Can someone advise on this? Absolute signs definitely not my

Question

Can someone advise on this? Absolute signs definitely not my fav...

Thanks. IF |X-2|=|Y-2|, X+Y=? 1. X is not equal to Y 2. X>2 AND Y<2.

gayathri · Answer

I am confused with this one

Stem: |x-2|=|y-2| => (x-2)^2 =(y-2)^2 x^2-4x+4 = y^2-4y+4 x^2 - y^2 = 4(x-y) (x+y)(x-y) = 4(x-y) => x+y = 4 So, got the value of x+y with just stem and not using the statements!

LRod · Answer

note to self - if stumped by absolutes - square them away!
good response.

thinker · Answer

Hi karovd,
According to the question:|X-2|=|Y-2|, we can get these four results:
(A) 2-X=Y-2 --> X+Y=4
(B) X-2=2-Y --> X+Y=4
(C) X-2=Y-2 --> X=Y
(D) 2-X=2-Y --> X=Y

Statement(1) tells us  X is not equal to Y , so we can eliminate (C) and (D), and we get (A) and (B), which is same result. Therefore, (1) is sufficient.
Statement(2) gives us obvious condition, and (B) fits that, so (2) is sufficient as well.

Hence, the answer should be D.

twixt · Answer

I pick A

Statement 1 :

X<>Y which means that -(X-2)=(Y-2) id est X+Y=4 (symmetrical)

Statement 2 :

As X and Y are both <2 it is not possible to reach X-2 >0 or Y-2 <0
So X=Y
question being value of X+Y i can not conclude...

karovd · Answer

Thanks to all. Thinker's solution is very clear, OA is D. Cheers. :lol:

twixt · Answer

Could you please guys develop the obvious condition froom statement 2 I do not understand how it helps to assess the value of X+Y...

artabro · Answer

Twixt: if x> 2 and y <2, this means that x (=2,3, 3.14159, 3.5, etc) canâ€™t never be equal to y (=-3, -4, -5.5, etc), so only x+y=4 is valid. The other option was x=y, which 2 denies.

ruhi · Answer

Guys, my two cents.

I did this prob in three simple steps.

stem: |x-2|=|y-2|
|x-2|>= |x|-|2| and |y-2|=|y|-|2|
therefore, |x|-|2|=|y|-|2|
or |x|=|y|....... condition 1

statement 1: x not equal to y. no prob. x=-1, y=1 . then condition 1 is satisfied. take x=2, y=-2, condition 1 is still satisfied. etc etc
hence, state. 1 is suff.

statement 2: x>2 and y<2. Let x=3 and y=-3, then also statement 1 is satisfied. hence, state. 2 is suff.

Hence, D.

ruhi · Answer

forgot to add, we get x+y=0 . since mod x =mod y. 

there is a typo in my previous post, |y-2|>=|y|-|2|

forumsmba · Answer

Hmm, I solved this one Gayathri's way and was wondering what I did wrong. 

After looking at another problem in the forum, I realized that whenever there is a quadratic equation, it is better to keep it in that form in DS, even though it might simplify into a linear equation.

As for this question (following Gayathri's solution):
(x-y)(x+y) = 4(x-y)

Here, the first instinct is to jump to x+y=4, However, lets substitute y=1:
(x-1)(x+1)=4(x-1)
x^2-4x+3=0
(x-1)(x-3)=0

Hence, there are 2 solutions x+y=2 & x+y=4

So, we need to look at A & B to solve the problem:
A) x not equal to y
 - from above we can then have one solution x+y=4;
B) x>2 , y<2
 - again from above x+y=4 is the only solution

-fm

PS: Lesson learnt: The reason a quadratic eqn that can be simplified to a linear equation, can still have 2 possible solutions instead of 1, is because we might be cancelling a term on both sides of "=" under the assumption "NOT EQUAL to 0"
(x-1)(x-3)=y(x-1)
you can cancel (x-1) only if you know x is not equal to 1!!!

gayathri · Answer

Well put forumsmba. That's the same mistakke I did in the other DS problem as well

Can someone advise on this? Absolute signs definitely not my

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