Question

A can complete a piece of work in 10 days, b in 15 days and c in 20 days. A and c worked together for two days and then a was replaced by b. In how many days, altogether, was the work completed ?

Answer 1

In 1 day A completes = 1/10 part of the work

In 1 day B completes = 1/15 part of the work

In 1 day C completes = 1/20 part of the work

In 1 day A and C together complete = (1/10) + (1/20)

                                             = (2+1)/20

                                             = 3/20 part of the work

∴ In 2 days A and C together complete = 2× 3/20

                                                                   = 3/10 part of the work

∴ The amount of work left = 1-3/10

                                            = 10-3/10

                                            = 7/10 part of the work      

In 1 day B and C together complete = (1/15)+(1/20)

                                                           = (4+3)/60

                                                           = 7/60 part of the work

7/60 part of the work is done within 1 day

∴ 1 part of the work is done in (60/7) days

∴ 7/10 part of the work is completed in = (60/7)×(7/10)

                                                                 = 6 days.

∴ B and C will together complete the remaining work in 6 days.

In 1 day A,B,C together do =(1/10)+(1/15)+(1/20)

                                           = (6+4+3)/60

                                           = 13/60 part of the work

13/60 part of the work is completed in 1 day

∴ 1 part of the work is completed in = (60/13)

                                                           = 4+(8/13) days

∴ They altogether complete the work in 4 whole of 8/13 days

If according to the question A and C worked together for 2 days AND then B replaced A then, the work would be completed within (6+2)

                                                                                               = 8 days

So, the answer is 8 days.

Answer 2

Answer:

The answer is 8 days.

Step-by-step explanation:

• A completes work in 10 days.

• B completes work in 15 days.

• C completes work in 20 days.

Work can be calculated by finding the product of time and efficiency.

The formula used to calculate for work is

                           Work= Time x Efficiency

1. Firstly we will find the LCM (Lowest Common Multiple) of A,B,C.

2. LCM of 10,15 and 20 is 60.

3. The efficiency of A,B and C is:

• The efficiency of A will be 60/10 = 6
• The efficiency of B will be 60/15 = 4
• The efficiency of C will be 60/20 = 3

4. A and C work for two days

  Therefore, they completed

⇒  (6+3) x 2 days = 18 units.

Therefore, the work left will be

⇒ 60 - 18 = 42 units.

5. Now A is replaced by B, (B+C) one day

⇒ 4+3 = 7 unit

6. Therefore, the remaining work will be completed in

⇒ 42/7 = 6 days               ..........(i)

Therefore, the total work completed will be 6 (i) +2 (as per the question) = 8 days.

Thus, altogether the work was completed in 8 days.