If y ≠ 1, is x = 1 ? (1) x^2 + y^2 = 1 (2) y = 1 - x

Question

If y ≠ 1, is x = 1 ?

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

MDK · Answer

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!

Bunuel · Answer

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?

fugitive · Answer

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

manpreetsingh86 · Answer

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

arnabs · Answer

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.

HieuNguyenVN · Answer

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.

anujpoonia · Answer

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   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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+1 Kudos if you find the reply useful.

Bunuel · Answer

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.

23a2012 · Answer

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

Bunuel · Answer

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.

Lucky2783 · Answer

(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

KarishmaB · Answer

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)

kablayi · Answer

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

u1983 · Answer

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

Case 01 , y=2, x=  \sqrt{(1- 2^2)}  =  \sqrt{(-3)} 
Case 02 , y=0, x=  \sqrt{(1- 0^2)}  =  \sqrt{(1)}  =  +1  or  -1 ....  Thus (NS)

(2) y = 1 – x  
Case 01 , y=2,  x= 1-2 = -1 
Case 02 , y=0,  x= 1-0 = 1 ....  Thus (NS)

(1) +(2) = =>

from (2)  x+y=1  
or  (x+y)^2= 1 
or  x^2+y^2+2*x*y =1 
or  1+2*x*y =1  ( substituting  x^2  +  y^2  =  1  from (1))
or  2*x*y =0

Now as y ≠ 1..... x has to be 0. Thus x is not 1. ......................Thus  sufficient. ..... Hence I would go for  option C .

GMATGuruNY · Answer

A very fast approach is to GRAPH the two statements.
x² + y² = r² is the equation for a circle that is centered at the origin and has a radius of r.
y = mx + b is the equation of a line with a slope of m and a y-intercept of b.

Statement 1: x² + y² = 1 
This is the equation for a circle that is centered at the origin and has a radius of 1:
 https://image.ibb.co/eQ5P1o/circle_with_a_radius_of_1.jpg 
The question prompt indicates that y≠1.
Thus, (x,y) can be any point on the circle other than (0,1).
Since it's possible that (x,y) = (1,0) or that (x.y) = any other point on the circle other than (0,1), INSUFFICIENT.

Statement 2: y = 1-x 
Rephrased in the form of y = mx + b:
y = -x + 1.
This is the equation of a line with a slope of -1 and a y-intercept of 1:
 https://image.ibb.co/dGCbE8/y_x_1.jpg 
The question prompt indicates that y≠1.
Thus, (x,y) can be any point on the line other than (0,1).
Since it's possible that (x,y) = (1,0) or that (x.y) = any other point on the line other than (0,1), INSUFFICIENT.

Statements combined: 
Overlaying the graphs, we get:
 https://image.ibb.co/cLiBgo/circle_with_a_radius_of_1_and_y_x_1.jpg 
Of the 2 points of intersection, only (1,0) is viable.
Thus, x=1.
SUFFICIENT.

C

dave13 · Answer

VeritasKarishma  where do you see Y here x(x - 1) = 0 ?

KarishmaB · Answer

You are using both statements which means you are given:
(1) x^2 + y^2 = 1 and
(2) y = 1 - x (this is where the relevant y is)

You get two values for x: 0 and 1.
If x = 0, y = 1 (not allowed)
If x = 1, y = 0 (allowed)

MathRevolution · Answer

Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 2 variables: Let the original condition in a DS question contain 1 variable. Now, 2 variables would generally require 2 equations for us to be able to solve for the value of the variable.

We know that each condition would usually give us an equation, and Since we need 2 equations to match the numbers of variables and equations in the original condition, the logical answer is C.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find is x = 1? where y ≠ 1.

Second and the third step of Variable Approach: From the original condition, we have 2 variables (x and y).To match the number of variables with the number of equations, we need 2 equations. Since conditions (1) and (2) will provide 1 equation , C would most likely be the answer.

Let’s take look at combined conditions.

Condition(1) tells us that \(x^2 + y^2 = 1\) .

Condition(2) tells us that y = 1 - x.

=> Substitute y = 1 - x in condition (1):

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

=> x^2 + 1 + x^2 - 2x = 1

=> 2x^2 - 2x = 0

=> 2x^2 - 2x = 0

=> x( x - 1) = 0

=> x = 0 or 1

=> When 'x = 0' , then y = 1 - x = 1 - 0 = 1 [Not possible as questions says y ≠ 1]

=> When 'x = 1' , then y = 1 - x = 1 - 1 = 0 Possible

Since the answer is a unique, both conditions combined together are sufficient by CMT 2.

Both conditions combined together are sufficient.

So, C is the correct answer.

Answer: C

SAVE TIME: By Variable Approach, when you know that we need 2 equations, we will directly check both conditions combined to be sufficient. We will not waste time in checking the conditions individually.

chuchosol03 · Answer

What about y= -1 here?

If y ≠ 1, is x = 1 ? (1) x^2 + y^2 = 1 (2) y = 1 - x

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