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If y≠1, is x=1?

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If y≠1, is x=1? [#permalink]

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If y≠1, is x=1?

(1) x^2 + y^2 = 1
(2) y = 1 - x
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Re: If y≠1, is x=1? [#permalink]

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New post 11 Oct 2013, 12:05
lucbesson wrote:
If y≠1, is x=1?

(1) \(X^2 + Y^2 = 1\)
(2) \(y = 1 - X\)


Hi Guys! I need your help!

My logic is following:
(1) not sufficient equation is virtually a circle around (0,0) point with the radius 1. If y ≠ 1, than x can be almost anything within following limits [-1;0) & (0;1]
(2) not sufficient: consider y = 0 => x=1; y = 3 => x=-2

(1)+(2) Substitute Y from (2) in (1):
\(X^2 + (1-X)^2=1\)
...
\(2X (X-1) = 0\)

Hense X= 0 or X = 1.

Not sufficient?

PLS Help! :)

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If y≠1, is x=1? [#permalink]

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lucbesson wrote:
lucbesson wrote:
If y≠1, is x=1?

(1) \(X^2 + Y^2 = 1\)
(2) \(y = 1 - X\)


Hi Guys! I need your help!

My logic is following:
(1) not sufficient equation is virtually a circle around (0,0) point with the radius 1. If y ≠ 1, than x can be almost anything within following limits [-1;0) & (0;1]
(2) not sufficient: consider y = 0 => x=1; y = 3 => x=-2

(1)+(2) Substitute Y from (2) in (1):
\(X^2 + (1-X)^2=1\)
...
\(2X (X-1) = 0\)

Hense X= 0 or X = 1.

Not sufficient?

PLS Help! :)


The point is that x=0 is not a valid solution. For x = 0, from y = 1 - x it follows that y = 1 but we are told in the stem that y ≠ 1. Thus x can only be 1. Sufficient.

Answer: C.

Does this make sense?
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Re: If y≠1, is x=1? [#permalink]

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statement 1
\(x^2 + y^2 = 1\)
Not sufficient here Y can be 0 or here x and y can have the values as \sqrt{1/2}.

statement 2
y = 1 - x.
Not sufficient, again Y can be 0 of Y can be 3 and x can be 2.

combining the 2,
\(x^2 + y^2 = 1\),
\(x^2 + y^2 + 2xy - 2xy = 1\),
\((x + y)^2 -2xy = 1\),

from second statement x + y = 1,
1 - 2xy = 1,
2xy = 0,
xy=0
either x or y is 0

if x is 0 then y = 1 (from statement 2), but this is not possible as the question stem.

hence y = 0 and x = 1.

Answer option c
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Re: If y is not equal to 1, is x = 1? [#permalink]

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vasili wrote:
If y is not equal to 1, is x = 1?

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

Quote:
Please help.



st.1 here different values of x and y can satisfy the equation \(x^2 + y^2 =1\). for example

x=\(\frac{1}{\sqrt{2}}\)
y=\(\frac{1}{\sqrt{2}}\)

as x is not equal to 1. hence answer to the original question is no.

also, x=1 and y=0 will satisfy this equation. as x=1, thus answer to the original question is yes.

x=-1 and y=0 will satisfy this equation. as x is not equal to 1, thus answer to the original question is no.

st. 2
y= 1-x again different values are possible. hence not sufficient

combining st.1 and st.2

put y=1-x in \(x^2+y^2=1\)

\(x^2 +1+x^2-2x=1\)
\(2(x^2-2x)=0\)
\(x(x-1)=0\)

x=0 or x=1

but x=0 is not possible, as y is not equal to 1. hence x=1. thus answer should be C

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Re: If y≠1, is x=1? [#permalink]

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New post 26 Nov 2014, 02:46
Bunuel wrote:
lucbesson wrote:
lucbesson wrote:
If y≠1, is x=1?

