Question

A and B together can do a work
in 12 days. B and C together do it
in 15 days. if A is twice
efficient as C , then the number
of days for B alone to finish
the work is
plz give me correct answer
Hrithik plz answer my question​

Answer 1

Step-by-step explanation:

Let the number of days that the work would take for A, B, and C, if each worked alone, be a, b, and c, respectively. We are given the following equations:

1/a + 1/b = 1/12, 1/b + 1/c = 1/15, c = 2a. Substituting c = 2a into the 2nd equation,

1/b + 1/2a =1/15.

Subtracting this new equation from the 1st equation, we get

(1/a + 1/b) - (1/b + 1/2a) = 1/12 - 1/15 or 1/a - 1/2a = 1/2a = (5 - 4)/60 = 1/60.

Therefore 2a = 60 or a = 30, and c = 2a = 60. Finally, from the 1st equation,

1/30 + 1/b = 1/12 or 2/60 + 1/b = 5/60 => 1/b = (5 - 2)/60 = 3/60 or b = 60/3 = 20.

B, working alone, will need 20 days to complete the work. Also, A and C, each working alone, will need 30 and 60 days, respectively. 

Hope This will Helps You

Answer 2

Answer:

Let A,B&C can complete the work in x,y&z days respectively, then

A's one day work=1/x

B's one day work=1/y

C's one day work=1/z

1/x+1/y=1/12…(1)

1/y+1/z=1/15…(2)

1/x=2/z…(3) putting in(1)

2/z+1/y=1/15…(4)

(4)-(1) gives

1/z=1/12–1/15

1/z=(5–4)/60

1/z . =1/60

z=60 days

From (3),

1/x=2/60

1/x=1/30

x=30days

From(1),

1/30+1/y=1/12

1/y=1/12–1/30

1/y=(5–2)/60

1/y=3/60

1/y=1/20

y=20 days