GMAT Club World Cup 2022 (DAY 6): If x^2 +y^2 <= 25, is x^2 < x ?

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

If (x)^2 + (y)^2 <= 25, we can also infer that:

0 <= (x)^2 + (y)^2 <= 25

The first statement says: y^2 > 9
Let's assume y^2 = 16
What if the sum of squares of x and y is 16? x^2 = 0. x = 0. x^2 is not less than x.
What if the sum of sq of x and y is 25? x^2 = 16. x = -4, or 4. x^2 > x
We can't say for certain. Hence, insufficient

The second statement says: x = y + 3.
What if y = -3? x^2 = 0. x = 0. x^2 is not less than x.
What if y = -1 or 1? x^2 = 16. x = -4, or 4. x^2 > x
We can't say for certain. Hence, insufficient

Combining both statements,
we get that y^2 > 9 and x = y + 3.
Therefore, y > 3, y < -3
Therefore, x > 6, x < -1

Assuming x = -1 (even though it's less than that), x^2 = 1, which helps us conclude that x^2 > x.

Hence, this is sufficient. The answer is option C.

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

the the correct answer is E
the hypothesis gives us 0<x<1 ??
statement 1: gives information on y>3 or y<-3 which means x could be >1 and x could be between 0 and 1
statement 2: gives y=x+3 , so x= y-3 so if y =3.5 x could be 0.5 and it includes 0<x<1 but if y= -3.5 then x=-6.5 and the hypothesis is wrong
statement 1 and 2, if if y =3.5, x=0.5 and it includes 0<x<1 but if y= -3.5 then x=-6.5 and x2>x
therefore, E is the answer

Correct answer : Choice E

we are trying to find if x(x-1) < 0
for example : if x = 0.5 , then yes x(x-1) is in fact less than 0. If x is 1, then NO x(x-1) = 0

From question stem and statement 1, we dont get to conclusively say whether x is a decimal or a whole number. x also satisfies multiple answers for which it can be YES and NO. Hence options A,D is out

From question stem and statement 2, we dont get to conclusively say whether x is a decimal or a whole number. x also satisfies multiple answers for which it can be YES and NO. Hence option B is out

From both statement 1 and 2 , it still unclear as x can take multiple values for which it can be YES and NO. Hence C is out.

So the correct answer is choice E

x^2+y^2 ≤ 25

if x^2<x ? this will be true for 0<x<1..

(1) y^2>9
from x^2+y^2 ≤ 25
x^2 ≤ 25 - y^2
that is we can deduce 9 < y^2 ≤ 25
and x^2 <= 9

Not sufficient to know value of x

st2 x=y+3.
y = x -3
Squaring both sides
y^2 = x^2 - 6x + 9
adding x^2 to both sides and simplifying
2x^2 - 6x -16 ≤ 0
or x^2 - 3x -8 ≤ 0
x ≤-1.7 or x ≤ 4.7
not sufficient
st 1 + st 2
x^2 ≤ 25 - y^2
9 < y^2 ≤ 25
and x^2 <= 9 .. eqn 1
x ≤-1.7 or x ≤ 4.7
but from eqn 1 we see that x<= 4.7 is not possible
hence x<= -1.7

therefore x^2 is not less than x
Hence combined statement is correct
Answer C

Given:
\(x^2 + y^2 \leq 25\)

Ques:
\(x^2 < x\)

\(x^2 - x < 0\)

\(x(x-1) < 0\)

In short the question is asking if 0 < x < 1

Statement 1

From this statement we know that -

\(y^2 - 9 > 0\)

y > 3 or y < -3

We do not have any information on x. Hence we can eliminate this statement

Statement 2

x = y + 3

Case 1

Say y = -2.5 ; x = -2.5 + 3 = 0.5

Also, \((-2.5)^2 + (0.5)^2 \leq 25\)

Is 0 < x < 1 -- Yes

Case 2

Say y = 0 ; x = 0 + 3 = 3

Also, \((3)^2 + (0)^2 \leq 25\)

Is 0 < x < 1 -- No

Combining

From Statement 1, we know that y > 3 or y < -3

So y+3 > 6 or y+3 < 0.

Therefore x > 6, this is not possible or x < 0, this is possible.

Is 0 < x < 1 -- No

IMO C

from 1 - value of y can be from 3 to 5; so value of x can be from 4 to 0

not sufficient

from 2 - value of x must be grater than 1 for \(x^2 +y^2 \leq 25\) ;
sufficient

Answer B

If x^2+y^2≤25, is x^2<x ?

x^2 - x < 0
x(x-1) < 0
=> if x is -ve i.e., x <0 then x-1 >0, but this is not possible
=> if x is +ve then x-1<0 i.e., x<1 => 0<x<1
Hence, the question is -> is 0<x<1 ?


(1) y^2>9
x^2+y^2≤25
x^2 ≤ 25 - y^2
=> x^2 < 16
=> -4<x<4

(2) x=y+3 => y =x-3
x^2+y^2 ≤ 25
x^2 + x^2-6x+9 ≤ 25
2x^2-6x-16≤0
x(x-3) ≤ 8

Combining both we still don't know whether 0<x<1.

Answer: E

IMO option C is the answer.

