A can do 3 dozens per hour and b can do 4 dozens per hour. At any time atleast one person is working. Both worked how many hours simultaneously to guarantee that 77 dozens are done in 14 hours?

ern5231,

I'm a little confused with the question, seems like there is something missing with regards to maximizing/minimizing the number of hours both A&B work together. In any case, we know that if A&B worked together for 11 hours, they would do 77 dozen, but since the question is asking to guarantee that the work be done in 14 hours, we know that somewhere in the process, from start to finish, either A or B will not work for part of it. So, here's my attempt at it:

Let the "A" be the number of hours A works alone, "B" be the number of hours B works alone, and "X" be the number of hours both A & B work together. Per the question there are no gaps while the work is being done, i.e., someone is always working. So, we can rephrase the question to read,

3A+4B+(3+4)X=77 dozen
3A+4B+7X=77.....(i)

From the question we also know that A+B+X=14 hours. Since A is the slower worker of the two, let's assume that A does not do any work by himself, but instead joins B at some point in time (still agrees with the question since "at any time atleast one person is working"). I am assuming so, because I am trying to solve for how long A&B have to work for at the least to guarantee that the work gets done in 14 hrs. So, I'm ignoring the slowest worker. Therefore, A works 0 hours by himself. So, we can rephrase the above equation as:

B+X=14 --> B=14-X.....(ii)

Combining the two equations,
3A+4(14-X)+7x=77
56-4X+7X=77 (as I assumed above, A will work 0 hours by himself)
3X=21
X=7

Therefore, A&B need to work for 7 hours at least to guarantee that the work gets done in 14 hours.

Hope that helped - like I said, I'm just taking a shot at it. By the way, what is the source for this question, also what's the official answer?

-Ash

First of all it is not the traditional work rate problem. It is more of solving the eqns.

in one hour both A and B can produce 7 dozens working together

7x=77=>x=11 but since work happened for 14 hrs therefore x cannot be 11

the general eqn would be
7x+3y+4z=77
where x+y+z=14
also x<>0 and x<11

7x+3y+4(14-x-y)=77
3x-y=21
x=7,y=0,z=7 is one soln
x=8,y=3,z=3
x=9,y=6,z=-1 not a soln
thus (7,0,7) and (8,3,3) are two possible soln

thus A and B have to work atleast 7 hrs to ensure that 77 dozens are produced in 14 hrs

Yes.

They must work 7 hours minimum simultaneously to guarentee the work mentioned to be done in 14 hours

ern5231 wrote:A can do 3 dozens per hour and b can do 4 dozens per hour. At any time atleast one person is working. Both worked how many hours simultaneously to guarantee that 77 dozens are done in 14 hours?

Let A be the time A works alone. And B be the time B works alone. Then each would have done 3A+4B amount of work working individually. But we are told the total amount of work is 77. So 77-(3A+4B) is amount they did together. When they work together we add their rates.
77-(3A+4B)/7 should give the time they work together. This time cannot be 14. It will be remainder after their individual times are subtracted.

PUtting all together

(77-(3A+4B))/7 =14-A-B

4A-3B=21

4(6)-3(1)=21

So A works 6 hours alone. B works 1 Hour alone They work 7 hours together.