ajit257 wrote:
If x and y are nonzero integers, is (x-1 + y-1)-1 > [(x-1)(y-1)]-1 ?
(1) x = 2y
(2) x + y > 0
I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.
Apologies again for giving out too much.
First of all the question should be:
If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > (x^(-1)*y^(-1))^(-1) ?Is
(x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}? --> is
(\frac{1}{x}+\frac{1}{y})^{-1}>(\frac{1}{xy})^{-1}? --> is
(\frac{x+y}{xy})^{-1}>xy --> is
\frac{xy}{x+y}>xy?
Now, from this point you can not divide both parts of the inequality by
xy and write
\frac{1}{x+y}>1 (as you did), because you don't know whether
xy is positive or negative: if
xy>0 then you should write
\frac{1}{x+y}>1 BUT if
xy<0 then you should flip the sign and write
\frac{1}{x+y}<1. But even if you knew that
xy>0 then the next step of writing
x+y<1 from
\frac{1}{x+y}>1 would still be incorrect for the same exact reason: you don't k now whether
x+y is positive or negative, hence you can not muliply both sides of the inequality by
x+y.
Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.Thus the question is boiled down to: is
\frac{xy}{x+y}>xy? Actually we can manipulate further but there is no need.
(1) x = 2y --> question becomes: is
\frac{2y^2}{3y}>2y^2? Now, as we know that
y is
nonzero then
2y^2>0 and we can divide both parts by it --> is
\frac{1}{3y}>1? As
y is an
integer (no matter positive or negative) then the answer to this question is always NO (if it's a positive integer then
\frac{1}{3y}<1 and if it's a negative integer then again:
\frac{1}{3y}<0<1). Sufficient.
(2) x + y > 0 --> if
x=y=1 then the answer will be NO but if
x=3 and
y=-1 then the answer will be YES. Not sufficient.
Answer: A.
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33% (02:44) correct
