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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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11 Oct 2013, 10:50
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If y≠1, is x=1?

(1) x^2 + y^2 = 1
(2) y = 1 - x
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Joined: 02 Sep 2009
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If y≠1, is x=1?  [#permalink]

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11 Oct 2013, 11:44
21
5
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.

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

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26 May 2014, 01:59
15
8
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.

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Re: If y≠1, is x=1?  [#permalink]

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11 Oct 2013, 11:05
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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Re: If y is not equal to 1, is x = 1?  [#permalink]

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26 Nov 2014, 01:37
4
3
vasili wrote:
If y is not equal to 1, is x = 1?

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

Quote:

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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26 Nov 2014, 01:46
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.

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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26 Nov 2014, 02:16
11
2
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 [ 4.56 KiB | Viewed 37059 times ]

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

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26 Nov 2014, 03:17
4
vasili wrote:
If y is not equal to 1, is x = 1?

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

Quote:

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.

-----------------------------------------------------------
+1 Kudos if you find the reply useful.
Attachments

File comment: Figure of circle and line

Figure circle and line.JPG [ 10.85 KiB | Viewed 37043 times ]

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Posts: 52210
If y≠1, is x=1?  [#permalink]

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26 Nov 2014, 04:41
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.

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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28 Nov 2014, 23:45
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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09 Apr 2015, 05:41
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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09 Apr 2015, 05:52
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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09 Apr 2015, 07:08
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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19 May 2015, 14:21
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:

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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20 May 2015, 04:43
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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20 May 2015, 22:17
1
1
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.

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Re: If y≠1, is x=1?  [#permalink]

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22 Mar 2017, 20:03
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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26 Aug 2017, 03:53
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]

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02 Oct 2017, 02:40
Got this question in GMATPrep EP1 and below is the approach I followed:
1. x^2 + y^2 = 1
Not given that x and y are integers. Hence, two cases possible:
a. If y = 0, x = 1 or x = -1
b. If y =1/sqrt2 , x = 1/sqrt2. Not sufficient

2. y = 1-x
x can take any value except for zero. Hence not sufficient.

(1) + (2),
x^2 + y^2 = 1 and y = 1-x
We know that (x+y)^2 = x^2 + y^2 + 2xy
1^2 = 1 + 2xy
2xy = 0
Given that y ≠ 1, for xy to be zero x has to be 1. Sufficient

Hence option C.

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If y ≠ 1, is x = 1 ?  [#permalink]

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09 May 2018, 14:00
If y ≠ 1, is x = 1 ?

(1) $$x^2$$ + $$y^2$$ = 1
(2) y = 1 – x
If y ≠ 1, is x = 1 ? &nbs [#permalink] 09 May 2018, 14:00

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