purvibagmar wrote:
mainbhiankit wrote:
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 optionC.
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 optionC.
What abouytfractions?
Could you please demonstrate fractions which would change the answer? Thank you!
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Statistics : Posted by Bunuel • on 18 Jul 2022, 08:00 • Replies 20 • Views 3928









