HPengineer wrote:If rs does not = 0, is 1/r + 1/s = 4

S1. r + s = 4rs

S2. r = s


My issue here is with statement two...

if r = s then why can i not insert this into the equation in the stem and re write as 1/r + 1/r = 4 solve for r and be done with it?

The question is asking whether (1/r + 1/s) = 4 or not.
If you replace r = s from the statement 2 into the equation, obviously you'll get a value of r which satisfies the equation. The solutions are in fact r = s = 1/2. But the question is not really asking you to do this! For this particular set of values of r and s the relation satisfies. But not for all r and s that satisfies the relation r = s. For instance, r = s = 1 does not satisfies the equation. Thus statement 2 is not sufficient to answer the question.

But for statement 1, (r + s) = 4rs => (r + s)/rs = 4 => (1/r + 1/s) = 4.
Therefore for any set of values of r and s that satisfies the condition given in statement 1, satisfies the equation given in question stem. Hence statement 1 is sufficient.

Hope that clears your confusion.