Last visit was: 03 Nov 2024, 15:51 It is currently 03 Nov 2024, 15:51
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
avatar
MDK
avatar
Current Student
Joined: 31 Oct 2012
Last visit: 06 Mar 2021
Posts: 40
Own Kudos:
565
 [267]
Given Kudos: 18
Status:Impossible is just an opinion
Location: Ukraine
Concentration: Strategy, Marketing
GMAT 1: 590 Q47 V24
GMAT 2: 650 Q47 V34
GMAT 3: 670 Q49 V31
GMAT 4: 690 Q48 V37
GPA: 3.8
WE:Marketing (Consumer Packaged Goods)
GMAT 1: 590 Q47 V24
GMAT 2: 650 Q47 V34
GMAT 3: 670 Q49 V31
GMAT 4: 690 Q48 V37
Posts: 40
Kudos: 565
 [267]
16
Kudos
Add Kudos
249
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 03 Nov 2024
Posts: 96,505
Own Kudos:
Given Kudos: 87,899
Products:
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 96,505
Kudos: 673,793
 [74]
56
Kudos
Add Kudos
17
Bookmarks
Bookmark this Post
User avatar
fugitive
Joined: 14 Apr 2013
Last visit: 05 Sep 2014
Posts: 7
Own Kudos:
62
 [57]
Given Kudos: 1
Location: United States
Concentration: Operations, Technology
WE:Programming (Computer Software)
Posts: 7
Kudos: 62
 [57]
40
Kudos
Add Kudos
17
Bookmarks
Bookmark this Post
User avatar
HieuNguyenVN
Joined: 21 Nov 2014
Last visit: 25 Oct 2024
Posts: 24
Own Kudos:
91
 [37]
Given Kudos: 31
Location: Viet Nam
GMAT 1: 760 Q50 V44
GMAT 2: 750 Q51 V40 (Online)
GMAT 1: 760 Q50 V44
GMAT 2: 750 Q51 V40 (Online)
Posts: 24
Kudos: 91
 [37]
32
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
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 117161 times ]

User avatar
manpreetsingh86
Joined: 13 Jun 2013
Last visit: 19 Dec 2022
Posts: 222
Own Kudos:
1,086
 [33]
Given Kudos: 14
Posts: 222
Kudos: 1,086
 [33]
20
Kudos
Add Kudos
13
Bookmarks
Bookmark this Post
vasili
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
General Discussion
avatar
MDK
avatar
Current Student
Joined: 31 Oct 2012
Last visit: 06 Mar 2021
Posts: 40
Own Kudos:
565
 [1]
Given Kudos: 18
Status:Impossible is just an opinion
Location: Ukraine
Concentration: Strategy, Marketing
GMAT 1: 590 Q47 V24
GMAT 2: 650 Q47 V34
GMAT 3: 670 Q49 V31
GMAT 4: 690 Q48 V37
GPA: 3.8
WE:Marketing (Consumer Packaged Goods)
GMAT 1: 590 Q47 V24
GMAT 2: 650 Q47 V34
GMAT 3: 670 Q49 V31
GMAT 4: 690 Q48 V37
Posts: 40
Kudos: 565
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
lucbesson
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! :)
User avatar
arnabs
Joined: 06 Aug 2013
Last visit: 29 Oct 2020
Posts: 45
Own Kudos:
Given Kudos: 17
Posts: 45
Kudos: 14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
lucbesson
lucbesson
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.
avatar
anujpoonia
Joined: 30 Oct 2014
Last visit: 22 Mar 2017
Posts: 1
Own Kudos:
16
 [10]
Given Kudos: 46
Location: India
GMAT 1: 670 Q48 V34
WE:Analyst (Computer Software)
GMAT 1: 670 Q48 V34
Posts: 1
Kudos: 16
 [10]
9
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
vasili
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.
Attachments

File comment: Figure of circle and line
Figure circle and line.JPG
Figure circle and line.JPG [ 10.85 KiB | Viewed 114173 times ]

