|
Author |
Message |
|
TAGS:
|
|
|
Director
Joined: 20 Sep 2006
Posts: 661
Followers: 2
Kudos [?]:
49
[0], given: 7
|
According to the graph above, when x = 3, y most nearly ? [#permalink]
 30 Aug 2008, 13:08
Question Stats:
0% (00:00) correct
100% (00:10) wrong based on 5 sessions
According to the graph above, when x = 3, y most nearly ? (A) –1 (B) -1/2 (C) 0 (D) 1/2 (E) 1 Should we solely rely on the graph assuming that it is drawn to scale????
Attachments
 Doc1.doc [31.5 KiB]
Downloaded 333 times
|
|
|
|
|
|
|
SVP
Joined: 07 Nov 2007
Posts: 1837
Location: New York
Followers: 20
Kudos [?]:
296
[0], given: 5
|
Re: PS geometery ... little confused. [#permalink]
30 Aug 2008, 13:22
rao_1857 wrote: According to the graph above, when x = 3, y most nearly ?
(A) –1
(B) -1/2
(C) 0
(D) 1/2
(E) 1
Should we solely rely on the graph assuming that it is drawn to scale???? What is the source of this question? By observation answer should be 1/2 or 1 its more close to 1.
_________________
Your attitude determines your altitude
Smiling wins more friends than frowning
Last edited by x2suresh on 30 Aug 2008, 13:26, edited 1 time in total.
|
|
|
|
|
|
Director
Joined: 23 Sep 2007
Posts: 806
Followers: 4
Kudos [?]:
67
[0], given: 0
|
Re: PS geometery ... little confused. [#permalink]
30 Aug 2008, 13:24
rao_1857 wrote: According to the graph above, when x = 3, y most nearly ?
(A) –1
(B) -1/2
(C) 0
(D) 1/2
(E) 1
Should we solely rely on the graph assuming that it is drawn to scale???? even if it is drawn to scale, it is hard to estimate between 1 and .5
Attachments
untitled.JPG [ 6.71 KiB | Viewed 1918 times ]
|
|
|
|
|
|
Intern
Joined: 02 Aug 2008
Posts: 29
Followers: 0
Kudos [?]:
2
[0], given: 0
|
Re: PS geometery ... little confused. [#permalink]
30 Aug 2008, 20:28
i doubt if the Q is a correct one. Its a parabola and from the diagram, you can make out 2 coordinates, (2,0) and (0,3); that takes you no where.
|
|
|
|
|
|
Intern
Joined: 03 Mar 2008
Posts: 48
Followers: 1
Kudos [?]:
3
[0], given: 0
|
Re: PS geometery ... little confused. [#permalink]
31 Aug 2008, 00:27
nishchals wrote: i doubt if the Q is a correct one. Its a parabola and from the diagram, you can make out 2 coordinates, (2,0) and (0,3); that takes you no where. pls look closer....(0,3) is not on the parabola...if it were than the problem was simple.........in that situation, solve parabola's equation and you would get that y coord will be 3/4 when x=3. The figure shows only one conformed point with coordinates ie the vertex (2,0). My guess for the answer is 1. However I don't have mathematical proof to support it. Logical proof is if (0,3) were on parabola....then (3, 3/4) is on parabola Hence in the case of (0, little less than 3)......the point should be (3, little more than 3/4)....closest answer is 1
|
|
|
|
|
|
Director
Joined: 20 Sep 2006
Posts: 661
Followers: 2
Kudos [?]:
49
[0], given: 7
|
Re: PS geometery ... little confused. [#permalink]
31 Aug 2008, 07:56
yeuapp .. I agree . there is not mathmatical way to solve this. by looking at it .. you can say its more close to 1.
src is 1000 ps.
|
|
|
|
|
|
Director
Joined: 14 Aug 2007
Posts: 745
Followers: 7
Kudos [?]:
66
[1] , given: 0
|
Re: PS geometery ... little confused. [#permalink]
16 Apr 2009, 00:17
1
This post received
KUDOS
proof:
vertex form of equation :
y = a (x-h)^2 + k
From the figure: upward parabola = a is +ve
vertex is at (2,0) => h = 2 k =0
y = a (x- 2)^2
point (0, 2.8) lies on this parabola => a = 0.7
y = 0.7 (x-2)^2
when x = 3, y =0.7 => ~1
HTH
|
|
|
|
|
|
Manager
Joined: 13 Jan 2009
Posts: 179
Schools: Harvard Business School, Stanford
Followers: 3
Kudos [?]:
14
[0], given: 9
|
Re: PS geometery ... little confused. [#permalink]
16 Apr 2009, 05:11
My guess is 1/2. but I don;t know how to solve it  It's just seems to me that way.
