A and B can together complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work ?

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 D

Explanation:

A+B 1 day work = 1/8
B+C 1 day work = 1/12
A+B+C 1 day work = 1/6

We can get A work by (A+B+C)-(B+C)
And C by (A+B+C)-(A+B)

So A 1 day work =
\begin{aligned}
\frac{1}{6}- \frac{1}{12} \\
= \frac{1}{12}
\end{aligned}

Similarly C 1 day work =
\begin{aligned}
\frac{1}{6}- \frac{1}{8} \\
= \frac{4-3}{24} \\
= \frac{1}{24}
\end{aligned}

So A and C 1 day work =

\begin{aligned}
\frac{1}{12} + \frac{1}{24} \\
= \frac{3}{24} \\
= \frac{1}{8}
\end{aligned}

So A and C can together do this work in 8 days

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}

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 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.