Bunuel
Official Solution:
Machine A and B working together at their constant rates can complete a certain task in 6 days. In how many days, working alone, can machine A complete the task?
Let A and B be the times needed for machines A and B to complete the task working alone, respectively. Thus, we have \(\frac{1}{A}+\frac{1}{B}=\frac{1}{6}\).
(1) The average (arithmetic mean) of the respective times A and B would each take to complete the task working alone is 12.5 days.
This implies that \(A+B=2*12.5=25\). However, since we do not know which machine is faster (we cannot differentiate between A and B), even if we substitute B with \(25-A\) into \(\frac{1}{A}+\frac{1}{B}=\frac{1}{6}\) to get \(\frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}\) and solve, we must get two different answers for A and B: \(A \lt B\) and \(A \gt B\). Not sufficient.
(2) It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task.
This implies that \(A=B+5\). Substituting this into the equation, we get \(\frac{1}{A}+\frac{1}{A-5}=\frac{1}{6}\). We can solve for \(A\) to get \(A=2\) or \(A=15\). However, \(A=2\) cannot be true since it would make \(B\) negative, so \(A = 15\) is the only valid solution. Sufficient.
Answer: B
while I did answer B.
I couldn't get to solve \(\frac{1}{A}+\frac{1}{A-5}=\frac{ 1}{6}\)
could you please explain how did we get A=2, A= 15? I understand that we will get a quadratic.
but the one I got is: \(A^2-5A-6=0 \)