On the coordinate plane is point (0, 0) closer to point (u, : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 21 Feb 2017, 07:40

# LIVE NOW:

Chat with Vanderbilt (Owen) Admission Director in Chat Room1 | Expecting Interview Invites from MIT Sloan Shortly - Join Chat Room3 for LIVE Updates

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# On the coordinate plane is point (0, 0) closer to point (u,

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 11 May 2008
Posts: 556
Followers: 8

Kudos [?]: 177 [0], given: 0

On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

### Show Tags

28 Aug 2008, 08:27
5
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

32% (02:10) correct 68% (01:29) wrong based on 230 sessions

### HideShow timer Statistics

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 May 2014, 23:50, edited 4 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 35

Kudos [?]: 877 [1] , given: 5

### Show Tags

28 Aug 2008, 08:45
1
KUDOS
1
This post was
BOOKMARKED
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE A CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = $$sqrt {u^2+v^2 }$$
Distance between (0, 0) and (u, v+1) = $$sqrt {u^2+v^2 +2v+1}$$

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
$$sqrt {u^2+v^2 +2v+1}$$ is alwasy less than $$sqrt {u^2+v^2 }$$
_________________

Smiling wins more friends than frowning

Last edited by x2suresh on 29 Aug 2008, 05:27, edited 1 time in total.
Current Student
Joined: 11 May 2008
Posts: 556
Followers: 8

Kudos [?]: 177 [0], given: 0

### Show Tags

28 Aug 2008, 08:55
hmmm . k .. suresh !! thanks...
Senior Manager
Joined: 29 Mar 2008
Posts: 348
Followers: 4

Kudos [?]: 74 [0], given: 0

### Show Tags

28 Aug 2008, 10:33
x2suresh wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = $$sqrt {u^2+v^2 }$$
Distance between (0, 0) and (u, v+1) = $$sqrt {u^2+v^2 +2v+1}$$

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
$$sqrt {u^2+v^2 +2v+1}$$ is alwasy less than $$sqrt {u^2+v^2 }$$

Why will S2 not work? What is the OA?
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 35

Kudos [?]: 877 [0], given: 5

### Show Tags

28 Aug 2008, 10:50
leonidas wrote:
x2suresh wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

1.v + u^2 = -1
2. v < 0

OA is A. but sorry, i didnt understand this ...

Boss don't post OA .. GIVE CHANCE TO OTHERS.

Distance between (0, 0) and (u, v) = $$sqrt {u^2+v^2 }$$
Distance between (0, 0) and (u, v+1) = $$sqrt {u^2+v^2 +2v+1}$$

1.v + u^2 = -1
--> u^2 = -1-v ( u^2 is alwasy positive.. So.. v must be -ve and <-1)
2v+1 --> -ve
$$sqrt {u^2+v^2 +2v+1}$$ is alwasy less than $$sqrt {u^2+v^2 }$$

Why will S2 not work? What is the OA?

Hey Nemo,

2v+1 --> -ve or +ve.
v<0 v=-1/4 2v+1>0
v=-2 2v+1<0
_________________

Smiling wins more friends than frowning

Senior Manager
Joined: 29 Mar 2008
Posts: 348
Followers: 4

Kudos [?]: 74 [0], given: 0

### Show Tags

28 Aug 2008, 12:26
x2suresh wrote:
leonidas wrote:

Why will S2 not work? What is the OA?

Hey Nemo,

2v+1 --> -ve or +ve.
v<0 v=-1/4 2v+1>0
v=-2 2v+1<0

Got it, didn't try a fraction
Thanks X2Suresh
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 556

Kudos [?]: 3659 [1] , given: 360

### Show Tags

29 Aug 2008, 00:44
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?
First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ?
Now, we can translate it to language of formulas: |v|<|v+1|
Eventually, we can write: v>-0.5

b) v + u^2 = -1 --> v=-1-u^2 --> v<-1

Now, we can restate our problem as following:

Does v>-0.5 ?
1. v<-1
2. v<0

_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Current Student
Joined: 11 May 2008
Posts: 556
Followers: 8

Kudos [?]: 177 [0], given: 0

### Show Tags

29 Aug 2008, 02:44
thanks , but
how did u get 2 from 1??
could not quite understand...

walker wrote:
Let's consider each statement carefully:

a) On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?
First of all, u does not influence on answer. Therefore, we can restate: On the coordinate line is point (0) closer to point (v) than to point (v + 1) ?
Now, we can translate it to language of formulas: |v|<|v+1|......(1)
Eventually, we can write: v>-0.5....(2)

b) v + u^2 = -1 --> v=-1-u^2 --> v<-1

Now, we can restate our problem as following:

Does v>-0.5 ?
1. v<-1
2. v<0

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 556

Kudos [?]: 3659 [0], given: 360

### Show Tags

29 Aug 2008, 03:13
Full solution:

|v|<|v+1|

1) v<-1: -v<-v-1 --> 0<-1 --> always false
2) -1<=v<=0: -v<v+1 --> v>-0.5 --> -0.5<v<=0
3) v>0: --> v<v+1 --> 0<1 always true.

