Question Detail

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 ?

\begin{aligned} 3\frac{1}{5} min \end{aligned}
\begin{aligned} 3\frac{2}{5} min \end{aligned}
\begin{aligned} 3\frac{3}{5} min \end{aligned}
\begin{aligned} 3\frac{4}{5} min \end{aligned}

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}

1. A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the work, how much time shall B take to do it

40 days
35 days
30 days
25 days

Answer: Option C

Explanation:

Suppose B takes x dáys to do the work.
As per question A will take
\begin{aligned}
2* \frac{3}{4} * x = \frac{3x}{2}days
\end{aligned}

(A+B)s 1 days work= 1/18
1/x + 2/3x = 1/18 or x = 30 days

2. A can finish a work in 18 days and B can do same work in half the time taken by A. then working together, what part of same work they can finish in a day

Answer: Option B

Explanation:

Please note in this question, we need to answer part of work for a day rather than complete work. It was worth mentioning here because many do mistake at this point in hurry to solve the question

So lets solve now,
A's 1 day work = 1/18
B's 1 day work = 1/9 [because B take half time than A]

(A+B)'s one day work =

\begin{aligned}
\left(\frac{1}{18}+\frac{1}{9} \right) \\
=\left(\frac{1+2}{18} \right) \\
= \frac{1}{6}
\end{aligned}

So in one day 1/6 work will be done.

3. 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 ?

10 hours
12 hours
16 hours
18 hours

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

4. A does a work in 10 days and B does the same work in 15 days. In how many days they together will do the same work ?

5 days
6 days
7 days
8 days

Answer: Option B

Explanation:

Firstly we will find 1 day work of both A and B, then by adding we can get collective days for them,
So,
A's 1 day work = 1/10
B's 1 day work = 1/15

(A+B)'s 1 day work =
\begin{aligned}
\left(\frac{1}{10}+\frac{1}{15} \right) \\
=\left(\frac{3+2}{30} \right) \\
= \frac{1}{6}
\end{aligned}

So together they can complete work in 6 days.

5. 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 ?

\begin{aligned} 3\frac{1}{5} min \end{aligned}
\begin{aligned} 3\frac{2}{5} min \end{aligned}
\begin{aligned} 3\frac{3}{5} min \end{aligned}
\begin{aligned} 3\frac{4}{5} min \end{aligned}

Answer: Option C

Explanation: