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GMAT Club World Cup 2022 (DAY 6): If x^2 +y^2 <= 25, is x^2 < x ?
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18 Jul 2022, 08:00
18 Jul 2022, 08:00
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If \(x^2 +y^2 \leq 25\), is \(x^2 < x\) ?
(1) \(y^2 > 9\).
(2) \(x = y + 3\).
(1) \(y^2 > 9\).
(2) \(x = y + 3\).
|
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Re: GMAT Club World Cup 2022 (DAY 6): If x^2 +y^2 <= 25, is x^2 < x ?
[#permalink]
18 Jul 2022, 08:50
18 Jul 2022, 08:50
2
Kudos
Bunuel wrote:
If \(x^2 +y^2 \leq 25\), is \(x^2 < x\) ?
(1) \(y^2 > 9\).
(2) \(x = y + 3\).
(1) \(y^2 > 9\).
(2) \(x = y + 3\).
|
First work on the question stem: \(x^2 < x\) = x(x-1)<0,
S1: \(y^2 > 9\) = y>3 or y<-3. If we substitute any eligible value within this range, it will give x(x-1)>0. Hence Sufficient
S2: \(x = y + 3\), y=x-3. Put it in original equation, we will get Xˆ2 + (x-3)ˆ2 < or equal to 25, xˆ2-x-8< or equal to 0. After solving we will get two value of X which gives x(x-1)>0. Hence Sufficient
Hence Option C is the answer
Re: GMAT Club World Cup 2022 (DAY 6): If x^2 +y^2 <= 25, is x^2 < x ?
[#permalink]
18 Jul 2022, 09:36
18 Jul 2022, 09:36
1
Kudos
Asked: If \(x^2 +y^2 \leq 25\), is \(x^2 < x\) ?
(1) \(y^2 > 9\).
x^2 + y^2 > x^2 + 9 <= 25
x^2 < =16
-4 <=x <=4
If x = .5 ; x^2 - x = .25 - .5 = -.25 < 0
But if x =3; x^2 - x = 9 - 3 = 6 >0
NOT SUFFICIENT
(2) \(x = y + 3\).
y = x-3
x^2 + (x-3)^2 <=25
x^2 + x^2 - 6x + 9 < =25
2x^2 - 6x - 16 <=0
x^2 - 3x - 8 <=0
(x-3/2)^2 -(8+9/4) <=0
(x-3/2)^2 - 41/4 <=0
\(-\frac{\sqrt{41}}{2}<=x<=\frac{\sqrt{41}}{2}\)
If x = .5 ; x^2 - x = .25 - .5 = -.25 < 0
But if x =3; x^2 - x = 9 - 3 = 6 >0
NOT SUFFICIENT
(1) + (2)
(1) \(y^2 > 9\).
(2) \(x = y + 3\).
(x-3)^2 > 9
(x-3)^2 - 3^2 > 0
x(x-6)>0
x > 6 or x<0
x^2 + 9 < x^2 + y^2 <= 25
x^2 < =16
-4 <=x <=4
-4<=x<0
x^2 - x = x(x-1) >0
SUFFICIENT
IMO C
(1) \(y^2 > 9\).
x^2 + y^2 > x^2 + 9 <= 25
x^2 < =16
-4 <=x <=4
If x = .5 ; x^2 - x = .25 - .5 = -.25 < 0
But if x =3; x^2 - x = 9 - 3 = 6 >0
NOT SUFFICIENT
(2) \(x = y + 3\).
y = x-3
x^2 + (x-3)^2 <=25
x^2 + x^2 - 6x + 9 < =25
2x^2 - 6x - 16 <=0
x^2 - 3x - 8 <=0
(x-3/2)^2 -(8+9/4) <=0
(x-3/2)^2 - 41/4 <=0
\(-\frac{\sqrt{41}}{2}<=x<=\frac{\sqrt{41}}{2}\)
If x = .5 ; x^2 - x = .25 - .5 = -.25 < 0
But if x =3; x^2 - x = 9 - 3 = 6 >0
NOT SUFFICIENT
(1) + (2)
(1) \(y^2 > 9\).
(2) \(x = y + 3\).
(x-3)^2 > 9
(x-3)^2 - 3^2 > 0
x(x-6)>0
x > 6 or x<0
x^2 + 9 < x^2 + y^2 <= 25
x^2 < =16
-4 <=x <=4
-4<=x<0
x^2 - x = x(x-1) >0
SUFFICIENT
IMO C
