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Is x^2+2x+1=y^2+2y+1?

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Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 03:57
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A
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Difficulty:

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Question Stats:

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Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0


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Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post Updated on: 06 Aug 2018, 04:38
1
chetan2u wrote:
Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0


New question!!!..
kudos to correct answer


+1 for D
Given:
\(x^2+2x+1=y^2+2y+1 ?\)
\((x+1)^2=(y+1)^2 ?\)
\(x+1=y+1 ?\)
\(x=y ?\)

We need to find if x=y.

Statement 1:
\(x\neq{y}\)
Hence sufficient
Statement 2:
x>y>0
Hence Sufficient
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Originally posted by Akash720 on 06 Aug 2018, 04:24.
Last edited by Akash720 on 06 Aug 2018, 04:38, edited 1 time in total.
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Re: Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 04:35
1
(1) \(x ≠ y\)

let: x = 3 and y = 4

so, \(x^2 + 2x + 1 ≠ y^2 + 2y + 1\)

now, let x = 0 and y = -2

so, \(x^2 + 2x + 1 = y^2 + 2y + 1\)

Insufficient.

(2) x>y>0

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

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

But, x>y>0

So, \(x^2 + 2x + 1 ≠ y^2 + 2y + 1\)

Sufficient.

Option B
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Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 04:43
2
chetan2u wrote:
Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0


New question!!!..
kudos to correct answer


Rephrasing question stem:- Is \((x+1)^2=(y+1)^2\) ?
The statement that says either \((x+1)^2\neq{(y+1)^2}\) or \((x+1)^2={(y+1)^2}\)would be sufficient.

St1:- \(x\neq{y}\)
Here we have three cases:
a) If x>y, then \((x+1)^2<(y+1)^2\) (say x=-2 and y=-3)
b) If x>y, then \((x+1)^2>(y+1)^2\) (say x=3 and y=2)
c) If x>y, then \((x+1)^2=(y+1)^2\) (say x=0, y=-2)
d)If x<y, then \((x+1)^2>(y+1)^2\) (say x=-3 and y=-2)
e)If x<y, then \((x+1)^2=(y+1)^2\) (say x=-2 and y=0)

Insufficient.

St2:- x>y>0 or x and y are positive numbers with x>y.

So, \((x+1)^2>(y+1)^2\) will hold true.
Sufficient.

Ans. (B)
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Re: Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 05:15
1
chetan2u wrote:
Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0


New question!!!..
kudos to correct answer


OA: B

Is \(x^2+2x+1=y^2+2y+1\)?
reducing the question \(x^2+2x+1-y^2-2y-1=0\)
\(x^2-y^2+ 2x-2y=0\)
\((x-y)(x+y)+2(x-y)=0\)
Question becomes Is\((x-y)(x+y+2)=0\)?

Statement 1 : \(x\neq{y}\)
As \({x-y}\neq{0}\), Questions become Is \((x+y+2)=0\)?
Now plugging in values ,

First taking \(x=-2,y=0\)

L.H.S \(: (x+y+2)=-2+0+2=0\)
R.H.S \(=0\)
L.H.S\(=\)R.H.S
Answer to Is \((x+y+2)=0\)? becomes yes

Now taking \(x=2,y=3\)

L.H.S \(: (x+y+2) = 2+3+2 =7\)
R.H.S\(= 0\)
\({L.H.S}\neq{R.H.S}\)
Answer to Is \((x+y+2)=0\)? becomes No

Statement 1 alone is not sufficient.

