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On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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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 M2211
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Originally posted by arjtryarjtry on 28 Aug 2008, 09:27.
Last edited by Bunuel on 14 May 2014, 00:50, edited 4 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.



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Re: wats this mean? [#permalink]
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Updated on: 29 Aug 2008, 06:27
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 = 1v ( 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 \(sqrt {u^2+v^2 }\)
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Originally posted by x2suresh on 28 Aug 2008, 09:45.
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Re: wats this mean? [#permalink]
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28 Aug 2008, 11:33
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 = 1v ( 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?
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Re: wats this mean? [#permalink]
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28 Aug 2008, 11:50
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 = 1v ( 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
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Re: wats this mean? [#permalink]
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28 Aug 2008, 13:26
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
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Re: wats this mean? [#permalink]
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29 Aug 2008, 01:44
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=1u^2 > v<1 Now, we can restate our problem as following: Does v>0.5 ? 1. v<1 2. v<0 Answer is obviously A.
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Re: wats this mean? [#permalink]
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29 Aug 2008, 03:44
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=1u^2 > v<1
Now, we can restate our problem as following:
Does v>0.5 ? 1. v<1 2. v<0
Answer is obviously A.



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Re: wats this mean? [#permalink]
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29 Aug 2008, 04:13
Full solution: v<v+1 1) v<1: v<v1 > 0<1 > always false 2) 1<=v<=0: v<v+1 > v>0.5 > 0.5<v<=03) v>0: > v<v+1 > 0<1 always true. Therefore, inequality is true when v>0.5Fast solution: large negative v: inequality is false large positive v: inequality is true switch point v=0.5 > v>0.5
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Re: wats this mean? [#permalink]
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29 Aug 2008, 09:11
Duh! I went for D but after reading your solutions I realized I forgot to test fractions.



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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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Updated on: 14 May 2014, 01:58
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
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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14 May 2014, 00:51
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
M2211 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 \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\). 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 \(\sqrt{(u0)^2+(v0)^2}<\sqrt{(u0)^2+(v+10)^2}\)? > 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) \(v + u^2 = 1\) > \(v=1u^2\leq{1}\) > so the answer to the question is NO. Sufficient. (2) \(v<0\). Not sufficient. Answer: A.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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24 Jun 2014, 22:49
How we could have represented this equation on graph ..... and solved by using graph technique
Please help



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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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04 Dec 2014, 03:27
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
M2211 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 \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\). 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 \(\sqrt{(u0)^2+(v0)^2}<\sqrt{(u0)^2+(v+10)^2}\)? > 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) \(v + u^2 = 1\) > \(v=1u^2\leq{1}\) > so the answer to the question is NO. Sufficient. (2) \(v<0\). Not sufficient. Answer: A. hi Bunuel, i could not understand the last step. v=1u^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??



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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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04 Dec 2014, 04:40
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
M2211 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 \(d=\sqrt{(x_1x_2)^2+(y_1y_2)^2}\). 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 \(\sqrt{(u0)^2+(v0)^2}<\sqrt{(u0)^2+(v+10)^2}\)? > 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) \(v + u^2 = 1\) > \(v=1u^2\leq{1}\) > so the answer to the question is NO. Sufficient. (2) \(v<0\). Not sufficient. Answer: A. hi Bunuel, i could not understand the last step. v=1u^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?? First of all please read Writing Mathematical Formulas on the Forum. As for your question, we need to find whether \(v>\frac{1}{2}\). (1) gives \(v=1u^2\). Since u^2 (the square of a number) must be nonnegative, 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.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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09 Mar 2017, 20:56
Bunuel wrote: GmatDestroyer2013 wrote: How we could have represented this equation on graph ..... and solved by using graph technique
Please help This question is not a good candidate for graphic approach. Hi Bunuel, What's wrong with graphing this? It worked for me but perhaps I am missing something? Prompt is asking basically asking if v > .5; graphically: are all possible u,v above the blue line? (1) States that the only solutions lie on the red line. Thus, all possible (u,v) are below the blue line. Sufficient. (2) States that all possible values are below the yellow/brown line. So, we can have some values above the blue line, some below the blue line. Insufficient.
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Re: On the coordinate plane is point (0, 0) closer to point (u, [#permalink]
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10 Mar 2017, 02:55
for positive value of v, (u,v) will be closer and for negative value of v, (u, v+1) will be closer
St 1: v+u^2 = 1. since u^ will always be positive, we can say that v is negative. for negative value of v, v+1 will always be closer to x axis as compared to v. ANSWER
St 2: v<0. Hence (u,v+1) will be closer. ANSWER
Option D




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