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Inequality

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by mehta003 » Fri Jul 08, 2011 2:15 pm
I agree that the answer is E.

if XY=0, x or y has to be 0. If X is 0, Y can e +1 or -1. If Y is 0, X can be +1 or -1 as well. There are always two solutions. We cannot conclude that X+Y=1 because it can also equal -1 as well
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by goalevan » Sat Jul 09, 2011 3:09 pm
The question stem gives the fact that x^2 + y^2 = 1, visually this is a circle with radius 1, centered at the origin. See the drawing below.

The question stem asks whether x + y = 1 also, or y = -x + 1. This is a line with -1 slope, y-intercept of 1, and x-intercept of 1.

The only two points that these two equations intersect are (0,1) and (1,0).

Rephrase the question: "Is (x,y) equal to either (0,1) or (1,0)?"

Statement 1) This statement says that visually, the point lies on one of the axes. Although both (0,1) and (1,0) satisfy this constraint, there are two additional points on the original circle that do not satisfy x + y = 1, namely (-1,0) and (0,-1).

Insufficient.

Statement 2) This statement says that the points lie on the x axis. There is still point (-1,0) that does not satisfy x + y = 1, but (1,0) satisfies.

Insufficient.

Statement 1 does not add any information to statement 2, so even combined they are insufficient.

E
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by ronnie1985 » Thu Mar 29, 2012 7:46 am
Given: x^2+y^2 = 1

To find: x+y = 1?

S1. xy = 0 => (x+y)^2 = 1 => x+y = +/- 1 Not sufficient.

S2. y = 0 => x^2 = 1 => x = +/-1, x+y = +/-1 Not Sufficient

Comb. Similar to S1. Not Sufficient

(E) is the answer.
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by kannan007 » Sat Jun 16, 2012 1:40 am
use plugig that satisfies first statement
x=-1/y=0 x+y =-1
x=1/y=0 x+y=1
so not sufficient

The same plugin helps to proove that 2nd statement is not sufficient:-)

The same plugin helps to proove that statement 1+statement 2 is not sufficient,hence answer E.
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by Lifetron » Sat Sep 01, 2012 7:50 pm
2 is clearly Insufficient !
1 comes down to (x+y)^2=+1 -> (x+y)= +1 or -1. So, Insufficient !

Take away : Square root of (+1) = +1 or -1
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by Ganesh hatwar » Fri Sep 21, 2012 11:25 pm
E

Uff held patience to apply the values of x = 1 and -1 or would have chosen C
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by coolanubhav2002 » Thu Nov 15, 2012 1:54 am
If x^2 + y^2 =1, is x + y =1?
(1) xy =0
(2) y = 0

IMO: E

the question says x^2+y^2=1 which means either of the one should be one and the other a zero, which means for if x+y=1,then one of them is 1 and the other a zero
Looking at statements:
1) it says xy=0, for this to be true either both can be zero or one of them 1 and the other zero, as both being zero can be ruled out as the question stem says the squaring of both x+y=1, but one of them could also be negative which makes it insufficient
2)It says y=0, nothing can be said about the other same reasons x could be 1 or -1,
Combined still insufficient same reasons x=-1 or 1

Answer has to be E
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by rajeshsinghgmat » Mon Apr 08, 2013 11:40 pm
E in answEr.

(x+y)^2=x^2+y^2+2*x*y
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by Java_85 » Fri Nov 01, 2013 2:06 pm
+1 E
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by scarpa220 » Wed Feb 12, 2014 10:31 am
This is the case where doing work before the statements is very useful. Notice there are four options. X=0, Y=1. X=1, Y=0. X=-1, Y=0 and X=0, Y=-1.
First statement:
This is not helpful because in all four of our cases xy=0.
(Note: there are other cases but these do not work with this first statement. ie: x=1/2, y=sqrt(3)/2.
Second statement:
With this statement we can eliminate cases where X=0
But y=-1 or 1. Not sufficient
Together:
We still do not know if y=1 or y=-1.

E
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by jaspreetsra » Sat Dec 27, 2014 3:05 am
If x^2 + y^2 = 1, is x + y = 1?

1. xy = 0
2. y = 0

IMO: E
Reason
(1) xy = 0

if x = 1
then 1^2 +0^2 = 1
if y = 1
then 0^2 + 1^2 = 1
i.e. x+y = 1+0 = 1
or 0+1 = 1
but if x = -1
then -1^2 + 0^2 = 1
x+y = -1+0 = -1 not sufficient
(2) y = 0
x^2 + y^2 = 1 => 1^2 + 0^2 = 1
x+y = 1+0 = 1
if x = -1
-1^2 + 0^2 = 1
x+y = -1+0 = -1 Not sufficient
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by nikhilgmat31 » Thu Aug 20, 2015 11:42 pm
pradeepkaushal9518 wrote:x^2+y^2=1

1.xy is zero hence either x=0 or y =0 but its not sufficent

2.y=0

so x^2+y^2=1

x^2+0=1
x^2=1
so X=1

so x+y=1

sufficeint So B
If X^2 =1
X can be +/- 1

x + y =-1 Not Sufficient.
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by nikhilgmat31 » Thu Aug 20, 2015 11:44 pm
I will go with E.

X + y can be +/- 1
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by nikhilgmat31 » Thu Aug 20, 2015 11:44 pm
I will go with E.

X + y can be +/- 1
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by akmba » Fri Aug 21, 2015 12:41 am
ANS : E

Hint : 1= -1x-1 or 1x1
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