Bunuel wrote:
Machines X and Y work at their respective constant rates. How many more hours does it take machine Y, working alone, to fill a production order of a certain size than it takes machine X, working alone?
Let \(x\) and \(y\) be the times needed for machines X and Y respectively working alone to fill a production order of this size.
Question: \(y-x=?\)
(1) Machines X and Y, working together, fill a production order of this size in 2/3 the time that machine X, working alone, does --> general relationship: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{total \ time}\) --> Total time needed for machines X and Y working together is \(total \ time=\frac{xy}{x+y}\) (general formula) --> given \(\frac{xy}{x+y}=x*\frac{2}{3}\) --> \(2x=y\). Not sufficient
(2) Machine Y, working alone, fills a production order of this size in twice the time that machine X, working alone, does --> \(2x=y\), the same info. Not sufficient
(1)+(2) Nothing new. Not Sufficient.
Answer: E.
Hope it helps.
P.S. Please post one question per topic. Also you can attach image files directly so no need to attach zip files.
Pardon me if this is a stupid approach. but still I want to get it clarified.
From 1, as you said, \frac{1}{x}+\frac{1}{y}=\frac{2x}{3}
threfore \frac{1}{y}=\frac{2x}{3}-\frac{1}{x}
this gives \frac{1}{y}=\frac{2x^2-3}{3x}
Now using the value y=2x, substitute in the above equation and u get 2x^2=\frac{9}{4}
hence x=\frac{3}{2}
with this we can even find y.
Hence answer is C.
What is wrong in this approach?