 It is currently 25 Nov 2020, 20:45 ### 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

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. # x^2-y^2<8, x+y>3  Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Intern Joined: 20 May 2018
Posts: 2
Followers: 0

Kudos [?]: 0 , given: 0 00:00

Question Stats: 46% (03:01) correct 53% (02:10) wrong based on 15 sessions
$$x^{2}-y^{2}< 8$$

$$x+y>3$$

If x and y are integers in the above inequalities and 0 < y < x, what is the greatest possible value of x?

[Reveal] Spoiler: OA
4
Founder  Joined: 18 Apr 2015
Posts: 13908
GRE 1: Q160 V160 Followers: 313

Kudos [?]: 3678 , given: 12922

Expert's post

The OA??
_________________
Intern Joined: 20 May 2018
Posts: 2
Followers: 0

Kudos [?]: 0 , given: 0

Carcass wrote:

The OA??

The answer is 4. I don't understand why it gets 4. Retired Moderator Joined: 07 Jun 2014
Posts: 4803
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 175

Kudos [?]: 3035  , given: 394

1
KUDOS
Expert's post
What is the source of this question. PS: OA means official answer.

I am not sure if this is a GRE question. THIS IS A BAD QUESTION.

condition 1: x+y>3
condition 2: $$x^{2}-y^{2}< 8$$

We start by picking numbers. So the difference between squares has to be less than 8.

condition 1: x+y>3
condition 2: $$x^{2}-y^{2}< 8$$

For example 1,3 now $$3^2-1^2 = 8$$ and $$3+1=4$$ condition 1 is statisfied condition 2 is not.

Next pair is 3,4. So $$4^2-3^2 = 7$$ and and $$3+4=7$$. This satisfies but this may not be the gratest value of x. Try the next pair. 5,4

$$5^2-4^2 = 9$$ SO not true.

$$6^2-5^2 = 11$$

$$7^2-6^2 = 13$$

Note that the difference between consecutive squares keep on increasing. So 3,4 is the only viable pair.

HERE is why this is a BAD question. $$y>x$$.

Say y=1000 x=500.

$$x+y =1500$$ holds

$$x^2-y^2=-750000 < 8$$ holds

Now here x can have infinitely many values. x=1 and y=1000 holds so does x=999 and y=1000. Also y can have any value.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test Intern Joined: 04 Jun 2018
Posts: 8
Followers: 0

Kudos [?]: 6  , given: 0

1
KUDOS
We have x+y>3 AND x^2-y^2<8. Therefore x-y< 8/3 (nearly equals 2.63) since x^2-y^2 = (x+y)(x-y).
x and y are integers, plus 0 < y < x then x-y equal z which is an integer, lower than 2.63. We only have (1,2)
Since the smallest number of x and y that product x-y = 2 is x=3 and y=1. But x^2-y^2=8 which is wrong.
Then only 1 case left which is x-y = 1. We can create a table of x and y from then to see the greatest value of x is 4 Intern Joined: 09 Jul 2018
Posts: 10
Followers: 1

Kudos [?]: 17  , given: 0

2
KUDOS
theAlbatross wrote:
$$x^{2}-y^{2}< 8$$

x+y>3

If x and y are integers in the above inequalities and 0 < y < x, what is the greatest possible value of x? _____

Since $$x$$ and $$y$$ are positive integers, $$x^2$$ and $$y^2$$ are PERFECT SQUARES, implying that $$x^2-y^2$$is equal to the difference of two perfect squares.
Make list of perfect squares:
1, 4, 9, 16, 25...
Since $$x^2 - y^2 < 8$$, the value of x will be maximized if $$x^2$$ and $$y^2$$ are the perfect squares in blue, which have a difference of 7.
Thus:
$$x^2 = 16$$, implying that the greatest possible value of $$x=4$$.

The inequality in red is not necessary to solve the problem.
_________________

GMAT and GRE Tutor
Over 1800 followers
GMATGuruNY at gmail
New York, NY
If you find one of my posts helpful, please take a moment to click on the "Kudos" icon.
Available for tutoring in NYC and long-distance.

Director  Joined: 22 Jun 2019
Posts: 517
Followers: 5

Kudos [?]: 107 , given: 161

Founder  Joined: 18 Apr 2015
Posts: 13908
GRE 1: Q160 V160 Followers: 313

Kudos [?]: 3678 , given: 12922

Expert's post
No.

The tag was wrong because the user used it. All and only the questions from powerprep are located here

https://greprepclub.com/forum/gre-power ... -3118.html

If you see a question with the same tag but NOT from the directory above, then it is NOT a pp question. Absolutely.

Regards
_________________
Director  Joined: 22 Jun 2019
Posts: 517
Followers: 5

Kudos [?]: 107 , given: 161

Carcass wrote:
No.

The tag was wrong because the user used it. All and only the questions from powerprep are located here

https://greprepclub.com/forum/gre-power ... -3118.html

If you see a question with the same tag but NOT from the directory above, then it is NOT a pp question. Absolutely.

Regards

Thanks, Although this question basically i found in Magoosh Very Hard section. Its' from Magoosh.

Attachment: Screenshot from 2019-11-05 13-50-44.png [ 68.83 KiB | Viewed 3083 times ]

_________________

New to the GRE, and GRE CLUB Forum?
Posting Rules: QUANTITATIVE | VERBAL

Questions' Banks and Collection:
ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides
3rd Party Resource's: All In One Resource's | All Quant Questions Collection | All Verbal Questions Collection | Manhattan 5lb All Questions Collection
Books: All GRE Best Books
Scores: Average GRE Score Required By Universities in the USA
Tests: All Free & Paid Practice Tests | GRE Prep Club Tests
Extra: Permutations, and Combination
Vocab: GRE Vocabulary Intern Joined: 30 Aug 2019
Posts: 6
Followers: 0

Kudos [?]: 7  , given: 0

1
KUDOS
Given : x^2-y^2<8
x+y>3

(x+y)*(x-y)<8
since x+y>3
case 1 : x+y = 4 -> x-y = 1 -> x=2.5 not an integer
case 2 : x+y = 5 -> x-y = 1 -> x=3 and y=2
case 3 : x+y = 6 -> x-y =1 ->x=3.5 not an integer
case 4 : x+y=7 -> x-y=1 -> x=4 and y=3.
As question asks for max value of x, hence 4 Director  Joined: 22 Jun 2019
Posts: 517
Followers: 5

Kudos [?]: 107  , given: 161

1
KUDOS
theAlbatross wrote:
$$x^{2}-y^{2}< 8$$

$$x+y>3$$

If x and y are integers in the above inequalities and 0 < y < x, what is the greatest possible value of x?

[Reveal] Spoiler: OA
7

OA answer should be 4 not 7.
_________________

New to the GRE, and GRE CLUB Forum?
Posting Rules: QUANTITATIVE | VERBAL

Questions' Banks and Collection:
ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides
3rd Party Resource's: All In One Resource's | All Quant Questions Collection | All Verbal Questions Collection | Manhattan 5lb All Questions Collection
Books: All GRE Best Books
Scores: Average GRE Score Required By Universities in the USA
Tests: All Free & Paid Practice Tests | GRE Prep Club Tests
Extra: Permutations, and Combination
Vocab: GRE Vocabulary Re: x^2-y^2<8, x+y>3   [#permalink] 06 Nov 2019, 11:48
Display posts from previous: Sort by

# x^2-y^2<8, x+y>3  Question banks Downloads My Bookmarks Reviews Important topics  Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.