Question

A and B can do a piece of work in 12 days, B and C in 15 days, C and A in 20 days. How long will

each take separately to do the same work?​

Answer 1

Answer :-

A takes 30 days, B takes 20 days and C takes 60 days if each works separately.

Solution :-

Let the A's one day work be 1/A

Let the B's one day work be 1/B

Let the C's one day work be 1/C

Number of days A and B needed to finish the work = 12

A and B one day work = 1/12

⇒ 1/A + 1/B = 1/12 ----(1)

Number of days B and C needed to finish the work = 15

B and C one day work = 1/15

⇒ 1/B + 1/C = 1/15 ---(2)

Number of days C and A needed to finish the work = 20

C and A one day work = 1/20

⇒ 1/C + 1/A = 1/20 ---(3)

Adding (1), (2) and (3)

⇒ 1/A + 1/B + 1/B + 1/C + 1/C + 1/A = 1/12 + 1/15 + 1/20

⇒ 2/A + 2/B + 2/C = (5 + 4 + 3)/60

⇒ 2(1/A + 1/B + 1/C) = 12/60

⇒ 1/A + 1/B + 1/C = 12/(60 * 2)

⇒ 1/A + 1/B + 1/C = 1/(5 * 2)

⇒ 1/A + 1/B + 1/C = 1/10 --(4)

i) 1/C = 1/10 - (1/A + 1/B)

[From (4)]

⇒ 1/C = 1/10 - 1/12

Since 1/A + 1/B = 1/12

⇒ 1/C = (6 - 5)/60

⇒ 1/C = 1/60

⇒ C = 60

ii) 1/B = 1/10 - (1/C + 1/A)

[From (4)]

⇒ 1/B = 1/10 - 1/20

Since 1/C + 1/A = 1/20

⇒ 1/B = (2 - 1)/20

⇒ 1/B = 1/20

⇒ B = 20

iii) 1/A = 1/10 - (1/B + 1/C)

[From (4)]

⇒ 1/A = 1/10 - 1/15

Since 1/B + 1/C = 1/15

⇒ 1/A = (3 - 2)/30

⇒ 1/A = 1/30

⇒ A = 30

Therefore A takes 30 days, B takes 20 days and C takes 60 days if each works separately.

Answer 2

A and B together do a work in 12 days.

Work done by A and B in one day = $\dfrac{1}{12}$ days

B and C together do a work in 15 days.

Work done by B and C in one day = $\dfrac{1}{15}$ days

C and A together do a work in 20 days.

Work done by C and A in one day = $\dfrac{1}{20}$ days.

We have to find how many days will each take separately to do the same work.

According to question,

Both A and B do a work in 12 days and in one day they do $\dfrac{1}{12}$ of the work.

=> $\dfrac{1}{A}$ + $\dfrac{1}{B}$ = $\dfrac{1}{12}$ _______ (eq 1)

Both B and C do work in 15 days and in one day they d $\dfrac{1}{15}$ of the work.

=> $\dfrac{1}{B}$ + $\dfrac{1}{C}$ = $\dfrac{1}{15}$ _______ (eq 2)

Both C and A do work in 20 days and in one day they d $\dfrac{1}{20}$ of the work.

=> $\dfrac{1}{C}$ + $\dfrac{1}{A}$ = $\dfrac{1}{20}$ _______ (eq 3)

Now,

Add (eq 1), (eq 2) and (eq 3)

=> $\frac{1}{A}$ + $\frac{1}{B}$ + $\frac{1}{B}$ + $\frac{1}{C}$ + $\frac{1}{C}$ + $\frac{1}{A}$ = $\frac{1}{12}$ + $\frac{1}{15}$ + $\frac{1}{20}$

=> $\dfrac{2}{A}$ + $\dfrac{2}{B}$ + $\dfrac{2}{C}$ = $\dfrac{5\:+\:4\:+\:3}{60}$

=> $2\bigg(\dfrac{1}{A} \: + \: \dfrac{1}{B} \: + \: \dfrac{1}{C} \bigg)$ = $\dfrac{12}{60}$

=> $\dfrac{1}{A}$ + $\dfrac{1}{B}$ + $\dfrac{1}{C}$ = $\dfrac{1}{5\:\times\:2}$

=> $\dfrac{1}{A}$ + $\dfrac{1}{B}$ + $\dfrac{1}{C}$ = $\dfrac{1}{10}$

A, B and C together do work in 10 days.

Now,

• A and B do work in 12 days.

So, wrok done by C = Work done by all three in one day together - Work done by A and B in one day

=> $\dfrac{1}{C}$ = $\dfrac{1}{10}$ - $\dfrac{1}{12}$

=> $\dfrac{1}{C}$ = $\dfrac{6\:-\:5}{60}$

=> $\dfrac{1}{C}$ = $\dfrac{1}{60}$

C do wrok in 60 days.

• B and C do work in 15 days

So, work done by A = Work done by all three in one day together - Work done by B and C in one day

=> $\dfrac{1}{A}$ = $\dfrac{1}{10}$ - $\dfrac{1}{15}$

=> $\dfrac{1}{C}$ = $\dfrac{3\:-\:2}{30}$

=> $\dfrac{1}{C}$ = $\dfrac{1}{30}$

A do work in 30 days.

• C and A do work in 20 days

So, work done by B = Work done by all three in one day together - Work done by C and A in one day

=> $\dfrac{1}{B}$ = $\dfrac{1}{10}$ - $\dfrac{1}{20}$

=> $\dfrac{1}{B}$ = $\dfrac{2\:-\:1}{20}$

=> $\dfrac{1}{B}$ = $\dfrac{1}{20}$

B do work in 20 days.