A man can do a piece of work in 5 days, but with the help of his son he can do it in 3 days. In what time can the son do it alone ?

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 B

Explanation:

C's 1 day's work =
$$\begin{aligned} \frac{1}{3}- \left(\frac{1}{6} +\frac{1}{8} \right) \\ =\left(\frac{1}{3} - \frac{7}{24} \right) \\ = \frac{1}{24} \\ A:B:C = \frac{1}{6}:\frac{1}{8}:\frac{1}{24} \\ = 4:3:1 \\ C's Share = \frac{1}{8}* 3200 \\ = 400 \end{aligned}$$
If you are confused how we multiplied 1/8, then please study ratio and proportion chapter, for small information, it is the C ratio divided by total ratio.

Answer: Option C

Explanation:

B's 5 days work =
$$\begin{aligned} \frac{1}{10}*5 = \frac{1}{2} \\ \text{Remaining work =} 1-\frac{1}{2} \\ = \frac{1}{2} \\ \text{A can finish work =}15*\frac{1}{2} \\ = 7.5 days \end{aligned}$$

Answer: Option B

Explanation:

As per question, A do twice the work as done by B.
So A:B = 2:1
Also (A+B) one day work = 1/18

To get days in which B will finish the work, lets calculate work done by B in 1 day =
$$\begin{aligned} =\left(\frac{1}{18}*\frac{1}{3} \right) \\ = \frac{1}{54} \end{aligned}$$
[Please note we multiplied by 1/3 as per B share and total of ratio is 1/3]

So B will finish the work in 54 days

Answer: Option B

Explanation:

Work done by A in 1 hour = 1/4

Work done by B and C in 1 hour = 1/3

Work done by A and C in 1 hour = 1/2

Work done by A,B and C in 1 hour = (1/4)+(1/3) = 7/12

Work done by B in 1 hour = (7/12)–(1/2) = 1/12

=> B alone can complete the work in 12 hour