Therefore, inequality is true when v>-0.5

Fast solution:

large negative v: inequality is false
large positive v: inequality is true
switch point v=-0.5 --> v>-0.5
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Senior Manager
Joined: 09 Oct 2007
Posts: 466
Followers: 1

Kudos [?]: 43 [0], given: 1

### Show Tags

29 Aug 2008, 08:11
Duh! I went for D but after reading your solutions I realized I forgot to test fractions.
Senior Manager
Joined: 17 Sep 2013
Posts: 394
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Followers: 19

Kudos [?]: 277 [0], given: 139

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

### Show Tags

13 May 2014, 14:15
An even better way to look at this one:

Is \sqrt{u^2+v^2} > \sqrt{u^2+ (v+1)^2}
\sqrt{u^2+v^2} > \sqrt{u^2+v^2+2v+1}

So \sqrt{u^2+v^2} is equal on both sides so the deciding factor is 2v+1
2v+1>0 or v>-1/2 then the right side of the inequality is the greater one

I sufficiently tells us...u^2= -1 - v...As u^2 can never be negative..we have
1. v > -1
2. v = -1..in both the cases we get the same answer..Suff

II. V can be between 0 and -1/2..or <-1/2 ...Insuff

A it is
_________________

Appreciate the efforts...KUDOS for all
Don't let an extra chromosome get you down..

Last edited by JusTLucK04 on 14 May 2014, 00:58, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 37058
Followers: 7238

Kudos [?]: 96205 [3] , given: 10716

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

### Show Tags

13 May 2014, 23:51
3
KUDOS
Expert's post
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$.

So basically the question asks whether the distance between the points $$(0, 0)$$ and $$(u, v)$$ is less than the distance between the points $$(0, 0)$$ and $$(u, v + 1)$$: is $$\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}$$? --> is $$\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}$$? --> is $$u^2+v^2<u^2+v^2+2v+1$$? --> is $$v>-\frac{1}{2}$$?

(1) $$v + u^2 = -1$$ --> $$v=-1-u^2\leq{-1}$$ --> so the answer to the question is NO. Sufficient.

(2) $$v<0$$. Not sufficient.

_________________
Manager
Joined: 17 Jul 2013
Posts: 110
Followers: 0

Kudos [?]: 6 [0], given: 67

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

### Show Tags

24 Jun 2014, 21:49
How we could have represented this equation on graph ..... and solved by using graph technique

Math Expert
Joined: 02 Sep 2009
Posts: 37058
Followers: 7238

Kudos [?]: 96205 [0], given: 10716

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

### Show Tags

25 Jun 2014, 01:44
GmatDestroyer2013 wrote:
How we could have represented this equation on graph ..... and solved by using graph technique

This question is not a good candidate for graphic approach.
_________________
Intern
Joined: 25 Jan 2014
Posts: 19
Concentration: Technology, General Management
Followers: 0

Kudos [?]: 0 [0], given: 75

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

### Show Tags

04 Dec 2014, 02:27
1
This post was
BOOKMARKED
Bunuel wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$.

So basically the question asks whether the distance between the points $$(0, 0)$$ and $$(u, v)$$ is less than the distance between the points $$(0, 0)$$ and $$(u, v + 1)$$: is $$\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}$$? --> is $$\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}$$? --> is $$u^2+v^2<u^2+v^2+2v+1$$? --> is $$v>-\frac{1}{2}$$?

(1) $$v + u^2 = -1$$ --> $$v=-1-u^2\leq{-1}$$ --> so the answer to the question is NO. Sufficient.

(2) $$v<0$$. Not sufficient.

hi Bunuel,

i could not understand the last step.

v=-1-u^2\leq{-1} --> so the answer to the question is NO.

I did not get how did you infer 'v' lesser than -1 from the last equation??
Math Expert
Joined: 02 Sep 2009
Posts: 37058
Followers: 7238

Kudos [?]: 96205 [0], given: 10716

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

### Show Tags

04 Dec 2014, 03:40
arshu27 wrote:
Bunuel wrote:
arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

(1) v + u^2 = -1

(2) v < 0

M22-11

On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?

The formula to calculate the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$.

So basically the question asks whether the distance between the points $$(0, 0)$$ and $$(u, v)$$ is less than the distance between the points $$(0, 0)$$ and $$(u, v + 1)$$: is $$\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}$$? --> is $$\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}$$? --> is $$u^2+v^2<u^2+v^2+2v+1$$? --> is $$v>-\frac{1}{2}$$?

(1) $$v + u^2 = -1$$ --> $$v=-1-u^2\leq{-1}$$ --> so the answer to the question is NO. Sufficient.

(2) $$v<0$$. Not sufficient.

hi Bunuel,

i could not understand the last step.

v=-1-u^2\leq{-1} --> so the answer to the question is NO.

I did not get how did you infer 'v' lesser than -1 from the last equation??

As for your question, we need to find whether $$v>-\frac{1}{2}$$.

(1) gives $$v=-1-u^2$$. Since u^2 (the square of a number) must be non-negative, then we have that $$v=-1-(nonnegative)\leq{-1}$$, therefore v is NOT greater than -1/2, so we have a definite NO answer to the question.

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13890
Followers: 589

Kudos [?]: 167 [0], given: 0

Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]

### Show Tags

19 Feb 2016, 12:59
Hello from the GMAT Club BumpBot!

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.
_________________
Re: On the coordinate plane is point (0, 0) closer to point (u,   [#permalink] 19 Feb 2016, 12:59
Similar topics Replies Last post
Similar
Topics:
5 In the xy-coordinate plane, line l passes through the point (-3, 0). 5 12 Nov 2014, 09:09
1 The coordinates of points A and C are (0, -3) and (3, 3), respectively 10 31 Jul 2011, 08:20
4 If, on a coordinate plane, point A has the coordinates 12 25 Jul 2010, 11:53
6 On the coordinate plane, is point (u,v) closer to point 18 21 Nov 2009, 07:25
1 In which quadrant of the coordinate plane does the point 4 14 Jul 2009, 06:22
Display posts from previous: Sort by