Question

A can do a piece of work in 20days while B can do it in 15days. With the help of C, they finish the work in 5days. In what time would C alone do it​

Answer 1

Answer:

ans : 12 days

Step-by-step explanation:

Work by A- 1/20

work by B- 1/15

Total work done by all - 1/5

work by C - 1/c

now add all = 1/20 + 1/15 + 1/c = 1/5

lcm is 60 .

4+3-. 12/60 = - 1/c

7-12/60 = -1/c

-5/60= -1/c

minus minus cancel.

cross multiplication

5c = 60

c= 60/5

c= 12

therefore the time taken by c alone to do a piece of work is 12 days.

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Answer 2

Step-by-step explanation:

AnswEr:

$\bold{\underline{\sf{\blue{\:\:Given\:\:}}}}$

A & B can do work in 12 Days.

B & C can do work in 15 Days.

C & A can do work in 20 Days.

Now,

• A & B can do work in 1 day = $\sf\dfrac{1}{12}$

$:\implies\sf\: A \: and \: B \:=\: \dfrac{1}{12}$ -------Eq(1)

• B & C can do work in 1 day = $\sf\dfrac{1}{15}$

$:\implies\sf\: B \: and \: C \:=\: \dfrac{1}{15}$ -------Eq(2)

• C & A can do work in 1 day = $\sf\dfrac{1}{20}$

$:\implies\sf\: C \: and \: A \:=\: \dfrac{1}{20}$ -------Eq(3)

From Equations (1) & (2)

$:\implies\sf\: A \: + \: B \:=\: \dfrac{1}{12}$

$:\implies\sf\: B \: + \: C \:=\: \dfrac{1}{15}$

$:\implies\sf\: A \: - \: C \:=\: \dfrac{1}{12} \:-\: \dfrac{1}{15}$

$\dag\:\small\bold{\underline{\sf{\red{Taking\:LCM}}}}$

$:\implies\sf\:\dfrac{5\:-4}{60}$

$:\implies\sf\: A \:-\: C\:=\: \dfrac{1}{60}$ -------Eq(4)

$\dag\:\small\bold{\underline{\sf{\red{Adding\:Equati\:(3)\:and\:4}}}}$

$:\implies\sf\: A \:+\: C = \dfrac{1}{20}$

$:\implies\sf\: A\:-\:C = \dfrac{1}{60}$

$:\implies\sf\: 2A = \dfrac{1}{20}\:-\: \dfrac{1}{60}$

$:\implies\sf\: \dfrac{3\:+\:60}{20}$

$:\implies\sf\: \cancel\dfrac{4}{60}$

$:\implies\sf\: \dfrac{1}{15}$

$:\implies\sf\: 2A = \dfrac{1}{15} \:+\:\dfrac{1}{15}$

$:\implies\sf\: A = \dfrac{2}{15} \times \; \dfrac{1}{2}$

$:\implies\sf\: A = \dfrac{1}{30}$

$\small\bold{\underline{\sf{\blue{Hence,\:A\:will\:take\;30\:days\:to\: complete\:the\:work.}}}}$