Is x - y + 1 greater than x + y - 1 ? (1) x > 0 (2) y < 0

Is x-y+1 greater than x+y-1 ?

Is \(x-y+1>x+y-1\)?

x's cancel out and we get:

Is \(2>2y\)?

Is \(y<1\)?

(1) x > 0. Not sufficient.
(2) y < 0. The answer to the question is YES. Sufficient.

Answer: B.
_________________

Check below links:

Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips

DS Inequalities Problems
PS Inequalities Problems

700+ Inequalities problems

inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

Hope it helps.
_________________

x-y+1 > x+y-1
1-y > y-1
2y < 2
y < 1

1) does not talk about x. Insufficient
2) implies y<1.
Sufficient

Answer : b
_________________

Hi Bunuel,

y<1 can also mean y=0 right.. how can we assume y<1 is equal to y<0?
please clarify if i am missing some point..

The question asks whether y is less than 1. The second statement says that y is a negative number, so it must be less than 1.

Does this make sense?
_________________

Is x-y+1>x+y-1?

On rearranging the variables,
we get 2>2y? or simply y<1?

Hence our new question stem becomes: Is y less than 1?

Statement 1: x > 0. No information about y. NOT SUFFICIENT
Statement 2: y < 0. From this, we can safely say that y will be less than 1
SUFFICIENT

Option B

Where can I practice more of these qsn types??I was not able to think of simplifying the qsn stem ,really need to work on Inequalities

Target question: Is x - y + 1 > x + y - 1 ?
This is a good candidate for rephrasing the target question.

Take: x - y + 1 > x + y - 1
Subtract x from both sides to get: -y + 1 > y - 1
Add y to both sides to get: 1 > 2y - 1
Add 1 to both sides to get: 2 > 2y
Divide both sides by 2 to get: 1 > y
REPHRASED target question: Is y < 1?

The video below has tips on rephrasing the target question

Statement 1: x > 0
Since we have no information about y, we cannot answer the REPHRASED target question with certainty
Statement 1 is NOT SUFFICIENT

Statement 2: y < 0
In other words, y is NEGATIVE
If is NEGATIVE, then y is definitely less than 1
So, the answer to the REPHRASED target question is YES, y IS less than 1
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: B

Cheers,
Brent

RELATED VIDEOS

Hi Brent, thank you for clear and well organized approach.

I solve this question thinking this way,

x+(-y+1) vs x+(y-1) who’s bigger?
(1) don’t know how’s y, thus insufficient
(2) think y<0 with few case of when x<0 or x>0, we see who’s the winner.

I got through this risky time consuming way, because I’ve been thought that we should not deduct or add in inequality when there is no infos about their signs.

For example, 5+x > 3+x
If x is 10, 15 > 13. If x is -10, -5 < -3. So things change over what x is...

I am so confused when every time I face this kind of problems ?

Please help my poor brain..??

Thank you in advance ?

Posted from my mobile device

When dealing with inequalities, we can add or subtract ANY values we want without altering the inequality.
Problems arise when we MULTIPLY or DIVIDE both sides of an inequality by a VARIABLE when we don't know whether that variable is positive or negative.
This is covered in the following video: https://www.gmatprepnow.com/module/gmat ... /video/979



Your calculations above are not correct.
If x = -10, then 5+x > 3+x becomes 5+(-10) > 3+(-10), which simplifies to be -5 > -7, which is correct.
In fact, the left side of the inequality will ALWAYS be greater, regardless of the value of x.

We I know this because we can take: 5+x > 3+x
And subtract x from both sides to get: 5 > 3
This tells us the left side of the inequality will ALWAYS be greater, regardless of the value of x.

Does that help?

Geez I’m so embarrassed.. How can I miscalculate a simple equation...?

Thank you so much for answering me. It helps me a lot ! You made it clear

Hi Bunuel,

Can we do such operations on inequalities when we are not sure of their signs?

Thanks in advance.

Yes. We are concerned about the sign of a variable when multiplying/dividing an inequality by it. However we can safely add/subtract a variable from both sides of an inequality regardless of its sign.

9. Inequalities

• Theory
  Inequalities Made Easy!
  Inequalities: Tips and hints
  Arithmetic with Inequalities
  Wavy Line Method Application - Complex Algebraic Inequalities
  Solving Quadratic Inequalities: Graphic Approach
  Inequalities - Quadratic Inequalities
  Graphic approach to problems with inequalities
  Inequalities trick
  INEQUATIONS(Inequalities) Part 1
  INEQUATIONS(Inequalities) Part 2
  
  • Questions
    Inequality and absolute value questions from my collection
    DS Questions
    PS Questions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
_________________

Asked: Is x-y+1 greater than x+y-1 ?

Q. x - y + 1 > x + y - 1?
Q. 2y < 2?
Q. y<1

(1) x > 0
NOT SUFFICIENT

(2) y < 0 -> y < 1
SUFFICIENT

IMO B

Bunuel, BrentGMATPrepNow, chetan2u

The question stem is asking whether y<1 and statement 2 says that y<0
Now we do not know what kind of classification of variable y is,
So can anyone explain how the OA is B ??

If y < 0, then we can be certain that y < 1.
In other words, we can be certain that the answer to the target question is YES.

Here's another way to look at it:
If y < 0, then that means y is negative, and if y is negative, we can be certain that y is less than 1.

Does that help?

Hi

The range asked in the question is already included in the statement, so it is sufficient.

An analogy could be.

Is Mr A from India?
Statement I: Mr A is from Delhi.
Sufficient

Had it been vice versa, it would not have been sufficient.

In this question too, say we had to find - Is y>1?
Statement I: y>0
But what if y is between 0 and 1…answer is NO
_________________

Thanks BrentGMATPrepNow
If we test with x=1 and y=0.9 then we have 1.1>0.9. Therefore why is not C. What am I missing here?

For which statement are you testing x=1 and y=0.9?
If you it's statement 2, then you are breaking the condition that y < 0

In general, if you are trying to show that a statement is not sufficient, then you need to find two cases such that the answers to the target question are different.
With the pair of values x=1 and y=0.9, you are only testing one set of values.