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08 Jan 2018, 07:38
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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:
35%
(medium)
Question Stats:
70% (01:23) correct
30%
(01:22)
wrong
based on 73
sessions
History
Date
Time
Result
Not Attempted Yet
If j is an integer, is k an integer?
(1) j = 2k
(2) The average (arithmetic mean) of j and k is not an integer.
(1) j = 2k
(2) The average (arithmetic mean) of j and k is not an integer.
08 Jan 2018, 09:24
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Bunuel
Statement 1: implies \(k=\frac{j}{2}\)
if \(j\) is even, \(k\) is an integer but if \(j\) is Odd then \(k\) is not an integer. Insufficient
Statement 2: implies \(j+k=2*\)non integer. we don't know the values of \(j\) & \(k\). Insufficient
Combining 1 & 2: \(2k+k=2*\)non integer
\(=>k=\frac{2}{3}*\)non integer. if non integer\(=\frac{3}{2}\), then \(k\) is an integer but if non integer \(= \frac{1}{2}\), then \(k\) is not an integer. Insufficient
Option E
08 Jan 2018, 10:54
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If j is an integer, is k an integer?
(1) j = 2k
(2) The average (arithmetic mean) of j and k is not an integer.
Hi,
1.) k=1/2 -----No
k=1------yes
bce
2.) (j+k)/2 isn't an integer
k=1, j=4
Yes K is an integer
k=1/2 j=1
No K isn't an integer.
CE
from 1 and 2 ----- (2k+k)/2 isn't an integer
3k/2 isn't an integer.
this can be valid for k= 1 and k=1/2 Hence E is the answer.
(1) j = 2k
(2) The average (arithmetic mean) of j and k is not an integer.
Hi,
1.) k=1/2 -----No
k=1------yes
bce
2.) (j+k)/2 isn't an integer
k=1, j=4
Yes K is an integer
k=1/2 j=1
No K isn't an integer.
CE
from 1 and 2 ----- (2k+k)/2 isn't an integer
3k/2 isn't an integer.
this can be valid for k= 1 and k=1/2 Hence E is the answer.