(1) \(X^2 + Y^2 = 1\)
(2) \(y = 1 - X\)


Hi Guys! I need your help!

My logic is following:
(1) not sufficient equation is virtually a circle around (0,0) point with the radius 1. If y ≠ 1, than x can be almost anything within following limits [-1;0) & (0;1]
(2) not sufficient: consider y = 0 => x=1; y = 3 => x=-2

(1)+(2) Substitute Y from (2) in (1):
\(X^2 + (1-X)^2=1\)
...
\(2X (X-1) = 0\)

Hense X= 0 or X = 1.

Not sufficient?

PLS Help! :)


The point is that x=0 is not a valid solution. For x=1, from y=1-x it follows that y=1 but we are told in the stem that y≠1. Thus x can only be 1. Sufficient.

Answer: C.

Does this make sense?



bunuel,
i have a doubt here...
x^2 + y^2 = 1
and
y = 1-x
=> x+y = 1 squaring both sides
=> x^2 + y^2 + 2xy = 1 substituting x^2 + y^2 = 1
=> 1 + 2xy = 1
=> 2xy = 0
=> xy = 0
now x can be anything on the basis of the above equatn
hence getting E. what am i missing

secondly, what might be the level of this question??

would really appreciate your help.
thanks.

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Re: If y is not equal to 1, is x = 1? [#permalink]

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New post 26 Nov 2014, 03:16
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My quick thought. Look at the picture, there are only two points satisfy the condition (1,0) and (0,1) but y is not equal 1 then only x = 1, y = 0 satisfy the conditions.
The answer is C, and you can come up with the answer in 30 seconds.
Attachments

is x=1.png
is x=1.png [ 4.56 KiB | Viewed 20065 times ]


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Re: If y is not equal to 1, is x = 1? [#permalink]

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vasili wrote:
If y is not equal to 1, is x = 1?

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

Quote:
Please help.


In the question it is given that y can take ANY VALUE except 1.

If we look at the first statement \(x^2 + y^2 = 1\), we are provided with a equation of a circle with center at (0,0) and radius 1. This circle will intersect x axis at points (1,0) and (-1,0) and y axis at (0,1) and (0,-1). Over the entire circumference of the circle there are infinite number of points that will satisfy statement 1 even if we exclude one specific point where value of y is 1. Therefore statement ONE alone is not sufficient.

Moving on to second statement \(y = 1 - x\), we are provided with a equation of a line passing through points (0,1) and (1,0). Yet again there are infinite number of points that will satisfy statement 2 even if we exclude one specific point where value of y is 1. Therefore statement TWO alone is not sufficient.

Now when we combine both first and second statement we get two specific points [i.e. (0,1) and (1,0)] where these curves meet. We are given that y is not equal to 1. Hence we are left with one unique value (1,0). Therefore answer option (c) is correct.

I have also uploaded the image of containing both the figures.

-----------------------------------------------------------
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Attachments

File comment: Figure of circle and line
Figure circle and line.JPG
Figure circle and line.JPG [ 10.85 KiB | Viewed 20049 times ]

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If y≠1, is x=1? [#permalink]

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New post 26 Nov 2014, 05:41
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arnabs wrote:
Bunuel wrote:
lucbesson wrote:

Hi Guys! I need your help!

My logic is following:
(1) not sufficient equation is virtually a circle around (0,0) point with the radius 1. If y ≠ 1, than x can be almost anything within following limits [-1;0) & (0;1]
(2) not sufficient: consider y = 0 => x=1; y = 3 => x=-2

(1)+(2) Substitute Y from (2) in (1):
\(X^2 + (1-X)^2=1\)
...
\(2X (X-1) = 0\)

Hense X= 0 or X = 1.

Not sufficient?

PLS Help! :)


The point is that x=0 is not a valid solution. For x=1, from y=1-x it follows that y=1 but we are told in the stem that y≠1. Thus x can only be 1. Sufficient.

Answer: C.

