|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
16 Sep 2014, 01:53
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
40% (02:22) correct
60%
(02:07)
wrong
based on 65
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x\) and \(y\) are positive integers and \(( \sqrt{x} )^y = y\), then \(x - y\) could equal which of the following?
A. -2
B. -1
C. 1
D. 2
E. 4
A. -2
B. -1
C. 1
D. 2
E. 4
16 Sep 2014, 01:53
Kudos
Bookmarks
Official Solution:
If \(x\) and \(y\) are positive integers and \(( \sqrt{x} )^y = y\), then \(x - y\) could equal which of the following?
A. -2
B. -1
C. 1
D. 2
E. 4
Since both \(x\) and \(y\) are restricted to positive integers, we should start with low positive integers.
If \(x = 1\) and \(y = 1\), then we have \(( \sqrt{1} )^1 = 1\), which is true. Thus, \(x - y\) could equal \(1 - 1 = 0\). However, 0 is not among our answer choices, so we must keep looking.
If \(x = 1\), then \(\sqrt{x}\) raised to any power is 1. Thus, no other values of \(y\) (besides 1) could equal that result, and we should look at other possible values of \(x\) and \(y\).
If \(x = 2\), then we have two possible values of \(y\):
\(y = 2\): \(( \sqrt{2} )^2 = 2\)
\(y = 4\): \(( \sqrt{2} )^4 = 4\)
In the first case, we have \(x - y = 2 - 2 = 0\) again. However, in the second case, we have \(x - y = 2 - 4 = -2\).
As it turns out, there are no other possible combinations of \(x\) and \(y\), but we don't need to prove that; we can stop here.
Answer: A
If \(x\) and \(y\) are positive integers and \(( \sqrt{x} )^y = y\), then \(x - y\) could equal which of the following?
A. -2
B. -1
C. 1
D. 2
E. 4
Since both \(x\) and \(y\) are restricted to positive integers, we should start with low positive integers.
If \(x = 1\) and \(y = 1\), then we have \(( \sqrt{1} )^1 = 1\), which is true. Thus, \(x - y\) could equal \(1 - 1 = 0\). However, 0 is not among our answer choices, so we must keep looking.
If \(x = 1\), then \(\sqrt{x}\) raised to any power is 1. Thus, no other values of \(y\) (besides 1) could equal that result, and we should look at other possible values of \(x\) and \(y\).
If \(x = 2\), then we have two possible values of \(y\):
\(y = 2\): \(( \sqrt{2} )^2 = 2\)
\(y = 4\): \(( \sqrt{2} )^4 = 4\)
In the first case, we have \(x - y = 2 - 2 = 0\) again. However, in the second case, we have \(x - y = 2 - 4 = -2\).
As it turns out, there are no other possible combinations of \(x\) and \(y\), but we don't need to prove that; we can stop here.
Answer: A
07 Jul 2015, 11:46
Kudos
Bookmarks
What are the arrows?
