Machines A and B each produce tablets

[DavidC696]

Machines A and B each produce tablets at their respective constant rates. Machine A has produced 30 tablets when Machine B is turned on. Both machines continue to run until Machine B’s total production catches up to Machine A’s total production. How many tablets does Machine A produce in the time that it takes Machine B to catch up?

(1) Machine A’s rate is twice the difference between the rates of the two machines.

(2) The sum of Machine A’s rate and Machine B’s rate is five times the difference between the two rates.

Each statement alone would allow Machine A to be faster than Machine B, however combined they tell you that is not the case;

If you solve it out with

Time = T
Machine A = Rate A
Machine B = Rate B

Statement one could mean A = 2(A-B) or A = 2(B-A); thus A = 2B or A = 2/3B and therefore 1 alone should not be enough
Statement two could mean A+B = 5(A-B) or A+B = 5(B-A); thus A =3/2B or A = 2/3B and therefore 2 alone is not enough

Combined they are enough because they allow you to triangulate A = 2/3B. Is that not correct? The test shows that each alone is enough.

[RonPurewal]

the problem statement implies that machine B is faster than machine A (otherwise B could never "catch up"; it would just fall farther and farther behind the head start that A already has).

algebraically, each statement gives two possibilities:
• rate A = (2/3)(rate B)
and
• another possibility in which rate A > rate B.
in each case, though, the second possibility can be rejected, because we know that machine B is the faster one. once these alternatives are rejected, you're left with rate A = (2/3)(rate B) as the sole possible ratio for each individual statement.