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.
For B to catch up to A, B's rate MUST be faster than A's rate.
The rate at which B will CATCH UP to A is equal to the DIFFERENCE between the two rates: B-A.
Statement 1: Machine A's rate is twice the difference between the rates of the two machines.
Test TWO cases.
Case 1: A's rate = 2 tablets per hour
Since A's rate is twice the difference between the two rates, B's rate = 3 tablets per hour, so that B-A = 3-2 = 1.
Time for B to catch up = w/(rate difference) = 30/1 = 30 hours.
Work produced by A in 30 hours = r*t = 2*30 = 60 tablets.
Case 2: A's rate = 4 tablets per hour
Since A's rate is twice the difference between the two rates, B's rate = 6 tablets per hour, so that B-A = 6-4 = 2.
Time for B to catch up = w/(rate difference) = 30/2 = 15 hours.
Work produced by A in 15 hours = r*t = 4*15 = 60 tablets.
Since Machine A produces the same number of tablets in each case, SUFFICIENT.
Statement 2: The sum of Machine A's rate and Machine B's rate is five times the difference between the two rates.
Both Case 1 and Case 2 also satisfy statement 2.
Case 1: (B+A)/(B-A) = (3+2)/(3-2) = 5.
Case 2: (B+A)/(B-A) = (6+4)/(6-4) = 5.
Thus -- like statement 1 -- statement 2 is SUFFICIENT.
The correct answer is
D.
An algebraic approach for Statement 1:
Let t = the time for B to catch up.
Work produced by A in t hours = At.
Since B = (3/2)A, work produced by B in t hours = Bt = (3/2)(At).
At the end of the t hours, A's work = B's work.
But BEFORE the t hours begin, A has already produced 30 tablets.
Implication:
During the t hours, A produces 30 fewer tablets than B.
Since A's work in t hours is equal to 30 less than B's work in t hours, we get:
At = (3/2)At - 30
30 = (1/2)(At)
60 = At.
Thus, the work produced by A in t hours = 60 tablets.
SUFFICIENT.
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