Question

Ab can do the work in 12 days B C in 15 days and AC in 20 days . in how many days they will finish the work together and separatley​

Answer 1

Solution :-

Given :
A and B and can do a piece of work in 12 days.
B and C do it in 15 days.
C and A do it in 20 days.

(A + B)'s one day work = 1/12
(B + C)'s one day work = 1/15
(C + A)'s one day work = 1/20

Find the (A + B + C)'s one day work :-

(A + B) + (B + C) + (C + A) = 1/12 + 1/15 + 1/20
=> 2(A + B + C) = (5 + 4 + 3)/60
=> A + B + C = 12/(30 × 2) = 1/5

A's one day work = (A + B + C) - ( B + C)
= 1/5 - 1/15
= (3 - 1)/15
= 2/15

B's one day work = (A + B + C) - (A + C)
= 1/5 - 1/20
= (4 - 1)/20
= 3/20

C's one day work = (A + B + C) - (A + B )
= 1/5 - 1/12
= (12 - 5)/60
= 7/60

Hence,
A, B and C can do the work in 5 days.
A, B and C finish the work alone in 7½ days , 6⅔ days and $8 \frac{4}{7}$ days respectively.

Answer 2

Step-by-step explanation:

Implementing given statement in terms of equations,we get,

A + B=1/12; as both do 1/12 work in 12 days. similarly,

B + C=1/15;

A + C=1/20;

now adding above equations we get,

==>2(A + B +C)=1/12 + 1/15 +1/20;

==>2(A + B +C)=(10+8+6)/120; as, lcm= 120,

==>2(A + B +C)=24/120;

==>(A + B +C)=24/240;

==>(A + B +C)=1/10;

hence they all work in 10 days to complete.