a woman in 1 day]

10 women 1 day work = 10/400 = 1/40

10 women will finish the work in 40 days

Answer: Option A

Explanation:

Ratio of times taken by A and B = 1:3
Means B will take 3 times which A will do in 1 time

If difference of time is 2 days, B takes 3 days
If difference of time is 10 days, B takes (3/2) * 10 =15 days

Answer: Option C

Explanation:

Do not be confused, Take this question same as that of work done question's. Like work done by 1st puncture in 1 minute and by second in 1 minute.
Lets Solve it:

1 minute work done by both the punctures =
$$\begin{aligned} \left(\frac{1}{9}+\frac{1}{6} \right) \\ =\left(\frac{5}{18} \right) \\ \end{aligned}$$

So both punctures will make the type flat in
$$\begin{aligned} \left(\frac{18}{5} \right)mins \\ = 3\frac{3}{5} mins \end{aligned}$$

Answer: Option D

Explanation:

In this type of questions, first we need to calculate 1 hours work, then their collective work as,

A's 1 hour work is 1/8
B's 1 hour work is 1/10

(A+B)'s 1 hour work = 1/8 + 1/10
= 9/40

So both will finish the work in 40/9 hours
= $$\begin{aligned} 4\frac{4}{9} \end{aligned}$$