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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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): \(X^2 + (1-X)^2=1\) ... \(2X (X-1) = 0\)

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.

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??

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

Show Tags

26 Nov 2014, 03:16

7

This post received KUDOS

2

This post was BOOKMARKED

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.

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

Show Tags

26 Nov 2014, 04:17

2

This post received KUDOS

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.

Attachments

File comment: Figure of circle and line

Figure circle and line.JPG [ 10.85 KiB | Viewed 20049 times ]

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.
_________________

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

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
_________________

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.
_________________

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.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...