Last visit was: 17 Jun 2024, 17:20 It is currently 17 Jun 2024, 17:20

Close

GRE Prep 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 GRE 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
Intern
Intern
Joined: 20 May 2018
Posts: 2
Own Kudos [?]: 8 [8]
Given Kudos: 0
Send PM
Most Helpful Expert Reply
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4810
Own Kudos [?]: 10689 [6]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Most Helpful Community Reply
Intern
Intern
Joined: 09 Jul 2018
Posts: 49
Own Kudos [?]: 77 [6]
Given Kudos: 0
Send PM
General Discussion
Verbal Expert
Joined: 18 Apr 2015
Posts: 28946
Own Kudos [?]: 33719 [0]
Given Kudos: 25365
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
Expert Reply
The answer should be seven.

The OA??
avatar
Intern
Intern
Joined: 20 May 2018
Posts: 2
Own Kudos [?]: 8 [0]
Given Kudos: 0
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
Carcass wrote:
The answer should be seven.

The OA??


The answer is 4. I don't understand why it gets 4.
avatar
Intern
Intern
Joined: 04 Jun 2018
Posts: 8
Own Kudos [?]: 12 [0]
Given Kudos: 0
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
1
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
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 651 [0]
Given Kudos: 161
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
Is this really a PP2 Question. ? So much harder.
Verbal Expert
Joined: 18 Apr 2015
Posts: 28946
Own Kudos [?]: 33719 [0]
Given Kudos: 25365
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
Expert Reply
No.

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

https://gre.myprepclub.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
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 651 [1]
Given Kudos: 161
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
1
Carcass wrote:
No.

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

https://gre.myprepclub.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
Screenshot from 2019-11-05 13-50-44.png [ 68.83 KiB | Viewed 12043 times ]
avatar
Intern
Intern
Joined: 30 Aug 2019
Posts: 6
Own Kudos [?]: 8 [0]
Given Kudos: 0
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
1
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
User avatar
Director
Director
Joined: 22 Jun 2019
Posts: 521
Own Kudos [?]: 651 [0]
Given Kudos: 161
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
1
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?



Show: :: OA
7


OA answer should be 4 not 7.
Retired Moderator
Joined: 09 Jan 2021
Posts: 576
Own Kudos [?]: 839 [1]
Given Kudos: 194
GRE 1: Q167 V156
GPA: 4
WE:Analyst (Investment Banking)
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
1
Hi,

I tried this one using trial and error.

x^2-y^2= (x+y)(x-y)
What we can try to understand here is we can minimize the value of x-y to get the maximum value of x+y. As x^2-y^2<8 the maximum value it can take is 7. Thus we can make x+y=7 and similarly minimize x-y=1. The two number, which we can use to satisfy the eqn are 4 & 3.

Thus x=4

Hope this helps!
avatar
Intern
Intern
Joined: 01 Oct 2021
Posts: 2
Own Kudos [?]: 2 [1]
Given Kudos: 0
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
1
In the question, its clearly given that y<x. Therefore, you cannot take values of y more than x. I think someone above in the comments has termed this question as a bad one, but I think the questions states things clearly.

1. x^2-y^2<8
2. x+y>3
3. 0<y<x (positive integers)

Solution:

x^2-y^2<8
(x+y) (x-y) < 8 (Using property)

- we are given than x+y >3, therefore we can conclude that (x-y) has to be = 1. (Since, if (x+y >3 and are positive integers, we take the value to be 4. If x+y = 4 , then x-y cannot be 2 since (x+y) (x-y) < 8, therefore x-y = 1 (Next less positive integer before 2)

If x-y=1, then x=y+1 giving us that x and y are consecutive positive integers.

We check for (x+y> 3) and take x=3 and y=2, this holds true and we check for next set till one of the conditions above do not satisfy.

We get, x=4 and y=3, which holds true. We then check x=5 and y=4, this does not hold x^2-y^2<8 as x^2 - y^2 = 9.

Therefore, x has a maximum value of 4.
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 4602
Own Kudos [?]: 69 [0]
Given Kudos: 0
Send PM
Re: x^2-y^2<8, x+y>3 [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep 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.
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1088 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne