Question

A, B and C can do a piece of work in 20, 20 and 40 days respectively.
They began the work together but C left 2 days before the completion of
work. In how many days would the work have been completed?​

Answer 1

Answer:

11 days

Explanation:

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days.

Number of days required for this initial work = 7 days.

Thus, the total numbers of days required = 4 + 7 = 11 days.

Answer 2

The work can be completed in  8.4 days

Explanation:

Given:

1. A, B and C can do a piece of work in 20,20 and40 days respectively.

2. They began the work together but C left 2 days before the completion of        

   work

To find:

How many days would the work have been completed

Solution:

==> A can do a work in 20 days

==> B can do a work in 20 days

==> C can do a work in 40 days

==> A's one-day efficiency is  $\frac{1}{20}$

==> B's one-day efficiency is  $\frac{1}{20}$

==> c's one-day efficiency is  $\frac{1}{40}$

==> Let, work done = x

==> A+B+C = $\frac{1}{20}+ \frac{1}{20}+ \frac{1}{40}$

==> Work done by A,B,C = $\frac{x}{20}+ \frac{x}{20}+ \frac{x}{40}$ = 1

==> C leaves before 2 days

==> Work done by A,B,C = $\frac{x}{20}+ \frac{x}{20}+ \frac{x-2}{40}$ = 1

==>Solve the equation

==> Work done =  $\frac{(2)x}{(2)20}+ \frac{(2)x}{(2)20}+ \frac{x-2}{40}$ = 1

==> Work done = $\frac{2x}{40}+ \frac{2x}{40}+ \frac{x-2}{40}$ = 1

==> Now, the denominator is the same

==> $\frac{2x+2x+x-2}{40}=1$

==> 2x+2x+x-2 =40

==> 5x-2=40

==> 5x = 40+2

==> 5x = 42

==> x = 42÷ 5

==> x = 8.4 days

==> The work can be completed in  8.4 days