Question

A and B can finish a work in 15 days from B and C

in 20 days and A and C can finish it in 30 days, then

in how many days will the three together finish the

work?​

Answer 1

Given:

• A and B can finish work in 15 days.
• B and C can finish work in 20days.
• A and C can finish work in 30days.

To find :

• In how many days will they together can finish work?

Solution:

We have three given i.e -

(A+B) can complete work in 15 days.

→ (A+B) one day's work = 1/15 part ••••eq.(i)

(B+C) can complete work in 20 days.

→ (B+C) one day's work = 1/20 part ••••eq.(ii)

(A+C) can complete work in 30 days.

→ (A+C) one day's work = 1/30 part ••••eq.(iii)

Now,we found that (A+B),(B+C) and (A+C) one day's work is 1/15,1/20 and 1/30 respectively.

So,(A+B),(B+C),(A+C) one day's work= Sum of eq.(i),(ii) and (iii) =>

(A+B)+(B+C)+(A+C) one day's work= $\frac{1}{15} + \frac{1}{20} + \frac{1}{30}$

→ (A+B)+(B+C)+(A+C) one day's work = $\frac{4 + 3 + 2}{60}$

→ (A+B)+(B+C)+(A+C) one day's work = $\dfrac{9}{60}$

→ (A+B)+(B+C)+(A+C) one day's work = $\dfrac{3}{20}$

→ (2A + 2B+2C)one day's work = $\dfrac{3}{20}$

→ 2(A+B+C) one day's work = $\dfrac{3}{20}$

→ (A+B+C) one day's work = $\dfrac{3}{20×2}$

→ (A+B+C) one day's work = $\dfrac{3}{40}$

→ (A+B+C) take to complete total work = $1÷\dfrac{3}{40}$

→ (A+B+C) take to complete total work = 40/3 days.

Therefore,

• $\large{\boxed{\sf{\green{A,B\:and\:C\:will\:take\:\dfrac{40}{3}\:days\:to\: complete\:whole\: work\: together}}}}$