Question

5. A and B can do a piece of work in 10 days, B and C in 15 days and C and A in 12 days.
In how many days can A, B and C finish it if they all work together? Also find the number
of days B will require to finish the work if he works alone.​

Answer 1

Given:

• (A+B) can do a piece work in 10 days ••••eq.(i)
• (B+C) can do a piece of work in 15 days ••••eq.(ii)
• (C+A) can do a piece of work in 12 days ••••eq.(iii)

To Find:

• In how may day can A, B and C finish work if they worked together?
• In how many days 'B' will complete the work alone?

Solution:

Given values are that (A+B),(B+C) and (C+A) can complete q piece of work in 10 days, 15 days and 12 days respectively.

Finding 1 day's work :

(A+B) can finish work in 10 days.

⁂ (A+B)'s one day work = 1/10 part ••••••eq.(iv)

(B+C) can finish work in 15 days.

⁂ (B+C)'s one day work = 1/15 part ••••••eq.(v)

(C+A) can finish work in 12 days.

⁂ (C+A)'s one day work = 1/12 part ••••••• eq.(vi)

_____________

Now, we found that (A+B)'s one day work, (B+C)'s one day work and (C+A)'s one day work is 1/10,1/15 and 1/12 respectively. Let think, can we find (A+B),(B+C) and (C+A)'s one day work together? Yes, definitely we can find it by adding eq.(iv),(v),(vi).

Finding (A+B),(B+C) and (C+A)'s one day work :

• (A+B)'s one day work = 1/10
• (B+C)'s one day work = 1/15
• (C+A)'s one day work = 1/12

⁂(A+B)+(B+C)+(C+A) one day's work = $\large{\sf{\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{12}}}$

$\implies\large{\sf{\dfrac{6+4+5}{60}}}$

$\implies\large{\sf{\dfrac{15}{60}}}$

$\implies\large{\sf{\dfrac{1}{4}}}$

$\implies\sf\small{(A+B)+(B+C)+(C+A)~ one~ day's ~work~=~ \dfrac{1}{4}}$

$\implies\sf\small{2(A+B+C)~one~day's~work\:=\:\dfrac{1}{4}}$

$\implies\sf\small{A+B+C\:one\: day's~work\:=\: \dfrac{1}{4×2}}$

$\implies\sf\small{A+B+C~one~day's~work~=~\dfrac{1}{8}}$

Therefore,

• $\large{\boxed{\sf{(A+B+C)~can~complete~whole~work~in~8~days}}}$

_______________

Now, we found that (A+B+C) can do 1/8 part in 1 day.

• We are given that(A+C) can do 1/12 part in 1 day.

Therefore,

(A+B+C) one day's work = 1/8

→(1/12 + B) one day's work = 1/8

→ B one day's work = $\large{\sf{\dfrac{1}{8}-\dfrac{1}{12}}}$

→ B one day's work = $\large{\sf{\dfrac{3-2}{24}}}$

→ B one day's work= $\large{\sf{\dfrac{1}{24}}}$

Therefore,

• $\large{\boxed{\sf{B~will~take~24~days~to~complete~whole~work~alone}}}$

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