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16 Sep 2014, 00:18
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16 Sep 2014, 00:18
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Official Solution:
Is \(x \gt y^2\)?
(1) \(x \gt y+5\).
Rewrite as \(x-y \gt 5\). This is clearly insufficient, for example: if \(x=1\) and \(y=-10\), then the answer is NO, but if \(x=10\) and \(y=1\), then the answer is YES. We have two different answers, so this is not sufficient.
(2) \(x^2-y^2=0\).
Rewrite as \((x-y)(x+y)=0\): so either \(x-y=0\) or \(x+y=0\). This is also insufficient: if \(x=1\) and \(y=1\), then the answer is NO, but if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. We have two different answers, so this is not sufficient.
(1)+(2) Since we know from (1) that \(x-y \gt 5\), we can conclude that \(x-y \neq 0\). Therefore, based on (2), we must have \(x+y=0\). This means that \(y=-x\), and we can rewrite the question as: is \(x \gt x^2\)?
Now, let's substitute \(y=-x\) in (1): \(x \gt -x + 5\). From this inequality, we can infer that \(x \gt \frac{5}{2}\). Since \(x \gt \frac{5}{2}\), it implies that \(x^2 \gt x\) (because for any \(x\) greater than 1, \(x^2 \gt x\)). Therefore, the answer to the question of whether \(x \gt y^2\) is NO. Sufficient.
Answer: C
Is \(x \gt y^2\)?
(1) \(x \gt y+5\).
Rewrite as \(x-y \gt 5\). This is clearly insufficient, for example: if \(x=1\) and \(y=-10\), then the answer is NO, but if \(x=10\) and \(y=1\), then the answer is YES. We have two different answers, so this is not sufficient.
(2) \(x^2-y^2=0\).
Rewrite as \((x-y)(x+y)=0\): so either \(x-y=0\) or \(x+y=0\). This is also insufficient: if \(x=1\) and \(y=1\), then the answer is NO, but if \(x=\frac{1}{2}\) and \(y=\frac{1}{2}\), then the answer is YES. We have two different answers, so this is not sufficient.
(1)+(2) Since we know from (1) that \(x-y \gt 5\), we can conclude that \(x-y \neq 0\). Therefore, based on (2), we must have \(x+y=0\). This means that \(y=-x\), and we can rewrite the question as: is \(x \gt x^2\)?
Now, let's substitute \(y=-x\) in (1): \(x \gt -x + 5\). From this inequality, we can infer that \(x \gt \frac{5}{2}\). Since \(x \gt \frac{5}{2}\), it implies that \(x^2 \gt x\) (because for any \(x\) greater than 1, \(x^2 \gt x\)). Therefore, the answer to the question of whether \(x \gt y^2\) is NO. Sufficient.
Answer: C
15 Nov 2014, 13:01
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Is \(x > y^2\)?
1) x > y+5
Instead of working with the inequality, assume that x = y+5 and keep in mind that x can have a slightly greater magnitude than that estimated (for instance +0,1) or a definitely greater one (for instance +5). Pick some values for y and compute x. If y = 1; x = 6 (precisely anything > 6) then x is greater than \(y^2\). Pick y = -6; x = -1 then \(y^2\) is greater than x. Not sufficient.
2) \(x^2 - y^2 = 0\) → x = ± y. Pick some values for x. If x = 5; y = ± 5 then x is smaller than \(y^2\). If x = 0,5; y = ± 0,5 then x is greater than \(y^2\). Not sufficient.
1+2) plug x=-y into the first inequality, you will get that y < -2,5. Say that y = -2,6 then x =2,6 and x is always smaller than \(y^2\). If two numbers are equal the question translates into: is \(x > x^2\)? Which happens only when 0 < x < 1.
Sufficient → answer C.
1) x > y+5
Instead of working with the inequality, assume that x = y+5 and keep in mind that x can have a slightly greater magnitude than that estimated (for instance +0,1) or a definitely greater one (for instance +5). Pick some values for y and compute x. If y = 1; x = 6 (precisely anything > 6) then x is greater than \(y^2\). Pick y = -6; x = -1 then \(y^2\) is greater than x. Not sufficient.
2) \(x^2 - y^2 = 0\) → x = ± y. Pick some values for x. If x = 5; y = ± 5 then x is smaller than \(y^2\). If x = 0,5; y = ± 0,5 then x is greater than \(y^2\). Not sufficient.
1+2) plug x=-y into the first inequality, you will get that y < -2,5. Say that y = -2,6 then x =2,6 and x is always smaller than \(y^2\). If two numbers are equal the question translates into: is \(x > x^2\)? Which happens only when 0 < x < 1.
Sufficient → answer C.