|
|
|
|
|
|
Intern
Joined: 16 Apr 2009
Posts: 9
Followers: 0
Kudos [?]:
1
[1] , given: 0
|
Re: PS geometery ... little confused. [#permalink]
23 Apr 2009, 23:59
1
This post received
KUDOS
alpha_plus_gamma wrote: proof:
vertex form of equation :
y = a (x-h)^2 + k
From the figure: upward parabola = a is +ve
vertex is at (2,0) => h = 2 k =0
y = a (x- 2)^2
point (0, 2.8) lies on this parabola => a = 0.7
y = 0.7 (x-2)^2
when x = 3, y =0.7 => ~1
HTH Hi alpha_plus_gamma , This is the bang on solution. This question features at many places. I wanted to ask one thing. How to figure out the starting eqn. Sudeep
|
|
|
|
|
|
Manager
Joined: 07 Apr 2009
Posts: 160
Followers: 2
Kudos [?]:
4
[0], given: 5
|
Re: PS geometery ... little confused. [#permalink]
24 Apr 2009, 00:58
bang on alpha, for simplification:
This is an equation of parabola .. y = a (x-h)^2 + k where h & k are values of vertex
A<0 if parabola opens downwards
a>0 if parabola opens upwards
here a is +ve
vertex is at (2,0) => h = 2 k =0
y = a (x- 2)^2
To find a, check other point on parabola , here (0, 2.8)
2.8= a(0+2)^2
2.8= a(4) a= .7
now
y = 0.7 (x-2)^2 put x= 3 then y = .7
|
|
|
|
|
|
Senior Manager
Joined: 19 Aug 2006
Posts: 261
Followers: 2
Kudos [?]:
6
[0], given: 0
|
Re: PS geometery ... little confused. [#permalink]
24 Apr 2009, 09:57
If drawn to scale, it's closer to 1 than to 0.5.
|
|
|
|
|
|
Manager
Joined: 23 Apr 2010
Posts: 140
Location: Tx
Schools: NYU,UCLA,BOOTH,STANFORD
Followers: 1
Kudos [?]:
2
[0], given: 36
|
Re: PS geometery ... little confused. [#permalink]
16 Aug 2010, 03:47
I really dont understand anything It should be like that but i dont know X1= 1/2 x2= 1 y1=2 y2=1 then (y2-y1)/(x2-x1)=M which is -2 in this case then -2= (y-y1)/(x-x1) y= -2x + 3 is equa. ıf we put there as X y= -6 +3 then y= -3 which is my answer....ı am confused where i am tripping up
_________________
This is not finished here...Watch me.....
|
|
|
|
|
|
Manager
Joined: 23 Apr 2010
Posts: 140
Location: Tx
Schools: NYU,UCLA,BOOTH,STANFORD
Followers: 1
Kudos [?]:
2
[0], given: 36
|
Re: PS geometery ... little confused. [#permalink]
16 Aug 2010, 04:08
Bunuel can you look at it pleasee
_________________
This is not finished here...Watch me.....
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 12096
Followers: 1876
Kudos [?]:
10096
[0], given: 959
|
Re: PS geometery ... little confused. [#permalink]
16 Aug 2010, 04:50
fatihaysu wrote: I really dont understand anything
It should be like that but i dont know
X1= 1/2
x2= 1
y1=2
y2=1
then
(y2-y1)/(x2-x1)=M which is -2 in this case
then
-2= (y-y1)/(x-x1)
y= -2x + 3 is equa.
ıf we put there as X y= -6 +3 then
y= -3 which is my answer....ı am confused where i am tripping up Frankly I don't quite understand what is tested here... But your approach is wrong because your method for finding equation of a graph is for straight line (linear equation) and obviously the curve shown on the diagram is not straight line. Is it a graph of quadratic equation? Are the intersection points with axes (0,3) and (2,0)? If the answer on both questions is YES then the equation of the graph would be y=\frac{3}{4}x^2-3x+3 and for x=1 --> y=\frac{3}{4}x^2-3x+3=\frac{3}{4}\approx{1}. I wouldn't worry about this question at all. P.S. Check the link about Coordinate Geometry in my signature for more on this issue.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!, 11 New DS set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Manager
Joined: 23 Apr 2010
Posts: 140
Location: Tx
Schools: NYU,UCLA,BOOTH,STANFORD
Followers: 1
Kudos [?]:
2
[0], given: 36
|
Re: PS geometery ... little confused. [#permalink]
16 Aug 2010, 06:26
Bunuel wrote: fatihaysu wrote: I really dont understand anything
It should be like that but i dont know
X1= 1/2
x2= 1
y1=2
y2=1
then
(y2-y1)/(x2-x1)=M which is -2 in this case
then
-2= (y-y1)/(x-x1)
y= -2x + 3 is equa.
ıf we put there as X y= -6 +3 then
y= -3 which is my answer....ı am confused where i am tripping up Frankly I don't quite understand what is tested here... But your approach is wrong because your method for finding equation of a graph is for straight line (linear equation) and obviously the curve shown on the diagram is not straight line. Is it a graph of quadratic equation? Are the intersection points with axes (0,3) and (2,0)? If the answer on both questions is YES then the equation of the graph would be y=\frac{3}{4}x^2-3x+3 and for x=1 --> y=\frac{3}{4}x^2-3x+3=\frac{3}{4}\approx{1}. I wouldn't worry about this question at all. P.S. Check the link about Coordinate Geometry in my signature for more on this issue. It was important for me that why my approach is wrong,i know now... Thank you Bunuel
_________________
This is not finished here...Watch me.....
|
|
|
|
|
|
|
Re: PS geometery ... little confused.
[#permalink]
16 Aug 2010, 06:26
|
|
|
|
|
|
|
|
|
|
|
close
