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29 Sep 2021, 01:00
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E
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Difficulty:
95%
(hard)
Question Stats:
37% (02:04) correct
63%
(01:48)
wrong
based on 138
sessions
History
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If x is the square of an integer, is y the square of an integer?
(1) xy is the square of an integer
(2) x/y is the square of an integer
(1) xy is the square of an integer
(2) x/y is the square of an integer
30 Sep 2021, 02:48
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Bunuel
1. This is a Y/N DS, hence we need a confirmed Y or N to mark a statement as sufficient
2. We are not told that \(x\) and \(y\) are distinct, hence we can assume same values to both
3. Similarly, we are not told that \(y\) is strictly an integer, hence we can assign it fractions
S1: xy is the square of an integer [Insufficient]
1. \(x = 4\) and \(y = \frac{1}{4}\) then \(xy = 1\) or the square of integer \(1\), but \(y\) is not the square of an integer
2. \(x = 4\) and \(y = 4\) then \(xy = 16\) or the square of integer \(4\), and \(y\) is also the square of an integer
S2: x/y is the square of an integer [Insufficient]
1. \(x = 4\) and \(y = \frac{1}{4}\) then \(\frac{x}{y} =\) \(16\) or the square of integer \(4\), but \(y\) is not the square of an integer
2. \(x = 4\) and \(y = 4\) then \(\frac{x}{y} = 1\) or the square of integer \(1\), and \(y\) is also the square of an integer
S1 + S2 [Insufficient]
1. \(x = 4\) and \(y = \frac{1}{4}\) then \(\frac{x}{y} = 16\) and \(xy = 1\) both of which are squares of integers, but \(y\) is not the square of an integer
2. \(x = 4\) and \(y = 4\) then \(\frac{x}{y} = 1\) and \(xy = 16\) both of which are squares of integers, and \(y\) is also the square of an integer
Ans. E
30 Sep 2021, 09:35
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Bunuel
Statement 1:
\(x = i^2\) and \(xy = j^2\), where i and j are integers. Then y = \((\frac{j}{i})^2\), but we don't know if \(\frac{j}{i}\) is an integer. Insufficient.
Statement 2:
\(x = i^2\) and \(\frac{x}{y} = k^2\), where i and j are integers. Then \(y = \frac{i^2}{k^2} = (\frac{i}{k})^2\). Yet we don't know if \(\frac{i}{k}\) is an integer, insufficient.
Combined:
Combined we still cannot confirm if y is an integer. For example, let \(x = 15^2\) and \(y = (\frac{3}{5})^2\). Multiplying x by y or dividing by y each gives us a square as a result, but y is still not an integer.
Ans: E