Does this make sense?



bunuel,
i have a doubt here...
x^2 + y^2 = 1
and
y = 1-x
=> x+y = 1 squaring both sides
=> x^2 + y^2 + 2xy = 1 substituting x^2 + y^2 = 1
=> 1 + 2xy = 1
=> 2xy = 0
=> xy = 0
now x can be anything on the basis of the above equatn
hence getting E. what am i missing

secondly, what might be the level of this question??

would really appreciate your help.
thanks.


The very post you are quoting answers the question.

xy = 0 means that x = 0 or y = 0. But x = 0 is not a valid solution. If x = 0, from y = 1 - x it follows that y=1 but we are told in the stem that y≠1. Thus y = 0 and from y = 1 - x it follows that x = 1.

As for the difficulty level of the question: stats in the original post say that it's a 700 level question.
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Re: If y≠1, is x=1? [#permalink]

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New post 29 Nov 2014, 00:45
vasili wrote:
If y is not equal to 1, is x = 1?

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


Hi
I am not sure if this is correct solution, someone kindly help

If y is not equal to 1, is x = 1?
1) x^2 + y^2 = 1
2) y = 1 - x

------------------------------------------------
Statement1
x^2 + y^2 = 1
y=0 , x=1 --> x=1 Yes
but y and x need not be integers so y = x = sqrt(0.5) --> x=1 No
------------------------------------------------
Statement2
y = 1 - x
x=1, Y=0 --> x=1 Yes
x=0.5 , Y = 0.5 --> x=1 No
--------------------------------------------------
1+2
Here I checked all ranges

Case1
If x=0 then Y=1 which can not be true
So x=1 and Y=0
x=1 Yes

Case2
x between 0 and 1
x=0.1 y=0.9
x=0.5 y=0.5
x=0.9 y=0.1
As we know number between 0 and 1 , then x2<x so in this case x^2 + y^2 <> 1
so we cant consider this range

Case3
x between 0 and -1
this will be same as case2 , because square will be same for + and -ve

Case4
x >=1 here again x^2 + y^2 <> 1
so we cant consider this range

x<=-1 here again x^2 + y^2 <> 1
so we cant consider this range

Hence only x=1 and Y=0 works
IMO C

Someone please confirm if it is correct.

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Re: If y≠1, is x=1? [#permalink]

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New post 09 Apr 2015, 06:41
MDK wrote:
If y≠1, is x=1?

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


But if we take the sqroot of both side in statment 1 we will get x+y=1

so, x=1,y=0 or x=4,y=-3 so statment 1 insuff

statment 2 also y could be equal to -3 when x=4

and y=0 when x=1 so statment 2 insuff

both statment 1and 2 insuff where x could be equal to 1 or 4 so the answer is E
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Re: If y≠1, is x=1? [#permalink]

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New post 09 Apr 2015, 06:52
23a2012 wrote:
MDK wrote:
If y≠1, is x=1?

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


But if we take the sqroot of both side in statment 1 we will get x+y=1

so, x=1,y=0 or x=4,y=-3 so statment 1 insuff

statment 2 also y could be equal to -3 when x=4

and y=0 when x=1 so statment 2 insuff

both statment 1and 2 insuff where x could be equal to 1 or 4 so the answer is E


Did you test whether x=4 and y=-3 satisfy x^2 + y^2 = 1?

If you take the square root from x^2 + y^2 = 1, you'd get \(\sqrt{x^2 + y^2}= 1\), not x + y = 1.
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Re: If y≠1, is x=1? [#permalink]

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New post 09 Apr 2015, 08:08
MDK wrote:
If y≠1, is x=1?

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



(1) x^2 + y^2 = 1
y^2= 1-x^2
X will be 0 when Y=-1 (cannot be 1)
X can be +/- 1 when Y=0.
insufficient.



