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# If x^2 +y^2 is less than 9, is x^2 less than x ? (1) y^2 is

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If x^2 +y^2 is less than 9, is x^2 less than x ? (1) y^2 is [#permalink]

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18 Apr 2007, 07:17
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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.

Kudos [?]: 344 [0], given: 0

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [0], given: 0

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18 Apr 2007, 11:42
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 763 times ]

Fig2_X divided by 3.gif [ 4.32 KiB | Viewed 762 times ]

Kudos [?]: 171 [0], given: 0

Current Student
Joined: 28 Dec 2004
Posts: 3345

Kudos [?]: 325 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

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18 Apr 2007, 12:01
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 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>>> 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).

Kudos [?]: 325 [0], given: 2

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [0], given: 0

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18 Apr 2007, 12:08
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 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>>> 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

Kudos [?]: 171 [0], given: 0

Current Student
Joined: 28 Dec 2004
Posts: 3345

Kudos [?]: 325 [0], given: 2

Location: New York City
Schools: Wharton'11 HBS'12

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18 Apr 2007, 12:12
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 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>>> 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

Kudos [?]: 325 [0], given: 2

Director
Joined: 26 Feb 2006
Posts: 900

Kudos [?]: 165 [0], given: 0

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19 Apr 2007, 12:03
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 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>>> 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

Kudos [?]: 165 [0], given: 0

19 Apr 2007, 12:03
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