Question Detail

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

1. 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.

2. 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 ?

4 days
5 days
6 days
7 days

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

3. 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.

4. 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}

5. A man can do a piece of work in 5 days, but with the help of his son he can do it in 3 days. In what time can the son do it alone ?

\begin{aligned} 7\frac{1}{2}days \end{aligned}
\begin{aligned} 6\frac{1}{2}days \end{aligned}
\begin{aligned} 5\frac{1}{2}days \end{aligned}
\begin{aligned} 4\frac{1}{2}days \end{aligned}

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}