Question

A, B and C can together do a piece of work in 10 days. A and B can complete it in 20

days and 30 days respectively. In how many days can C alone complete the same work?

plz help fast

Answer 1

A can complete the work in 10 days.

⇒ A’s 1day’s work = 1/10

B can complete the work in 20 days.

⇒ B’s 1 day’s work = 1/20

C can complete the work in 15 days.

⇒ C’s 1day’s work = 1/15

(A + B)’s 1 day’s work = (1/10) + (1/20) = 3/20 (A + B implies A and B)

(A + C)’s 1 day’s work = (1/10) + (1/15) = 1/6 (A + C implies A and C)

According to the question, A is assisted by B and C on alternate days.

⇒ Their 2 day’s work = (3/20) + (1/6) = 19/60

[∵ (A + B) work on 1st day and (A + C) work on 2nd day]

⇒ Their (3 × 2 =) 6 day’s work = 3 × (19/60) = 57/60

Remaining work = 1 - (57/60) = 3/60 = 1/20

On 7th day (A + B) will work.

(A + B) do (3/20)th work in 1 day

⇒ Time taken by (A + B) to do remaining work = (1/20)/(3/20) = (1/3) days

∴ The work will get completed in = days

Answer 2

Answer:

60 days

Step-by-step explanation:

A can do work in 20 days

B can do work in 30 days

A, B and C can do work in 10 days

LCM of 20, 30 and 10 is 60

Efficiency of A = 60/20 = 3

Efficiency of B = 60/ 30 = 2

Efficiency of A, B and C together = 60/10 = 6

Efficiency of C = Combined efficieny - A - B

6 - 3 - 2 = 1

C alone can do work in = 60 / 1 = 60 days