Statement 2 : \(x>y>0\)

This means \((x-y)\) =some +ve number, \(x+y+2\) will also be some +ve number.
So \((x-y)(x+y+2)>0\)
Answer to question Is \((x-y)(x+y+2)=0\)? is always No
Statement 2 alone is sufficient.
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Re: Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 05:19
Is \(x^2+2x+1=y^2+2y+1\) , \((x+1)^2=(y+1)^2\) , \((x+1)^2-(y+1)^2=0\) , \((x+1-y-1)(x+1+y+1)=0\) , \((x-y)(x+y+2)=0\), x=y or x+y=-2

(1) \(x\neq{y}\), but we don't know whether x+y=-2 - Not Suff
(2) x>y>0 - means that \(x\neq{y}\) and \(x+y>0\) - Suff

Answer B.
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Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 08:12
1
chetan2u wrote:
Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0



\(x^2+2x+1=y^2+2y+1\)
\((x+1)^2=(y+1)^2\)
\(|x + 1|=|y + 1|\)

2 cases to open modulus

\(x+1=y+1\)..........x=y

\(x+1=-y-1\).........x+y=-2

The question becomes:

x=y or x+y=-2

(1) \(x\neq{y}\)

It could be that x + y =-2

Let x =1 & y =-3............Answer is Yes

or

Any any other values

Let x =1 & y =-2............Answer is No

Insufficient

\((2) x>y>0\)

It is clear that \(x\neq{y}\) and psotive numbers can't produce negative number (-2). Therefore it is certain answer with NO.

Answer: B
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Re: Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 09:32
PKN wrote:
chetan2u wrote:
Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0


New question!!!..
kudos to correct answer


Rephrasing question stem:- Is \((x+1)^2=(y+1)^2\) ?
The statement that says either \((x+1)^2\neq{(y+1)^2}\) or \((x+1)^2={(y+1)^2}\)would be sufficient.

St1:- \(x\neq{y}\)
Here we have three cases:
a) If x>y, then \((x+1)^2<(y+1)^2\) (say x=-2 and y=-3)
b) If x>y, then \((x+1)^2>(y+1)^2\) (say x=3 and y=2)
c) If x>y, then \((x+1)^2=(y+1)^2\) (say x=0, y=-2)
d)If x<y, then \((x+1)^2>(y+1)^2\) (say x=-3 and y=-2)
e)If x<y, then \((x+1)^2=(y+1)^2\) (say x=-2 and y=0)

Insufficient.

St2:- x>y>0 or x and y are positive numbers with x>y.

So, \((x+1)^2>(y+1)^2\) will hold true.
Sufficient.

Ans. (B)



hey PKN can you please explain
how from here \(x^2+2x+1=y^2+2y+1\) you got this \((x+1)^2=(y+1)^2\) ?
thank you :-)
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Re: Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 09:38
1
dave13 wrote:
PKN wrote:
chetan2u wrote:
Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0


New question!!!..
kudos to correct answer


Rephrasing question stem:- Is \((x+1)^2=(y+1)^2\) ?
The statement that says either \((x+1)^2\neq{(y+1)^2}\) or \((x+1)^2={(y+1)^2}\)would be sufficient.

St1:- \(x\neq{y}\)
Here we have three cases:
a) If x>y, then \((x+1)^2<(y+1)^2\) (say x=-2 and y=-3)
b) If x>y, then \((x+1)^2>(y+1)^2\) (say x=3 and y=2)
c) If x>y, then \((x+1)^2=(y+1)^2\) (say x=0, y=-2)
d)If x<y, then \((x+1)^2>(y+1)^2\) (say x=-3 and y=-2)
e)If x<y, then \((x+1)^2=(y+1)^2\) (say x=-2 and y=0)

Insufficient.

St2:- x>y>0 or x and y are positive numbers with x>y.

So, \((x+1)^2>(y+1)^2\) will hold true.
Sufficient.

Ans. (B)



hey PKN can you please explain
how from here \(x^2+2x+1=y^2+2y+1\) you got this \((x+1)^2=(y+1)^2\) ?
thank you :-)


I hope you know the basic algebra formula:-

\((a+b)^2=a^2+2ab+b^2\)

Similarly, \((x+1)^2=x^2+2*x*1+1^2=x^2+2x+1\)

Hope it helps.
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Re: Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 09:57
1
1
1
It's (B).

Be careful when you go from here:

\((x+1)^2=(y+1)^2\)?

to here:

\(x+1=y+1\)?

It actually isn't quite that simple. For instance, if \(a^2 = b^2\), that doesn't mean that \(a = b\). a could be -2 and b could be positive 2.

It actually simplifies like this:

\(|x + 1| = |y + 1|\)?

