Question

A, B and C can do a piece of work in
24, 32 and 64 days respectively. They
starts working, A left the work after 6
days while B left the work before 6 days
from the completion of work. In how
many days work will be finished?
A. 20
B. 18
C. 15
D. None of these​

Answer 1

The work will be finished in 20 days, if A leaves after the first 6 days,  B leaves before the last 6 days and C does continue to work till the end.

Step-by-step explanation:

Step 1:

It is given that,

A, B & C can do a piece of work in 24, 32 and 64 days respectively.

So, The fraction of work done by each of them in 1 day:

A = $\frac{1}{24}$

B = $\frac{1}{32}$

C = $\frac{1}{64}$

A leaves after 6 days of the start of the work therefore, A, B & C each have worked for 6 days, i.e.,

The fraction of work done by each of them in 6 days is:

A = $\frac{6}{24}$ = ¼

B = $\frac{6}{32} = \frac{3}{16}$

C = $\frac{6}{64} = \frac{3}{32}$

∴ Total fraction of work done in first 6 days = $\frac{1}{4} + \frac{3}{16} + \frac{3}{32} = \frac{8+6+3}{32} = \frac{17}{32}$

And

The remaining fraction of work = $1 - \frac{17}{32} = \frac{15}{32}$

Step 2:

Now, B leaves 6 days before the completion of the work, which means that C works alone for 6 days.

So, the fraction of work that C does alone in last 6 days is =  $\frac{6}{64} = \frac{3}{32}$

∴ The fraction of work done before B has left =  $\frac{15}{32} - \frac{3}{32} = \frac{12}{32}$ $= \frac{3}{8}$

Let's assume B and C did the $\frac{3}{8}$th of the work in "x" days.

So, we have

$\frac{x}{32}$ work  is done by B and $\frac{x}{64}$ work is done by C.

Therefore, we can write the equation as,

$\frac{x}{32} + \frac{x}{64} = \frac{3}{8}$

⇒ x * $\frac{2+1}{64} = \frac{3}{8}$

⇒ x = $\frac{64}{8}$

⇒ x = 8 days

Thus,

The total days required to finish the work is,

= 6 + 8 + 6

= 20 days

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Answer 2

The correct answer is A. 20 days.

The efficiency of each one is given as follows:

A's = $\frac{1}{24}$

B's = $\frac{1}{32}$

C's = $\frac{1}{64}$

Now A did work for 6 days so total work done by A = $\frac{6}{24}$ = $\frac{1}{4}$

Let the total work be finished in x days then

Work done by B = $\frac{x-6}{32}$

Work done by C = $\frac{x}{64}$

Total work done = work done by A + work done by B + work done by C

1 = $\frac{1}{4}$ + $\frac{x-6}{32}$ + $\frac{x}{64}$

1 = $\frac{16+2x - 12+x }{64}$

64 = 3x + 4

3x = 60

x = 20 days