Question Detail

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.

1. 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}

2. A is twice as good as workman as B and together they finish a piece of work in 18 days. In how many days will B alone finish the work.

27 days
54 days
56 days
68 days

Answer: Option B

Explanation:

As per question, A do twice the work as done by B.
So A:B = 2:1
Also (A+B) one day work = 1/18

To get days in which B will finish the work, lets calculate work done by B in 1 day =
\begin{aligned}
=\left(\frac{1}{18}*\frac{1}{3} \right) \\
= \frac{1}{54}
\end{aligned}
[Please note we multiplied by 1/3 as per B share and total of ratio is 1/3]

So B will finish the work in 54 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:

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:

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}