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If x^2 +y^2 is less than 9, is x^2 less than x ? (1) y^2 is [#permalink]
 18 Apr 2007, 08:17
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If x^2 +y^2 is less than 9, is x^2 less than x ?
(1) y^2 is greater than 8.
(2) x is 3 greater than y.
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I would like to say (E) 
I prefer to solve it with the XY Plan.
x^2 < x ?
<=> x^2 - x < 0 ?
<=> x*(x-1) < 0 ?
<=> 0 < x < 1 ?
x^2 +y^2 < 9
<=> x^2 + y^2 < 3^2 >>> In the XY plan, this represents a surface delimited by the circle with 0(0,0) as center and 3 as radius.
So, 0 < x < 1 creates a part of this circle's surface.
From 1
y^2 > 8
<=> y > 2*qrt(2) or y < -2*qrt(2)
Creating 2 horizontal lines (Fig 1), one at -2*qrt(2) and one at 2*qrt(2), we observe especially for negative values of x that we have a surface out of 0 < x < 1.
Also, we have a part of the surface created by these 2 lines that are in the 0 < x < 1.
INSUFF.
From 2
x=3*y
<=> y = 1/3*x >>> it's a line passing by 0(0,0) (Fig 2).
All negative values of x will create negative values of y. In this case, we are not in the section 0 < x < 1.
Also, we have a part of the line that passes throught 0 < x < 1.
INSUFF.
Both (1) & (2)
Here, looking the 2 conditions and so 2 graphs, there is no common area within the circle.... So, it's pretty weird and we cannot conclude...
I tend to say (E).
Attachments
Fig1_Y power 2 above 8.gif [ 4.32 KiB | Viewed 114 times ]
Fig2_X divided by 3.gif [ 4.32 KiB | Viewed 114 times ]
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Current Student
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statement 1 simply means that y^2 is greater than 8.. or X is less than 1..we dont know if X is positive or negative..therefore insuff..
statement 2..x is 3 greater than y..to me means x=y+3?
not knowing the sign of y or x ..we cant really say anything ..
together..we know that y has to be negative..then we know that x is 3+y; which means x is positive..and is less than 1..this is sufficient..
C it is..
Fig wrote: I would like to say (E) I prefer to solve it with the XY Plan. x^2 < x ? <=> x^2 - x < 0 ? <=> x*(x-1) < 0 ? <=> 0 < x < 1 ? x^2 +y^2 < 9 <=> x^2 + y^2 <3>>> In the XY plan, this represents a surface delimited by the circle with 0(0,0) as center and 3 as radius. So, 0 < x < 1 creates a part of this circle's surface. From 1y^2 > 8 <y> 2*qrt(2) or y < -2*qrt(2) Creating 2 horizontal lines (Fig 1), one at -2*qrt(2) and one at 2*qrt(2), we observe especially for negative values of x that we have a surface out of 0 < x < 1. Also, we have a part of the surface created by these 2 lines that are in the 0 < x < 1. INSUFF. From 2x=3*y <y>>> it's a line passing by 0(0,0) (Fig 2). All negative values of x will create negative values of y. In this case, we are not in the section 0 < x < 1. Also, we have a part of the line that passes throught 0 < x < 1. INSUFF. Both (1) & (2)Here, looking the 2 conditions and so 2 graphs, there is no common area within the circle.... So, it's pretty weird and we cannot conclude... I tend to say (E). |
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fresinha12 wrote: statement 1 simply means that y^2 is greater than 8.. or X is less than 1..we dont know if X is positive or negative..therefore insuff.. statement 2..x is 3 greater than y..to me means x=y+3? not knowing the sign of y or x ..we cant really say anything .. together..we know that y has to be negative..then we know that x is 3+y; which means x is positive..and is less than 1..this is sufficient.. C it is.. Fig wrote: I would like to say (E) I prefer to solve it with the XY Plan. x^2 < x ? <=> x^2 - x < 0 ? <=> x*(x-1) < 0 ? <=> 0 < x < 1 ? x^2 +y^2 < 9 <=> x^2 + y^2 <3>>> In the XY plan, this represents a surface delimited by the circle with 0(0,0) as center and 3 as radius. So, 0 < x < 1 creates a part of this circle's surface. From 1y^2 > 8 <y> 2*qrt(2) or y < -2*qrt(2) Creating 2 horizontal lines (Fig 1), one at -2*qrt(2) and one at 2*qrt(2), we observe especially for negative values of x that we have a surface out of 0 < x < 1. Also, we have a part of the surface created by these 2 lines that are in the 0 < x < 1. INSUFF. From 2x=3*y <y>>> it's a line passing by 0(0,0) (Fig 2). All negative values of x will create negative values of y. In this case, we are not in the section 0 < x < 1. Also, we have a part of the line that passes throught 0 < x < 1. INSUFF. Both (1) & (2)Here, looking the 2 conditions and so 2 graphs, there is no common area within the circle.... So, it's pretty weird and we cannot conclude... I tend to say (E). Yes... Probably, it's not 3 times as I understood it but rather 3 upper than y.
