A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ?

Answer: Option A

Explanation:

In this type of question, where we have one person work and together work done. Then we can easily get the other person work just by subtracting them. As,

Son's one day work =
\begin{aligned}
\left(\frac{1}{3}-\frac{1}{5} \right) \\
=\left(\frac{5-3}{15} \right) \\
= \frac{2}{15}
\end{aligned}

So son will do whole work in 15/2 days
which is =

\begin{aligned} 7\frac{1}{2}days \end{aligned}

Answer: Option C

Explanation:

A's 1 day work = 1/16
B's 1 day work = 1/12

As they are working on alternate day's
So their 2 days work = (1/16)+(1/12)
= 7/48

[here is a small technique, Total work done will be 1, right, then multiply numerator till denominator, as 7*6 = 42, 7*7 = 49, as 7*7 is more than 48, so we will consider 7*6, means 6 pairs ]

Work done in 6 pairs = 6*(7/48) = 7/8

Remaining work = 1-7/8 = 1/8

On 13th day it will A turn,
then remaining work = (1/8)-(1/16) = 1/16

On 14th day it is B turn,

1/12 work done by B in 1 day

1/16 work will be done in (12*1/16) = 3/4 day

So total days =
\begin{aligned} 13\frac{3}{4} \end{aligned}

It may be a bit typical question, but if are not getting it in first try then give it a second try. Even not, then comment for explanation for this. We will be happy to help you.

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 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