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04 Nov 2019, 03:26
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E
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:
48% (02:07) correct
52%
(01:54)
wrong
based on 347
sessions
History
Date
Time
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x and y are positive integers. Is \(\sqrt{x}\)*\(\sqrt{y}\) an integer?
(1) \(\frac{x}{y} = \frac{1}{n^2}\)
(2) \(\sqrt[3]{x}*\sqrt[3]{y}\) is an integer
Are You Up For the Challenge: 700 Level Questions
(1) \(\frac{x}{y} = \frac{1}{n^2}\)
(2) \(\sqrt[3]{x}*\sqrt[3]{y}\) is an integer
Are You Up For the Challenge: 700 Level Questions
04 Nov 2019, 05:39
Kudos
Bookmarks
x and y are positive integers. Is \(\sqrt{x}\)*\(\sqrt{y}\) an integer?
(1) \(\frac{x}{y} = \frac{1}{n^2}\)
(2) \(\sqrt[3]{x}*\sqrt[3]{y}\) is an integer
n could be anything so not only statements are individually insufficient but are insufficient combined too..
combined..
x=8, y=8.....
\(\frac{x}{y} =1= \frac{1}{n^2}...n=1.. or....n= -1\)
\(\sqrt[3]{x}*\sqrt[3]{y}=2*2\) is an integer
\(\sqrt{x}\)*\(\sqrt{y}=8\)...Yes
x=8, y=64.....
\(\frac{x}{y} =\frac{8}{64}= \frac{1}{n^2}...n^2=8.. ....n= -2\sqrt{2}...n=2\sqrt{2}\)
\(\sqrt[3]{x}*\sqrt[3]{y}=2*4\) is an integer
\(\sqrt{x}\)*\(\sqrt{y}=\sqrt{8*64}=16\sqrt{2}\)...No
E
(1) \(\frac{x}{y} = \frac{1}{n^2}\)
(2) \(\sqrt[3]{x}*\sqrt[3]{y}\) is an integer
n could be anything so not only statements are individually insufficient but are insufficient combined too..
combined..
x=8, y=8.....
\(\frac{x}{y} =1= \frac{1}{n^2}...n=1.. or....n= -1\)
\(\sqrt[3]{x}*\sqrt[3]{y}=2*2\) is an integer
\(\sqrt{x}\)*\(\sqrt{y}=8\)...Yes
x=8, y=64.....
\(\frac{x}{y} =\frac{8}{64}= \frac{1}{n^2}...n^2=8.. ....n= -2\sqrt{2}...n=2\sqrt{2}\)
\(\sqrt[3]{x}*\sqrt[3]{y}=2*4\) is an integer
\(\sqrt{x}\)*\(\sqrt{y}=\sqrt{8*64}=16\sqrt{2}\)...No
E
14 Feb 2020, 00:37
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Bunuel
Solution:
Step 1: Analyse Question Stem
- • \(x\) and \(y\) are positive integers.
- o \(x\) can be \(1, 2, 3…so\ on\), and \(y\) can be \(1, 2, 3…so\ on\)
- • \(√x*√y=√(x*y),\) we need to check whether \(x*y\) is a perfect square or not.
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: \(\frac{x}{y} = \frac{1}{n^2}\)
- • \(x*n^2 = y*1\)
- o \(y = n^2*x\)
- • \(√x*√(n^2*x)=n^2*√x*√x=n^2*x\)
• Note: - After this step, most of the students will consider statement 1 to be sufficient, but is it so? let’s see
- o If n is an integer, then the above expression is also an integer.
- If \(n = 2\) and \(x = 3\), then \(n^2*x=2^2*3=4*3=12\), which is an integer
- If \( n = 0.3\) and \(x = 3\), then \(0.3^2*3=0.09*3=0.27\), which is not an integer.
Statement 2: \(∛x*∛y=∛(x*y)\)
- • \(x*y\) is a perfect cube.
- o But is \(x*y\) a perfect square?
- o Example 1: Take \(x= 1\) and \(y= 1\), then \(x*y= 1\) is a perfect cube and \(1\) is also a perfect square.
o Example 2: Take \(x=2\) and \(y=4\), then \(x*y=8\) is a perfect cube, but it is not a perfect square.
- We don’t know whether \(x*y\) is a perfect square or not.
Step 3: Analyse Statements by combining
From statement 1: \(√x*√y=n^2*x\)
From statement 2: \(x*y\) is a perfect cube.
With both the statements combined, we can neither infer if \(n\) is an integer nor can we infer if \(x*y\) is a perfect square.
Hence, the correct answer is Option E.
