Math, asked by riya246822, 5 months ago

12. A and B can do a work in 8 days, B and C in
12 days, and A and C in 16 days. In what time
could they do it, all working together?​

Answers

Answered by EnchantedGirl
12

Given:-

  • A and B can do a work in 8 days
  • B and C in  12 days
  • A and C in 16 days

To find:-

  • In what time  could they do it, all working together?​

Solution:-

According to question,

(A+B)'s 1 day work = 1/8

(B+C)'s 1 day work = 1/12

(A+C)'s 1 day work = 1/16

Then, 1 day work of (A+B) , (B+C) &(A+C) will be:-

⇒[(A+B)+(B+C)+(A+C)]

⇒1/8  + 1/12 + 1/16

⇒(6 + 4 + 3)/48 = 13/48

And,

⇒[(A+B)+(B+C)+(A+C)]

⇒A+B + B+C + A+C

⇒2(A+B+C)

Therefore,

⇒2(A + B + C)’s 1 day work = 13/48

Then,

(A + B + C)’s 1 day work:-

⇒ 13/48 × 1/2

⇒ 96/13 days

7.384 days.

Hence,

All working together can finish it in 7.384 days.

_______________

Answered by Anonymous
19

Given : -

Given A and B together can do a piece of work in 8 days.

B and C  together can do the same piece of work in 12 days.

A and C together can do the same piece of work in 16 days.

Concept : -

Work done on each day = (total no. of units of work ) / (total no. of days required).

Solution : -

Let the total work be of  48 units (∵ L.C.M of 8 , 12 , 16 )

Then,

Work done by A and B together each day = 48/8 = 6 units per day.

Work done by B and C together each day = 48/12 = 4 units per day.

Work done by A and C together each day = 48/16 = 3 units per day.

Now when all together are working together,

work done on each day =  ( 6 + 4 + 3 )/2 = 13/2 units per day.

No. of days required to complete by all together (D) = [ 48 / ( 13/2 ) ]

  ∴  D = 96/13 days (or) 7.384 days .

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