Question

A and B can do a piece of work in 30 days.B and C can do it in 37.5 days.C and A can do it in 50 days.How many days will they finish,if A,B,C work together​

Answer 1

Answer:

Let the number of days taken by A, B & C to complete the work individually be a, b & c respectively.

Given, A and B can do a piece of work in 30 days

=> 1/a + 1/b = 1/30

Given, B and C can do the same work 24 days

=> 1/b + 1/c = 1/24

Given, A and C can do the same work in 20 days

=> 1/c + 1/a = 1/20

Adding all the three equations,

=> 2(1/a +1/b + 1/c) = 1/30 + 1/24 + 1/20 = (20+25+30)/600 = 75/600 = 1/8

=> 1/a + 1/b + 1/c = 1/16

=> When A, B & C work together, the amount of work done in one day is 1/16

(1/a + 1/b + 1/c) - (1/b + 1/c) = 1/16 - 1/24 = (3–2)/48 = 1/48

=> 1/a = 1/48

=> The work done by A individually in one day = 1/48

Given that A, B & C work together for 10 days,

=> Work done by them together for 10 days = 10 * (1/16) = 10/16 = 5/8

=> The remaining work to be done by A alone = 1–5/8 = 3/8

The number of days taken by A alone to complete the remaining work = (3/8)/(1/48) = 18 days

Answer 2

Given:

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

B and C can do it in 37.5 days.

C and A can do it in 50 days.

To Find:

No. of days taken by A,B,C work together​?

Step-by-step explanation:

• Given  A and B can do piece of work in 30 days.

        work done by A and B in 1 day = $\frac{1}{30}$

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

        work done by B and C in 1 day = $\frac{1}{37.5} =\frac{2}{75}$

• Given  A and C can do piece of work in 30 days.

        work done by A and C in 1 day = $\frac{1}{50}$

• Now add all three

       Work done by 2(A+B+C) in 1 day=

                 $\frac{1}{30}+\frac{2}{75}+\frac{1}{75}\\=\frac{12}{150}$

• Work done by A+B+C in 1 day=

                $=\frac{1}{2}\times\frac{12}{150}\\=\frac{1}{25}$

                 

     Thus A+B+C can finish work in 25 days.

Hence, A+B+C can finish work in 25 days