1/b+1/s = 1/12
1/b=2/s
2b=s
we can say
3b/2b^2 = 1/12
solve
we get b= 18
so s = 36
IMO B

Working together at their respective constant rates, Bob and Sam can mow a lawn in 12 hours

Bob's work rate is \(\frac{1 (job)}{Bob (hours)}\)
Sam's work rate is \(\frac{1 (job)}{Sam (hours)}\)
Their combined work rate is \(\frac{1 (job)}{Bob (hours)} + \frac{1 (job)}{Sam (hours)} = \frac{1 (job)}{12 (hours)}\)
(see: https://gmatclub.com/forum/combined-rat ... l#p1237427)
Simplify by eliminating units \(\frac{1}{B} + \frac{1}{S}= \frac{1}{12}\)

If Bob’s rate is twice Sam’s rate
So \(\frac{1}{B} = \frac{2}{S}\) (completes two jobs in the time it takes S to complete 1)

New equation
\(\frac{2}{S} + \frac{1}{S} = \frac{1}{12}\)
\(\frac{3}{S}=\frac{1}{12}\)
Cross multiply
S=36

how many hours would it take Bob, working alone, to mow the lawn?
\(\frac{1}{B}+\frac{1}{36}=\frac{1}{12}\)
\(\frac{36}{36B} + \frac{B}{36B}=\frac{1}{12}\)
\(\frac{36+B}{36B}=\frac{1}{12}\)
Cross multiply
12(36+B)=36B
36+B=3B
36=2B
18=B

and answer is B

B+S t is = 12 to complete Work = 1, so their combined rate is 1/12
B's rate is twice that of S, so B=2r, S=r (note this means B does twice as much work as S for t=1)
combined rates: 1/12 = 3r ---> 1/36 = r
B working alone is 2(1/36) = 1/18 rate, so Work(1) / 1/18 ---> t = 18

I always put on the top of my sheet the formula W=R*t

then I say:

Names-------W-------------R------------t

B+S---------1--------------1/12---------12h

B------------1--------------2x------------?

S------------1---------------x------------?

then, as we can sum up rates, 2x + x = 1/12 ---> x= 1/36
thus, B's rate = 1/18. Therefore, reversing the rate I will obtain the time. 18h -->(B)

in 12 hours Bob mows 2/3 lawn
12/(2/3)=18 hours for Bob to mow lawn alone.
18
B

Note: Bob’s rate is twice Sam’s rate.

Thus if bob needs x days to complete a work , sam needs 2x days to do so.

1/x + 1/2x = 1/12

3/2x = 1/12

x = 18.

B is the correct answer.