Question Detail

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}

1. A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C

Rs. 300
Rs. 400
Rs. 500
Rs. 600

Answer: Option B

Explanation:

C's 1 day's work =
\begin{aligned}
\frac{1}{3}- \left(\frac{1}{6} +\frac{1}{8} \right) \\
=\left(\frac{1}{3} - \frac{7}{24} \right) \\
= \frac{1}{24} \\
A:B:C = \frac{1}{6}:\frac{1}{8}:\frac{1}{24} \\
= 4:3:1 \\
C's Share = \frac{1}{8}* 3200 \\
= 400
\end{aligned}
If you are confused how we multiplied 1/8, then please study ratio and proportion chapter, for small information, it is the C ratio divided by total ratio.

2. 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

3. 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}

4. 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.

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}