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. 4 men and 6 women finish a job in 8 days, while 3 men and 7 women finish it in 10 days. In how many days will 10 women working together finish it ?

30 days
40 days
50 days
60 days

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

2. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do a job. How long should it take both A and B, working together to do same job.

\begin{aligned} \frac{4}{9} \end{aligned}
\begin{aligned} 2\frac{4}{9} \end{aligned}
\begin{aligned} 3\frac{4}{9} \end{aligned}
\begin{aligned} 4\frac{4}{9} \end{aligned}

Answer: Option D

Explanation:

In this type of questions, first we need to calculate 1 hours work, then their collective work as,

A's 1 hour work is 1/8
B's 1 hour work is 1/10

(A+B)'s 1 hour work = 1/8 + 1/10
= 9/40

So both will finish the work in 40/9 hours
= \begin{aligned} 4\frac{4}{9} \end{aligned}

3. 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.

4. A is thrice as good a workman as B and takes 10 days less to do a piece of work than B takes. B alone can do the whole work in

15 days
10 days
9 days
8 days

Answer: Option A

Explanation:

Ratio of times taken by A and B = 1:3
Means B will take 3 times which A will do in 1 time

If difference of time is 2 days, B takes 3 days
If difference of time is 10 days, B takes (3/2) * 10 =15 days

5. 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