Question Detail

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}

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 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

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. 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

5. A alone can complete a work in 16 days and B alone can do in 12 days. Starting with A, they work on alternate days. The total work will be completed in

\begin{aligned} 13\frac{1}{4} \end{aligned}
\begin{aligned} 13\frac{1}{2} \end{aligned}
\begin{aligned} 13\frac{3}{4} \end{aligned}
\begin{aligned} 13\frac{4}{4} \end{aligned}

Answer: Option C

Explanation:

A's 1 day work = 1/16
B's 1 day work = 1/12

As they are working on alternate day's
So their 2 days work = (1/16)+(1/12)
= 7/48

[here is a small technique, Total work done will be 1, right, then multiply numerator till denominator, as 7*6 = 42, 7*7 = 49, as 7*7 is more than 48, so we will consider 7*6, means 6 pairs ]

Work done in 6 pairs = 6*(7/48) = 7/8

Remaining work = 1-7/8 = 1/8

On 13th day it will A turn,
then remaining work = (1/8)-(1/16) = 1/16

On 14th day it is B turn,

1/12 work done by B in 1 day

1/16 work will be done in (12*1/16) = 3/4 day

So total days =
\begin{aligned} 13\frac{3}{4} \end{aligned}

It may be a bit typical question, but if are not getting it in first try then give it a second try. Even not, then comment for explanation for this. We will be happy to help you.