Needgmat wrote: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.
OAE
Say machine A and machine B, each produces x units. Thus, when machine B runs, machine A produces (x - 30) units.
S1: Say the rate of machine A = r, and the rate of machine B = R
Since machine B catches the production of machine A, it implies that R > r.
As per the statement: Machine A's rate is twice the difference between the rates of the two machines, r = 2(R - r); since the difference should be positive, we must take it as (R - r) as R > r.
r = 2(R - r) => R = 3r/2
We know that the time machine A runs equals to that machine B runs.
=> (x-30)/r = x/R
=> (x-30)/r = 2x/3r; replaing the value of R =3r/2
=> 3x-90 = 2x => x =90. Thus machine A produces 90-30 = 60 tablets. A unique answer. Sufficient.
S2: The sum of Machine A's rate and Machine B's rate is five times the difference between the two rates.
=> r+R=5(R-r)
=> R=3r/2; it is the same relationship, we derived for S1, thus S2 itself is also sufficient.
OA:
D
-Jay
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