Which can be true in two situations: if x and y are equal, or if one or both of them is negative (for instance, x = -1.5 and y = -0.5, or x = -10 and y = 8.)

If they're both positive, and they aren't equal, you know the answer to the question is 'no'.
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Re: Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 06 Aug 2018, 20:25
we can simplify the question as
|x+1|=|y+1|
Statemment1
consider x=-2 and y =0, then we have |x+1|=|y+1| = 1
Consider x=1 and y =2 , then we have |x+1|≠|y+1|
Hence insufficient
Statement 2
X>Y>0, we can deduce that both are positive. And hence we can never have |x+1|=|y+1| for any values of x and y
Hence sufficient
Therefore IMO the answer should be B
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Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 07 Aug 2018, 02:23
Mo2men wrote:
chetan2u wrote:
Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0



\(x^2+2x+1=y^2+2y+1\)
\((x+1)^2=(y+1)^2\)
\(|x + 1|=|y + 1|\)

2 cases to open modulus

\(x+1=y+1\)..........x=y

\(x+1=-y-1\).........x+y=-2

The question becomes:

x=y or x+y=-2

(1) \(x\neq{y}\)

It could be that x + y =-2

Let x =1 & y =-3............Answer is Yes

or

Any any other values

Let x =1 & y =-2............Answer is No

Insufficient

\((2) x>y>0\)

It is clear that \(x\neq{y}\) and psotive numbers can't produce negative number (-2). Therefore it is certain answer with NO.

Answer: B



hi Mo2men

can you please explain how you opened modulus in second case
\(|x + 1|=|y + 1|\)

\(x+1=-y-1\).........x+y=-2

i thought we need to open moduls on both sides like this \(-x-1=-y-1\) why it is not correct opening ? :) :?

pushpitkc maybe you can help with this :-)

thank you :-)
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Re: Is x^2+2x+1=y^2+2y+1?  [#permalink]

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New post 08 Aug 2018, 03:36
1
dave13 wrote:
Mo2men wrote:
chetan2u wrote:
Is \(x^2+2x+1=y^2+2y+1\)?
(1) \(x\neq{y}\)
(2) x>y>0



\(x^2+2x+1=y^2+2y+1\)
\((x+1)^2=(y+1)^2\)
\(|x + 1|=|y + 1|\)

2 cases to open modulus

\(x+1=y+1\)..........x=y

\(x+1=-y-1\).........x+y=-2

The question becomes:

x=y or x+y=-2

(1) \(x\neq{y}\)

It could be that x + y =-2

Let x =1 & y =-3............Answer is Yes

or

Any any other values

Let x =1 & y =-2............Answer is No

Insufficient

\((2) x>y>0\)

It is clear that \(x\neq{y}\) and psotive numbers can't produce negative number (-2). Therefore it is certain answer with NO.

Answer: B



hi Mo2men

can you please explain how you opened modulus in second case
\(|x + 1|=|y + 1|\)

\(x+1=-y-1\).........x+y=-2

i thought we need to open moduls on both sides like this \(-x-1=-y-1\) why it is not correct opening ? :) :?



Hi Dave,

The modulus lesson is one of the important lessons that you need to take it step by step to learn sufficiently. Loot at chetan2u signuatre as he has created nice lesson in absolute value besides GC math book or any other source you like.

I will try gladly to help

When You have both side with absolute value:

1- You will fix one side and change the alter the sign of other side. and therefore,

2- you must consider two cases

Let's apply in the above:

\(|x + 1|=|y + 1|\)

Case 1:

1- Fix the the LHS and
2- Fix the sign of RHS to be positive which means it will be as it is.

x + 1 = y + 1............X = Y

Case 2:

1- Fix the the LHS and
2- Change the sign of RHS to be Negative.

x + 1 = -(y + 1)............x + 1 = -y -1..............x + y = -2

What you did wrong is the you multiplied the whole equation in negative sign.

I hope it hlpes
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Re: Is x^2+2x+1=y^2+2y+1? &nbs [#permalink] 08 Aug 2018, 03:36
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