So, we have a line y = x-3 and then make sens to have only a few part of it, at bottom of the circle, that is only between 0 < x < 1 limit and y < -2*sqrt(2).
C it is
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Current Student
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nice graphs by the way
Fig wrote: fresinha12 wrote: statement 1 simply means that y^2 is greater than 8.. or X is less than 1..we dont know if X is positive or negative..therefore insuff.. statement 2..x is 3 greater than y..to me means x=y+3? not knowing the sign of y or x ..we cant really say anything .. together..we know that y has to be negative..then we know that x is 3+y; which means x is positive..and is less than 1..this is sufficient.. C it is.. Fig wrote: I would like to say (E) I prefer to solve it with the XY Plan. x^2 < x ? <=> x^2 - x < 0 ? <=> x*(x-1) < 0 ? <=> 0 < x < 1 ? x^2 +y^2 < 9 <=> x^2 + y^2 <3>>> In the XY plan, this represents a surface delimited by the circle with 0(0,0) as center and 3 as radius. So, 0 < x < 1 creates a part of this circle's surface. From 1y^2 > 8 <y> 2*qrt(2) or y < -2*qrt(2) Creating 2 horizontal lines (Fig 1), one at -2*qrt(2) and one at 2*qrt(2), we observe especially for negative values of x that we have a surface out of 0 < x < 1. Also, we have a part of the surface created by these 2 lines that are in the 0 < x < 1. INSUFF. From 2x=3*y <y>>> it's a line passing by 0(0,0) (Fig 2). All negative values of x will create negative values of y. In this case, we are not in the section 0 < x < 1. Also, we have a part of the line that passes throught 0 < x < 1. INSUFF. Both (1) & (2)Here, looking the 2 conditions and so 2 graphs, there is no common area within the circle.... So, it's pretty weird and we cannot conclude... I tend to say (E). Yes... Probably, it's not 3 times as I understood it but rather 3 upper than y. So, we have a line y = x-3 and then make sens to have only a few part of it, at bottom of the circle, that is only between 0 < x < 1 limit and y < -2*sqrt(2). C it is 
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Director
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good one. i was drawn to B but agree with C.
fresinha12 wrote: nice graphs by the way Fig wrote: fresinha12 wrote: statement 1 simply means that y^2 is greater than 8.. or X is less than 1..we dont know if X is positive or negative..therefore insuff.. statement 2..x is 3 greater than y..to me means x=y+3? not knowing the sign of y or x ..we cant really say anything .. together..we know that y has to be negative..then we know that x is 3+y; which means x is positive..and is less than 1..this is sufficient.. C it is.. Fig wrote: I would like to say (E) I prefer to solve it with the XY Plan. x^2 < x ? <=> x^2 - x < 0 ? <=> x*(x-1) < 0 ? <=> 0 < x < 1 ? x^2 +y^2 < 9 <=> x^2 + y^2 <3>>> In the XY plan, this represents a surface delimited by the circle with 0(0,0) as center and 3 as radius. So, 0 < x < 1 creates a part of this circle's surface. From 1y^2 > 8 <y> 2*qrt(2) or y < -2*qrt(2) Creating 2 horizontal lines (Fig 1), one at -2*qrt(2) and one at 2*qrt(2), we observe especially for negative values of x that we have a surface out of 0 < x < 1. Also, we have a part of the surface created by these 2 lines that are in the 0 < x < 1. INSUFF. From 2x=3*y <y>>> it's a line passing by 0(0,0) (Fig 2). All negative values of x will create negative values of y. In this case, we are not in the section 0 < x < 1. Also, we have a part of the line that passes throught 0 < x < 1. INSUFF. Both (1) & (2)Here, looking the 2 conditions and so 2 graphs, there is no common area within the circle.... So, it's pretty weird and we cannot conclude... I tend to say (E). Yes... Probably, it's not 3 times as I understood it but rather 3 upper than y. So, we have a line y = x-3 and then make sens to have only a few part of it, at bottom of the circle, that is only between 0 < x < 1 limit and y < -2*sqrt(2). C it is |
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