Question Detail

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.

1. 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

2. 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}

3. A completes 80% of a work in 20 days. Then B also joins and A and B together finish the remaining work in 3 days. How long does it need for B if he alone completes the work?

\begin{aligned} 35\frac{1}{2} \end{aligned}
\begin{aligned} 36\frac{1}{2} \end{aligned}
\begin{aligned} 37\frac{1}{2} \end{aligned}
\begin{aligned} 38\frac{1}{2} \end{aligned}

Answer: Option C

Explanation:

Work done by A in 20 days = 80/100 = 8/10 = 4/5

Work done by A in 1 day = (4/5) / 20 = 4/100 = 1/25 --- (1)

Work done by A and B in 3 days = 20/100 = 1/5 (Because remaining 20% is done in 3 days by A and B)

Work done by A and B in 1 day = 1/15 ---(2)

Work done by B in 1 day = 1/15 – 1/25 = 2/75

=> B can complete the work in 75/2 days = 37 (1/2) days

4. 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

5. To complete a work A and B takes 8 days, B and C takes 12 days, A,B and C takes 6 days. How much time A and C will take

24 days
16 days
12 days
8 days

Answer: Option D

Explanation:

A+B 1 day work = 1/8
B+C 1 day work = 1/12
A+B+C 1 day work = 1/6

We can get A work by (A+B+C)-(B+C)
And C by (A+B+C)-(A+B)

So A 1 day work =
\begin{aligned}
\frac{1}{6}- \frac{1}{12} \\
= \frac{1}{12}
\end{aligned}

Similarly C 1 day work =
\begin{aligned}
\frac{1}{6}- \frac{1}{8} \\
= \frac{4-3}{24} \\
= \frac{1}{24}
\end{aligned}

So A and C 1 day work =

\begin{aligned}
\frac{1}{12} + \frac{1}{24} \\
= \frac{3}{24} \\
= \frac{1}{8}
\end{aligned}

So A and C can together do this work in 8 days