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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
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Bunuel
If \(x^2 +y^2 \leq 25\), is \(x^2 < x\) ?

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

 


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purvibagmar

mainbhiankit
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.
­What abouyt fractions? 
­Could you please demonstrate fractions which would change the answer? Thank you!
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