Classic problem of 2 machines working together. We'll need the basic Work formula (Rate * Time = Work) and also keep in mind that when working together, Rates are additive!

If we want A to operate the least amount of hours, then B is going to have to operate the whole time, so B's share of production in 8 hrs is:

W = R * T
W = 700 cans/hr * 8 hr = 5,600 cans

That means we need machine A to pick up the slack for 12,000 - 5,600 = 6,400 cans.

So how long does A have to operate to make 6,400 cans?

W = R * T
6,400 cans = 1000 cans/hr * T
[spoiler]T = 6.4 hrs
E[/spoiler]


You could also do this in 1 step/1 equation, but maybe it's a bit messier to set up?
W (total) = R(A) * T(A) + R(B) * T(B)
12,000 = 1,000 * T(A) + 700 * 8
Same result, [spoiler]T = 6.4; E[/spoiler]

ritumaheshwari02 wrote:Operating alone at their respective constant rates, machine A produces 1,000 cans each hour and machine B produces 700 cans each hour. If the two machines together must produce 12,000 cans within a certain 8-hour period, what is the least number of hours that machine A must operate?

A. 4.8
B. 5.2
C. 5.6
D. 6
E. 6.4

Back solve.....

In Which of the following type question....Always start with D/E choice.

Lets select D

in 6 Hrs A will produce 6000 Cans
B can work a max of 8Hrs. So B can produce 8*700=5600 Cans

Total cans produced = 6000+5600 = 11600
Since this is less than 12000
A has to work more...

The only option more than 6Hrs is

E