Question

8. A and B can do a piece of work in 6 days, B and C do it in 10 days and C and A do it in 15 day
(a) In how many days will A, B and C finish it working together?
(6) In how many days will A and B finish their working alone?
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Answer 1

$\Large\sf{\underline{\underline{\color{red}{GIVEN:-}}}}$

• A and B can do a piece of work in 6 days.
• B and C do the same work in 10 days.
• C and A do it in 15 days.

$\Large\sf{\underline{\underline{\color{red}{TO\ FIND:-}}}}$

(a) The number of days will A, B and C finish it working together.

(b) The number of days will A and B finish their working alone.

$\Large\sf{\underline{\underline{\color{red}{SOLUTION:-}}}}$

$\bigstar$ A and B can do a piece of work in 6 days.

→ Work done by A and B in one day = 1/6

→ A + B = 1/6

$\bigstar$ B and C do it in 10 days.

→ Work done by B and C in 1 day = 1/10

→ B + C = 1/10

$\bigstar$ C and A do it in 15 days.

→ Work done by C and A in 1 day = 1/15

→ C + A = 1/15

____________________________

(a) Work done by A, B and C in one day :

$\sf{\implies 2(A+B+C)=\bigg(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}\bigg)}$

$\sf{\implies 2(A+B+C)=\dfrac{25+15+10}{150}}$

$\sf{\implies 2(A+B+C)=\dfrac{50}{150}}$

$\sf{\implies A+B+C=\dfrac{1}{6}}$

Therefore, A, B and C will finish the work in 6 days.

____________________________

(b) From above,

A + B + C = 1/6

$\bold{\dag}$ Work done by A :

$\sf{=A+B+C-(B+C)}$

$\sf{= \dfrac{1}{6}-\dfrac{1}{10}}$

$\sf{=\dfrac{10-6}{60}}$

$\sf{=\dfrac{4}{60}}$

$\sf{=\dfrac{1}{15}}$

So, A can alone do the work in 15 days.

$\bold{\dag}$ Work done by B :

$\sf{=A+B+C-(A+C)}$

$\sf{=\dfrac{1}{6}-\dfrac{1}{15}}$

$\sf{=\dfrac{5-2}{30}}$

$\sf{=\dfrac{3}{30}}$

$\sf{=\dfrac{1}{10}}$

So, B can alone do the work in 10 days.

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♕ Peace out ♕

Answer 2

Step-by-step explanation:

GIVEN:−

A and B can do a piece of work in 6 days.

B and C do the same work in 10 days.

C and A do it in 15 days.

\Large\sf{\underline{\underline{\color{red}{TO\ FIND:-}}}}

TO FIND:−

(a) The number of days will A, B and C finish it working together.

(b) The number of days will A and B finish their working alone.

\Large\sf{\underline{\underline{\color{red}{SOLUTION:-}}}}

SOLUTION:−

\bigstar★ A and B can do a piece of work in 6 days.

→ Work done by A and B in one day = 1/6

→ A + B = 1/6

\bigstar★ B and C do it in 10 days.

→ Work done by B and C in 1 day = 1/10

→ B + C = 1/10

\bigstar★ C and A do it in 15 days.

→ Work done by C and A in 1 day = 1/15

→ C + A = 1/15

____________________________

(a) Work done by A, B and C in one day :

\sf{\implies 2(A+B+C)=\bigg(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}\bigg)}⟹2(A+B+C)=(

6

1

+

10

1

+

15

1

)

\sf{\implies 2(A+B+C)=\dfrac{25+15+10}{150}}⟹2(A+B+C)=

150

25+15+10

\sf{\implies 2(A+B+C)=\dfrac{50}{150}}⟹2(A+B+C)=

150

50

\sf{\implies A+B+C=\dfrac{1}{6}}⟹A+B+C=

6

1

Therefore, A, B and C will finish the work in 6 days.

____________________________

(b) From above,

A + B + C = 1/6

\bold{\dag}† Work done by A :

\sf{=A+B+C-(B+C)}=A+B+C−(B+C)

\sf{= \dfrac{1}{6}-\dfrac{1}{10}}=

6

1

−

10

1

\sf{=\dfrac{10-6}{60}}=

60

10−6

\sf{=\dfrac{4}{60}}=

60

4

\sf{=\dfrac{1}{15}}=

15

1

So, A can alone do the work in 15 days.

\bold{\dag}† Work done by B :

\sf{=A+B+C-(A+C)}=A+B+C−(A+C)

\sf{=\dfrac{1}{6}-\dfrac{1}{15}}=

6

1

−

15

1

\sf{=\dfrac{5-2}{30}}=

30

5−2

\sf{=\dfrac{3}{30}}=

30

3

\sf{=\dfrac{1}{10}}=

10

1

So, B can alone do the w