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31 Jan 2012, 17:24
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
38% (02:44) correct
62%
(02:51)
wrong
based on 868
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If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?
(1) x + y = 8z +1
(2) x – y = 2z – 1
(1) x + y = 8z +1
(2) x – y = 2z – 1
OA is B. But I got stuck after a while. Please see below my solution.
From question stem
z = \sqrt{x}
i.e. x = \(z^2\) ---------------------------------------------(1)
Considering Statement 1
y = 8z-x+1 -----------------------------------------------------(2)
Putting some values of x and z
When x =2 , z =4
Substituting these values in Equation 2 we will get y = 13. But this cannot be applied as x >y. So I tried another values which are
x = 25 then z =5
Substituting these values in Equation 2 we will get y = 16. Therefore , YES x and y are perfect squares.
I tried one more value which was
when x = 36, z =6
Substituting these values in Equation 2 we will get y = 13. Therefore , NO x and y are NOT perfect squares.
Two conclusions, therefore this statement alone is NOT sufficient.
Considering Statement 2
After simplifying we will get y = \((z-1)^2\)
Therefore z-1 = \sqrt{y}
But from statement 1 we can say z = \sqrt{x}
Therefore, \sqrt{x} - 1 = \sqrt{y}
\sqrt{x} = \sqrt{y} + 1
I am stuck after this. Can someone please help?
From question stem
z = \sqrt{x}
i.e. x = \(z^2\) ---------------------------------------------(1)
Considering Statement 1
y = 8z-x+1 -----------------------------------------------------(2)
Putting some values of x and z
When x =2 , z =4
Substituting these values in Equation 2 we will get y = 13. But this cannot be applied as x >y. So I tried another values which are
x = 25 then z =5
Substituting these values in Equation 2 we will get y = 16. Therefore , YES x and y are perfect squares.
I tried one more value which was
when x = 36, z =6
Substituting these values in Equation 2 we will get y = 13. Therefore , NO x and y are NOT perfect squares.
Two conclusions, therefore this statement alone is NOT sufficient.
Considering Statement 2
After simplifying we will get y = \((z-1)^2\)
Therefore z-1 = \sqrt{y}
But from statement 1 we can say z = \sqrt{x}
Therefore, \sqrt{x} - 1 = \sqrt{y}
\sqrt{x} = \sqrt{y} + 1
I am stuck after this. Can someone please help?
31 Jan 2012, 17:41
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If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?
If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers. So, the question becomes: is \(\sqrt{y}=\sqrt{x}-1\)? --> square both sides: is \(y=x-2\sqrt{x}+1\)?
(1) x + y = 8z +1 --> \(x+y=8\sqrt{x}+1\) --> \(y=8\sqrt{x}+1-x\). Now, question becomes is \(8\sqrt{x}+1-x=x-2\sqrt{x}+1\)? --> is \(10\sqrt{x}=2x\)? --> is \(5\sqrt{x}=x\)? --> is \(\sqrt{x}=5\)? --> is \(x=25\)? But we don't know that, hence insufficient.
(2) x – y = 2z – 1 --> \(y=x-2\sqrt{x}+1\), which is exactly what we needed to know. Sufficient.
Answer: B.
Hope it's clear.
If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers. So, the question becomes: is \(\sqrt{y}=\sqrt{x}-1\)? --> square both sides: is \(y=x-2\sqrt{x}+1\)?
(1) x + y = 8z +1 --> \(x+y=8\sqrt{x}+1\) --> \(y=8\sqrt{x}+1-x\). Now, question becomes is \(8\sqrt{x}+1-x=x-2\sqrt{x}+1\)? --> is \(10\sqrt{x}=2x\)? --> is \(5\sqrt{x}=x\)? --> is \(\sqrt{x}=5\)? --> is \(x=25\)? But we don't know that, hence insufficient.
(2) x – y = 2z – 1 --> \(y=x-2\sqrt{x}+1\), which is exactly what we needed to know. Sufficient.
Answer: B.
Hope it's clear.
General Discussion
10 Mar 2012, 03:13
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Isn't this question a good candidate for plugging in simple numbers? I plugged in x=4 and y=1 and got the answer pretty quickly.
