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. 5 men and 2 boys working together can do four times as much work as a man and a boy. Working capacity of man and boy is in the ratio

Answer: Option C

Explanation:

Let 1 man 1 day work = x
1 boy 1 day work = y

then 5x + 2y = 4(x+y)
=> x = 2y
=> x/y = 2/1
=> x:y = 2:1

2. A can do a piece of work in 15 days and B alone can do it in 10 days. B works at it for 5 days and then leaves. A alone can finish the remaining work in

5 days
6 days
7.5 days
8.5 days

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}

3. A piece of work can be done by 6 men and 5 women in 6 days or 3 men and 4 women in 10 days. It can be done by 9 men and 15 women in how many days ?

3 days
4 days
5 days
6 days

Answer: Option A

Explanation:

To calculate the answer we need to get 1 man per day work and 1 woman per day work.

Let 1 man 1 day work =x
and 1 woman 1 days work = y.
=> 6x+5y = 1/6
and 3x+4y = 1/10
On solving, we get x = 1/54 and y = 1/90

(9 men + 15 women)'s 1 days work =
(9/54) + (15/90) = 1/3

9 men and 15 women will finish the work in 3 days

4. A can do a job in 16 days, B can do same job in 12 days. With the help of C they did the job in 4 days. C alone can do the same job in how many days ?

\begin{aligned} 6\frac{1}{2}days \end{aligned}
\begin{aligned} 7\frac{1}{2}days \end{aligned}
\begin{aligned} 8\frac{3}{5}days \end{aligned}
\begin{aligned} 9\frac{3}{5}days \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}

5. 10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?

6 days
7 days
8 days
9 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.