Statement 1 Alone Not Sufficient, for example x can be 0, 1 or 2.

Statement 2 Alone, Not sufficient, for example x can be 0, 1 or 2.

Statement 1 and Statement 2 Combined.
if y is +ve, then x is at least 6, which contradicts the statement in the question.
I can conclude that y is then -ve and that y must be less than -3.
I can also conclude that x must be -ve.
Therefore I can conclude that the answer to is \(x^2 < x\) ? is FALSE using Statement 1 and Statement 2.
Therefore option C is the answer IMO.

Question: Is \(x^2 < x\) ?

In other words, Is 0 < x <1 ?

S1:
"\(y^2 > 9\)."

Clearly, insufficient. With "\(y^2 > 9\)." x could be anything that suffice the equation x^2+y^2<=25.
INSUFFICIENT.

S2:
"\(x = y + 3\)."
Again x can be anything:
When y=1, x=4 x^2+y^2=17<25, here x > 1
also when y=-2.5 x=0.5 x^2+y^2=6.5<25 here 0<x<1
INSUFFICIENT

S1+S2:
From S1, we have y < -3 OR y > 3
Thus from S2, we have x < 0 OR x > 6.
Clearly, x is NOT in between 0 and 1.
Thus together statements are
SUFFICIENT

ANS C

\(x^2 < x\) => \(0 < x < 1\)

(1) \(y^2 > 9\). =>\( -3 < y < 3 \) =>\( -2 < x < 2 \) => x can be -1 or 0.9 Insufficient
(2) \(x = y + 3\). => x can be -1, 0.9 Insufficient

Both
\(x^2 -6x - 16 \leq 0\) => \(-1.5 \leq x \leq 4.7\) .... Insufficient

Ans E

[quote="Bunuel"]If \(x^2 +y^2 \leq 25\), is \(x^2 < x\) ?

(1) \(y^2 > 9\).
(2) \(x = y + 3\).

1)Given x^2+y^2<=25,y^2>9
We have to prove if x^2<x
x^2<x - this is only possible if 0<x<1
Now if y^2>9 means y<-3 or y>3 if y=4 then x can be +3 ,x>0 --1
or x can be -3,x<0--2
Or if y=4 and x=.5 , 0<x<1--3

In all these cases(4,3) ,(4,-3),(4,.5) x^2+y^2<=25 is getting satisfied
Thus for the cases 1,2 x^2>x while for case 3 x^2<x
Thus statement 1 is insufficient.
2)x=y+3 ,if we consider y=1 then x=4
for x^2+y^2<=25 . x^2>x is this case
if we consider y=-2.5 then x=0.5 for x^2+y^2<=25
Thus statement 2 is insufficient.
Combining statement 1 and 2
y^2>9 => y>3 or y<-3
Now x=y+3
Then x>6 or x<0
as 0<x<1 does not fall in this range ,thus x2<x is not true and combining both the conditions are sufficient.
C is the answer

1) Insufficient
2) Sufficient

Ans - B

Solution attached

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Answer: C

If x^2+y^2≤25, is x^2<x ?

x^2 < x
=> x(x-1) < 0
0 < x < 1 ?

x^2 + y^2 <= 25
=> points will cover the area of circle with centre (0,0) and radius 5.

(1) y^2>9.
=> y < -3 or y > 3
If we consider the circle, clearly we will have x as points both inside and outside the range of 0 and 1 satisfying the conditions.
y = 3.1, x = 0.5 or 1.1, we get two results.
Insufficient.

(2) x=y+3.
x = 0.5, y = -2.5; satisfies given condition
x = 1.5, y = -1.5; satisfies given condition
x can be both inside and outside the range.
Insufficient.

Both statements together.
When y > 3, x > 6, and x^2 > x
When y < -3, x < 0, and x^2 > x
So, x^2 will be greater than x.
Sufficient.

Statement 1: Not sufficient. It only talks about Y, we can't say anything about X.X can be 1, 0.5 or 2 = all have different values between x2 and x

Statement 2: not sufficient, same as above.

Combining together: y2> 9 that means y>3 or y <-3. When used with the second statement: x can be greater than 6 (x2 greater than x) or x can be less than 0 (again,x2 greater than x). This is enough to answer the question.

Correct choice C.

Imo C

We have to prove: x^2< x
Solving further: 0<x<1

Given x2+y2≤25 ----(1)
Statement1: y2>9, which means y>3 or y<-3 ---- (2)
Using (1) and (2)

x2≤25- y2
when y>3 or y<-3 then x2<16 or -4 <x< 4 and the range of x will decrease further more.
Insufficient as x can take any values within (0,1) and outside.


Statement2: x=y+3 ----(3)
Putting eq 3 in eq 1
x^2+(x-3)^2 <=25
x2-8- 3x<=0
Solving for x we can get an minimum range as 4.5<x <-1.5
Insufficient

Combining both the statements
From statement 1 we have y>3 or y<-3
So from statement 2 we have when y>3 then x>6 and when y<-3 then x<0 ---(4)
x>6 is not possible from statement 1 and 2, while x<0 is possible.
So we are getting a definite 'No' as x doesn't lie between (0,1)
Sufficient