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25 Jan 2020, 23:52
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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:
42% (01:20) correct
58%
(01:06)
wrong
based on 55
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(\frac{m^2}{n} \)>0?
1) \(m-2n>1\)
2) \(2<n<7\)
1) \(m-2n>1\)
2) \(2<n<7\)
GMATbuster's Collection
26 Jan 2020, 00:16
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C
Statement 1 is insufficient because we do not know the value of n.n can be +ve or -ve.
Statement 2 is insufficient because we do not know either m is 0 or +ve. Because squares are non negative
1 + 2 provides the solution.
Posted from my mobile device
Statement 1 is insufficient because we do not know the value of n.n can be +ve or -ve.
Statement 2 is insufficient because we do not know either m is 0 or +ve. Because squares are non negative
1 + 2 provides the solution.
Posted from my mobile device
26 Jan 2020, 01:52
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#1
\(m-2n>1\)
value of n could be +/-ve and then we get both yes & no to \(\frac{m^2}{n} \)>0?
insufficient
#2
\(2<n<7\)
value of m not know insufficient
from1&2
we can say \(\frac{m^2}{n} \)>0 sufficient
IMO C
\(m-2n>1\)
value of n could be +/-ve and then we get both yes & no to \(\frac{m^2}{n} \)>0?
insufficient
#2
\(2<n<7\)
value of m not know insufficient
from1&2
we can say \(\frac{m^2}{n} \)>0 sufficient
IMO C
gmatbusters
Is \(\frac{m^2}{n} \)>0?
1) \(m-2n>1\)
2) \(2<n<7\)
1) \(m-2n>1\)
2) \(2<n<7\)
GMATbuster's Collection