User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 03 Nov 2024
Posts: 96,505
Own Kudos:
673,793
 [4]
Given Kudos: 87,899
Products:
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 96,505
Kudos: 673,793
 [4]
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
arnabs
Bunuel
lucbesson

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.
User avatar
23a2012
Joined: 03 Oct 2013
Last visit: 14 Jun 2015
Posts: 64
Own Kudos:
Given Kudos: 144
Status:Kitchener
Location: Canada
Concentration: Finance, Finance
GPA: 2.9
WE:Education (Education)
Posts: 64
Kudos: 47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MDK
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 03 Nov 2024
Posts: 96,505
Own Kudos:
Given Kudos: 87,899
Products:
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 96,505
Kudos: 673,793
Kudos
Add Kudos
Bookmarks
Bookmark this Post
23a2012
MDK
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.
User avatar
Lucky2783
Joined: 07 Aug 2011
Last visit: 08 May 2020
Posts: 420
Own Kudos:
1,857
 [1]
Given Kudos: 75
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27
GMAT 1: 630 Q49 V27
Posts: 420
Kudos: 1,857
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
MDK
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
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 03 Nov 2024
Posts: 15,427
Own Kudos:
69,248
 [7]
Given Kudos: 446
Location: Pune, India
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 15,427
Kudos: 69,248
 [7]
6
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
MDK
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)
avatar
kablayi
Joined: 05 Mar 2015
Last visit: 08 Feb 2023
Posts: 26
Own Kudos:
19
 [1]
Given Kudos: 111
Location: Azerbaijan
GMAT 1: 530 Q42 V21
GMAT 2: 600 Q42 V31
GMAT 3: 700 Q47 V38
GMAT 1: 530 Q42 V21
GMAT 2: 600 Q42 V31
GMAT 3: 700 Q47 V38
Posts: 26
Kudos: 19
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
u1983
User avatar
Current Student
Joined: 24 Aug 2016
Last visit: 06 Jun 2021
Posts: 726
Own Kudos:
809
 [4]
Given Kudos: 97
GMAT 1: 540 Q49 V16
GMAT 2: 680 Q49 V33
Products:
GMAT 1: 540 Q49 V16
GMAT 2: 680 Q49 V33
Posts: 726
Kudos: 809
 [4]
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
FillFM
If y ≠ 1, is x = 1 ?


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


(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[/m]
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.
User avatar
GMATGuruNY
Joined: 04 Aug 2010
Last visit: 03 Nov 2024
Posts: 1,333
Own Kudos:
3,352
 [5]
Given Kudos: 9
Schools:Dartmouth College
Expert reply
Posts: 1,333
Kudos: 3,352
 [5]
4
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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:

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:

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:

Of the 2 points of intersection, only (1,0) is viable.
Thus, x=1.
SUFFICIENT.

User avatar
dave13
Joined: 09 Mar 2016
Last visit: 22 Oct 2024
Posts: 1,134
Own Kudos:
Given Kudos: 3,851
Posts: 1,134
Kudos: 1,047
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasKarishma
MDK
If y≠1, is x=1?


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)

VeritasKarishma where do you see Y here x(x - 1) = 0 ?
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 03 Nov 2024
Posts: 15,427
Own Kudos:
69,248
 [1]
Given Kudos: 446
Location: Pune, India
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 15,427
Kudos: 69,248
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
dave13
VeritasKarishma
MDK
If y≠1, is x=1?


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)

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

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)
User avatar
MathRevolution
User avatar
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Last visit: 27 Sep 2022
Posts: 10,126
Own Kudos:
17,599
 [1]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Expert reply
GMAT 1: 760 Q51 V42
Posts: 10,126
Kudos: 17,599
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
chuchosol03
Joined: 07 Aug 2022
Last visit: 26 Feb 2024
Posts: 1
Posts: 1
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
lucbesson
lucbesson
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?

What about y= -1 here?
 1   2   
Moderator:
Math Expert
96505 posts