Gmatboat wrote:
Fighter1095 wrote:
If printing press A, running alone at its constant rate, produces 27,000 newspapers in six hours, how long does it take printing press B, running alone at its constant rate, to produce 27,000 newspapers?
So, r(A) = (27k/6)
Let t be the time which we need to find
r(B) = (27k/t).
Statement1:
3r(A) = 2r(B)If we substitute the above statements,
We get the value of t --> Sufficient(A).
Hence, D
Hi
Fighter1095, great logic! It was helpful. Anyway, there is a small error in the highlighted portion. It should actually be 2r(A) = 3r(B), because the rate of "A" is "faster" than the rate of "B", so "B" produces 2/3 of what "A" produces.
Dear
Gmatboatthe quibble is what the highlighted portion represents
What if it is the time? then the initial expression is utterly valid.
3r(A) = 2r(B) = > Time B = 1.5 Time A, substitute Time B = 1.5 *6h = 9h
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CEdward wrote:
Not sure if this was the most efficient way to think about this. I struggle with math and it took me 15 minutes to get this even though it is very straightforward. But I am challenging myself to think things through rather than look at the answers.
Statement 1 - Sufficient
RateA = 27000 newspapers/6 hours = 4500 papers/1hr
We now know that RateB is 3/2x greater than RateA.
3/2 x 4500papers/hr = 6750 papers/hr
Statement 2 - Sufficient
TotalWork = (RateA x 1hr) + (RateB x 1hr)
7500 = 4500papers/hr + RateB
RateB = 7500papers/4500 papers/hr
Dear
CEdwardI presume that you meant RateB = \(\frac{2}{3}\)RateA
Then \(\frac{2}{3}\)*\(\frac{27000}{6}\) =\(\frac{3000}{h}\)
Thus, B time: 27000 = 3000*T = > 9h
First time I had saw the question, I made unnecessary computation. But when you realize that it is DS question and that you can form the equation with 1 variable, you do not need to calculate anything. That is valid for the 1st and 2nd statements.
Hope it helps