Question

A and B can do a piece of work in 40 days,B and C in 30 days,and C and A in 24 days.How long will it take them to do the work together?A and B can do a piece of work in 40 days B and C in 30 days and C and a in 24 days how long will it take them to do the work together ​

Answer 1

Answer:

Step-by-step explanation:

Given :

A and B can do a piece of work in 40 days,

B and C can do a piece of work in 30 days &

C and A can do a piece of work in 24 days.

To find :

How long will it take them to do the work together?

Solution :

Statement 1 :

A and B can do a piece of work in 40 days.

_

⇒ Work done by A in 40 days + Work done by B in 40 days = Complete work

Let Work done by A in 1 day be "a"

Let Work done by B in 1 day be "b"

So,

⇒ $40a+40b=1$

⇒ $a+b=\frac{1}{40}$ ...(i)

_

Statement 2 :

B and C can do a piece of work in 30 days.

_

⇒ Work done by B in 30 days + Work done by C in 30 days = Complete work.

Let Work done by C in 1 day be "c"

So,

⇒ $30b+30c=1$

⇒ $b+c=\frac{1}{30}$ ...(ii)

_

Statement 3 :

C and A can do a piece of work in 24 days.

_

⇒ Work done by C in 24 days + Work done by A in 24 days = Complete work

So,

⇒ $24c+24a=1$

⇒ $c+a=\frac{1}{24}$ ...(iii)

_

By adding all 3 equations,.

We get,

⇒ $(a+b)+(b+c)+(c+a)=\frac{1}{40}+ \frac{1}{30}+ \frac{1}{24}$

⇒ $2a+2b+2c = \frac{3}{120} +\frac{4}{120} + \frac{5}{120} = \frac{12}{120}$

⇒ $2a+2b+2c=\frac{12}{120} = \frac{1}{10}$

⇒ $2(a+b+c) = \frac{1}{10}$

⇒ $a+b+c = \frac{1}{20}$

∴ If A , B & C works for 1 day, they cn complete $\frac{1}{20}^{th}$ of the whole work,.

By multiplying both the sides by 20,

We get,

⇒ $20(a+b+c) =20(\frac{1}{20})$

⇒ $20a+20b+20c =1$

∴ It takes 20 days to complete the whole work,.

Answer 2

Answer:

Step-by-step explanation:

Given :

A and B can do a piece of work in 40 days,

B and C can do a piece of work in 30 days &

C and A can do a piece of work in 24 days.

To find :

How long will it take them to do the work together?

Solution :

Statement 1 :

A and B can do a piece of work in 40 days.

_

⇒ Work done by A in 40 days + Work done by B in 40 days = Complete work

Let Work done by A in 1 day be "a"

Let Work done by B in 1 day be "b"

So,

⇒ 40a+40b=140a+40b=1

⇒ a+b=\frac{1}{40}a+b=

40

1

...(i)

_

Statement 2 :

B and C can do a piece of work in 30 days.

_

⇒ Work done by B in 30 days + Work done by C in 30 days = Complete work.

Let Work done by C in 1 day be "c"

So,

⇒ 30b+30c=130b+30c=1

⇒ b+c=\frac{1}{30}b+c=

30

1

...(ii)

_

Statement 3 :

C and A can do a piece of work in 24 days.

_

⇒ Work done by C in 24 days + Work done by A in 24 days = Complete work

So,

⇒ 24c+24a=124c+24a=1

⇒ c+a=\frac{1}{24}c+a=

24

1

...(iii)

_

By adding all 3 equations,.

We get,

⇒ (a+b)+(b+c)+(c+a)=\frac{1}{40}+ \frac{1}{30}+ \frac{1}{24}(a+b)+(b+c)+(c+a)=

40

1

+

30

1

+

24

1

⇒ 2a+2b+2c = \frac{3}{120} +\frac{4}{120} + \frac{5}{120} = \frac{12}{120}2a+2b+2c=

120

3

+

120

4

+

120

5

=

120

12

⇒ 2a+2b+2c=\frac{12}{120} = \frac{1}{10}2a+2b+2c=

120

12

=

10

1

⇒ 2(a+b+c) = \frac{1}{10}2(a+b+c)=

10

1

⇒ a+b+c = \frac{1}{20}a+b+c=

20

1

∴ If A , B & C works for 1 day, they cn complete \frac{1}{20}^{th}

20

1

th

of the whole work,.

By multiplying both the sides by 20,

We get,

⇒ 20(a+b+c) =20(\frac{1}{20})20(a+b+c)=20(

20

1

)

⇒ 20a+20b+20c =120a+20b+20c=1

∴ It takes 20 days to complete the whole work,.