A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?

Answer: Option B

Explanation:

1 woman's 1 day's work = 1/70
1 Child's 1 day's work = 1/140
5 Women and 10 children 1 day work =
\begin{aligned}
\left(\frac{5}{70}+\frac{10}{140}\right) \\
= \frac{1}{7}
\end{aligned}

So 5 women and 10 children will finish the work in 7 days.

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 C

Explanation:

(A+B)'s 1 day work = 1/4

A's 1 day work = 1/12

B's 1 day work =
\begin{aligned}
\left( \frac{1}{4} - \frac{1}{12} \right) \\
= \frac{3-1}{12} \\
= \frac{1}{6} \\
\end{aligned}

So B alone can complete the work in 6 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 question we having, A's work, B's work and A+B+C work. We need to calculate C's work.
We can do it by,
(A+B+C)'s work - (A's work + B's work).

Let's solve it now:

C's 1 day work =
\begin{aligned}
\frac{1}{4}- \left(\frac{1}{16} +\frac{1}{12} \right) \\
=\left(\frac{1}{4} - \frac{7}{48} \right) \\

= \frac{5}{48}
\end{aligned}

So C can alone finish the job in 48/5 days,
Which is =
\begin{aligned} 9\frac{3}{5}days \end{aligned}