(2) y = 1 - x
Y can be 0 in which case X=1 , Y can be 2 in which case X=-1 .
not sufficient.

x^2 + (1-x)^2 = 1
2x(x-1)=0
X = 0 or 1

Since Y cannot be 1 so X cannot be 0 .
hence X=1

Ans: C
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Re: If y≠1, is x=1? [#permalink]

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New post 19 May 2015, 15:21
manpreetsingh86 wrote:
vasili wrote:
If y is not equal to 1, is x = 1?

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

Quote:
Please help.



st.1 here different values of x and y can satisfy the equation \(x^2 + y^2 =1\). for example

x=\(\frac{1}{\sqrt{2}}\)
y=\(\frac{1}{\sqrt{2}}\)

as x is not equal to 1. hence answer to the original question is no.

also, x=1 and y=0 will satisfy this equation. as x=1, thus answer to the original question is yes.

x=-1 and y=0 will satisfy this equation. as x is not equal to 1, thus answer to the original question is no.

st. 2
y= 1-x again different values are possible. hence not sufficient

combining st.1 and st.2

put y=1-x in \(x^2+y^2=1\)

\(x^2 +1+x^2-2x=1\)
\(2(x^2-2x)=0\)
\(x(x-1)=0\)

x=0 or x=1

but x=0 is not possible, as y is not equal to 1. hence x=1. thus answer should be C

Hi manpreetsingh86! The correct version of the bold part is 2(x^2-x)=0. In your solution 2 in the brackets is wrong since you factor out 2 already. Please correct
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Re: If y≠1, is x=1? [#permalink]

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New post 20 May 2015, 05:43
Quick soln

Statement 1: Circle with infinite values for x when y is not 1. reject.

Statement 2: Line with infinite values for x when y is not 1. reject

Combines statement: Only two point of intersection (0,1) and (1,0). Thus solvable.

As pointed in the above posts the easiest way is the graphical way in most of these type of questions. I think we should not even think on the track of solving equations.
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Re: If y≠1, is x=1? [#permalink]

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New post 20 May 2015, 23:17
MDK wrote:
If y≠1, is x=1?

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



Something to think about here:

y is "not 1" is an unusual information. There is no reason for which it needs to be "not 1" such as (y - 1) in denominator.

(1) x^2 + y^2 = 1
x can take many values since this is an equation in two variables. y will take corresponding value(s). What we need to check is whether x can be 1. The only time the statement may be sufficient is if x is 1 only when y = 1.
Put x = 1, you get y = 0.
So x may or may not be 1. Insufficient


(2) y = 1 - x
Similarly, x and y may take infinite different value. When x = 1, y = 0 here. So again, x may or may not be 1. Not sufficient.

Using both, x^2 + (1-x)^2 = 1
x(x - 1) = 0
Either x is 0 or 1. But if x is 0, y is 1 which is not allowed. So x must be 1.

Answer (C)
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Re: If y≠1, is x=1? [#permalink]

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Re: If y≠1, is x=1? [#permalink]

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New post 22 Mar 2017, 21:03
MDK wrote:
If y≠1, is x=1?

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



1) y=0, x=1 then yes
y=(sqroot0.5)^2, x=y=(sqroot0.5)^2, then no

2) Clearly insuff. 4=1-(-3) then no, 0=1-1, then yes

1)&2) (1-x)^2 + x^2 = 1
x^2-x=0
x=1 or x=0 but y cannot be 1 so from y=1-x, x must be 1.

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Re: If y≠1, is x=1? [#permalink]

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New post 26 Aug 2017, 04:53
If y ≠ 1, is x = 1?

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

(2) y=1-x


It is obvious that neither statement alone is sufficient. Now, let's look two statements together.

Think about it x+y = 1 but x^2 + y^2 = 1. Can you think of a pair of two numbers whose sum is 1. But when you square those numbers and add together the results, you again get 1? The only pair 1 and 0

Since y can't be 1, x must be 1

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Re: If y≠1, is x=1?   [#permalink] 26 Aug 2017, 04:53
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