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If y≠1, is x=1? [#permalink]
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11 Oct 2013, 11:50
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If y≠1, is x=1? (1) x^2 + y^2 = 1 (2) y = 1  x
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Re: If y≠1, is x=1? [#permalink]
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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): \(X^2 + (1X)^2=1\) ... \(2X (X1) = 0\) 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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Re: If y≠1, is x=1? [#permalink]
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26 May 2014, 02:59
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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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26 Nov 2014, 02:37
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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= 1x again different values are possible. hence not sufficient combining st.1 and st.2 put y=1x in \(x^2+y^2=1\) \(x^2 +1+x^22x=1\) \(2(x^22x)=0\) \(x(x1)=0\) 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, 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): \(X^2 + (1X)^2=1\) ... \(2X (X1) = 0\) 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=1x 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 = 1x => 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, 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.
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Re: If y is not equal to 1, is x = 1? [#permalink]
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26 Nov 2014, 04:17
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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.  +1 Kudos if you find the reply useful.
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Figure circle and line.JPG [ 10.85 KiB  Viewed 26861 times ]



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If y≠1, is x=1? [#permalink]
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26 Nov 2014, 05:41
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): \(X^2 + (1X)^2=1\) ... \(2X (X1) = 0\) 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=1x 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 = 1x => 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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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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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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Re: If y≠1, is x=1? [#permalink]
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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= 1x^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 + (1x)^2 = 1 2x(x1)=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, 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= 1x again different values are possible. hence not sufficient combining st.1 and st.2 put y=1x in \(x^2+y^2=1\) \(x^2 +1+x^22x=1\) \( 2(x^22x)=0\) \(x(x1)=0\) 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^2x)=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, 05:43
Quick solnStatement 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. A s 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, 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 + (1x)^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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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=11, then yes 1)&2) (1x)^2 + x^2 = 1 x^2x=0 x=1 or x=0 but y cannot be 1 so from y=1x, x must be 1.



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Re: If y≠1, is x=1? [#permalink]
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26 Aug 2017, 04:53
If y ≠ 1, is x = 1?
(1) x^2 + y^2 = 1
(2) y=1x
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, 03:40
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 = 1x x can take any value except for zero. Hence not sufficient. (1) + (2), x^2 + y^2 = 1 and y = 1x 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. Kudos if it helps



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If y ≠ 1, is x = 1 ? [#permalink]
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09 May 2018, 15:00
If y ≠ 1, is x = 1 ?
(1) \(x^2\) + \(y^2\) = 1 (2) y = 1 – x




If y ≠ 1, is x = 1 ?
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