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If m is not equal to k, is |x^2 - y^2| > 0? 13 Feb 2016, 11:12
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C
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75% (hard)

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50% (02:20) correct 50% (02:07) wrong based on 135 sessions

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If \(m\neq{k}\), is \(|x^2 - y^2| > 0\)?

(1) \((x+m)^2 = (y+m)^2\)
(2) \((x+k)^2 = (y+k)^2\)
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C
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If m is not equal to k, is |x^2 - y^2| > 0? 13 Feb 2016, 23:55
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TeamGMATIFY
If \(m\neq{k}\), is \(|x^2 - y^2| > 0\)?

(1) \((x+m)^2 = (y+m)^2\)
(2) \((x+k)^2 = (y+k)^2\)
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Hi,
the Q should make it to 600 level Qs..

lets see the info from Q stem


\(|x^2 - y^2| > 0\)...
|x^2 - y^2| is always positive so will always be equal to or greater than 0..
here if we can say that x and y are not of same numeric value irrespective of sign, the answer will be YES..
and If their mod are same, answer is NO as it will be equal to 0..

lets check the statements



(1) \((x+m)^2 = (y+m)^2\)
\(x^2 + 2xm + m^2 = y^2 + 2ym + m^2\)..
\(x^2-y^2+2xm-2ym=0\)..
\((x-y)(x+y)+2m(x-y)=0\)..
\((x+y+2m)(x-y)=0\)..
either x=y, or x+y=-2m...
so either x=y or both may not be equal..
insuff..


(2) \((x+k)^2 = (y+k)^2\)
same as (1)..
\((x+y+2k)(x-y)=0\)..
either x=y, or x+y=-2k...
so either x=y or both may not be equal..
insuff..

Combined..
we have x+y= -2m=-2k, BUT \(m\neq{k}\)..
so only possiblity common in two statements is x=y..
so ans is NO..
SUFF
C
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If m is not equal to k, is |x^2 - y^2| > 0? 15 Feb 2016, 11:23
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TeamGMATIFY
If \(m\neq{k}\), is \(|x^2 - y^2| > 0\)?

(1) \((x+m)^2 = (y+m)^2\)
(2) \((x+k)^2 = (y+k)^2\)
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I have a small doubt here

can we write (X+M)^2 = IX+MI or distance between X & M.... :?
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If m is not equal to k, is |x^2 - y^2| > 0? 15 Feb 2016, 11:48
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rohit8865
TeamGMATIFY
If \(m\neq{k}\), is \(|x^2 - y^2| > 0\)?

(1) \((x+m)^2 = (y+m)^2\)
(2) \((x+k)^2 = (y+k)^2\)

I have a small doubt here

can we write (X+M)^2 = IX+MI or distance between X & M.... :?
Show more

Not unless x=m=0 or any other similar condition. The only thing common (generally) between (x+m)^2 and |x+m| is that both the expressions will yield a non negative value.

Hope this helps.
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If m is not equal to k, is |x^2 - y^2| > 0? 16 Feb 2016, 02:09
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If m≠k , is |x ^2 −y ^2 |>0 ?

(1) (x+m)^2 =(y+m) ^2
(2) (x+k) ^2 =(y+k) ^2


When you modify the original condition and the question, they become x^2=/y^2?. Then, there are 4 variables(m,k,x,y), which should match with the number of equations. So you need 4 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer.
When 1) & 2), they become 1) x+m=y+m, x+m=-y-m, x+k=y+k, x+k=-y-k. That is, from x=y, x+y=-2m=-2k, m is not k, which only satisfies x=y. Then, the question is no to x^2=/y^2?, which is sufficient. Therefore, the answer is C.


 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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If m is not equal to k, is |x^2 - y^2| > 0? 25 May 2021, 00:48
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