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 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. A can do a certain job in 25 days which B alone can do in 20 days. A started the work and was joined by B after 10 days. The number of days taken in completing the wotk were ?

\begin{aligned} 14\frac{2}{3}kmph \end{aligned}
\begin{aligned} 15\frac{2}{3}kmph \end{aligned}
\begin{aligned} 16\frac{2}{3}kmph \end{aligned}
\begin{aligned} 17\frac{2}{3}kmph \end{aligned}

Answer: Option C

Explanation:

Work done by A in l0 days = (1/25) *10 = 2/5
Remaining work = 1 - (2/5) = 3/5
(A+B)s 1 days work = (1/25) + (1/20) = 9/100

9/100 work is done by them in 1 day.
hence 3/5 work will be done by them in (3/5)*(100/9)
= 20/3days.

Total time taken = (10 + 20/3) = 16*(2/3) days

4. 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.

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}