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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}

Similar Questions :

1. A does half as much work as B in three-fourth of the time. If together they take 18 days to complete the work, how much time shall B take to do it

  • 40 days
  • 35 days
  • 30 days
  • 25 days

Answer: Option C

Explanation:

Suppose B takes x dáys to do the work.
As per question A will take
\begin{aligned}
2* \frac{3}{4} * x = \frac{3x}{2}days
\end{aligned}

(A+B)s 1 days work= 1/18
1/x + 2/3x = 1/18 or x = 30 days

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 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}

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.

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