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If \(x \gt 1\) and \(y \gt 1\), is \(x \gt y\)?
(1) \(\sqrt{x} \gt y\)
(2) \(\sqrt{y} \lt x\)
(1) \(\sqrt{x} \gt y\)
(2) \(\sqrt{y} \lt x\)
28 Jan 2015, 11:53
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Official Solution:
If \(x \gt 1\) and \(y \gt 1\), is \(x \gt y\)?
Notice that we are told that both \(x\) and \(y\) are greater than 1.
(1) \(\sqrt{x} \gt y\).
Since \(x > 1\), then \(x > \sqrt{x}\). Therefore, \(x > \sqrt{x} > y\) and thus \(x > y\). Sufficient.
(2) \(\sqrt{y} \lt x\).
Since \(y > 1\), then \( \sqrt{y} < y\). However, \(x\) can be between \(\sqrt{y}\) and \(y\), on the number line: \(\sqrt{y}< x < y\) as well as to the right of \(y\), on the number line: \(\sqrt{y}< y < x\). For example, if \(y=4\) and \(x=3\) then the answer to the question is NO but if \(y=4\) and \(x=100\) then the answer to the question is YES. Not sufficient.
Answer: A
If \(x \gt 1\) and \(y \gt 1\), is \(x \gt y\)?
Notice that we are told that both \(x\) and \(y\) are greater than 1.
(1) \(\sqrt{x} \gt y\).
Since \(x > 1\), then \(x > \sqrt{x}\). Therefore, \(x > \sqrt{x} > y\) and thus \(x > y\). Sufficient.
(2) \(\sqrt{y} \lt x\).
Since \(y > 1\), then \( \sqrt{y} < y\). However, \(x\) can be between \(\sqrt{y}\) and \(y\), on the number line: \(\sqrt{y}< x < y\) as well as to the right of \(y\), on the number line: \(\sqrt{y}< y < x\). For example, if \(y=4\) and \(x=3\) then the answer to the question is NO but if \(y=4\) and \(x=100\) then the answer to the question is YES. Not sufficient.
Answer: A
10 Sep 2017, 09:38
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Given : x>1 and y>1, so here we don't have to worry about fractions, negative numbers and value 0 and 1
Now find if x>y
Condition 1:
\(\sqrt{x}\) >y
=> x> \(y^2\)
Now as we know \(y^2\) >y (if y is not equal to 0 and 1)
so => x>\(y^2\) > y
=> x>y . Answer. Sufficient
by putting values:
\(\sqrt{x}\) >y
So we have to find value of \(\sqrt{x}\) is "greater" than y.
Assume y=2 .. => 2.1 > 2
So \(\sqrt{x}\) = 2.1 => x = 4.41 => 4.41 >2 => x>y Sufficient
Condition 2:
\(\sqrt{y}\) < x
=> y <\(x^2\)
=> 3<(3)^2 => but here x=y so Answer x> y answer to main question is No
=> 2< (4)^2 =>Here x> y answer to our main question is yes
=> 9 < (4)^2 => Here x<y so answer to our main question is No
Not sufficient
Answer: A
Now find if x>y
Condition 1:
\(\sqrt{x}\) >y
=> x> \(y^2\)
Now as we know \(y^2\) >y (if y is not equal to 0 and 1)
so => x>\(y^2\) > y
=> x>y . Answer. Sufficient
by putting values:
\(\sqrt{x}\) >y
So we have to find value of \(\sqrt{x}\) is "greater" than y.
Assume y=2 .. => 2.1 > 2
So \(\sqrt{x}\) = 2.1 => x = 4.41 => 4.41 >2 => x>y Sufficient
Condition 2:
\(\sqrt{y}\) < x
=> y <\(x^2\)
=> 3<(3)^2 => but here x=y so Answer x> y answer to main question is No
=> 2< (4)^2 =>Here x> y answer to our main question is yes
=> 9 < (4)^2 => Here x<y so answer to our main question is No
Not sufficient
Answer: A
