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13 May 2020, 09:51
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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:
55%
(hard)
Question Stats:
54% (01:40) correct
46%
(01:08)
wrong
based on 52
sessions
History
Date
Time
Result
Not Attempted Yet
13 May 2020, 20:38
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Bunuel
(1) \(\frac{1}{x} < \frac{1}{y}\)
If both x and y are positive...x>y..Yes
If x is negative and y positive..y>x..
If both are negative...x>y
Insuff
(2) x + y = 10
One of them is surely positive, but nothing about which one.
If y=6, and x=4..y>x
But if x=6 and y=4...x>y
Combined..
Both x and y can be positive..x=6 and y=4...x>y
But x=-1 and y=11...y>x
E
14 May 2020, 01:54
Kudos
Bookmarks
Bunuel
Solution
Step 1: Analyse Question Stem
- • We need to find out if \(x < y\)
Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE
Statement 1: 1/x < 1/y
- • Let us consider following two cases:
• Case 1: \(\frac{1}{x} = -\frac{1}{2}\) and \(\frac{1}{y} = \frac{1}{2}\), so that \(\frac{1}{x }< \frac{1}{y}\)
- o In this case, \(x = -2\) and \(y = 2 \)
o Hence, \(x < y\)
- o In this case, \(x = 2\) and \(y = \frac{1}{2}\)
o Hence, \(x > y\)
Statement 2: x + y = 10
- • Let us consider the following two cases:
• Case 1: \(x = 1\) and \(y = 9\) so that \(x+y = 10\)
- o In this case \(x < y\)
- o In this case \(x > y\)
Step 3: Analyse Statements by Combining
From statement 1: \(\frac{1}{x} < \frac{1}{y}\)
From statement 2: \(x + y = 10\)
On combining the both statements, out of many possibilities that exist, let us consider below two possible values of x and y such that \(\frac{1}{x} < \frac{1}{y} \)and \(x+y = 10\)
- • Case 1: \(x = 9\) and \(y = 1 \)
- o \(9 +1 = 10\) and \(\frac{1}{9 }< 1\)
o Here, x > y
- o \(-9+19 = 10\) and \(-\frac{1}{9} < \frac{1}{19}\)
o Here, x < y
Thus, the correct answer is Option E
