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Bunuel
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If e is a nonzero integer, is 1/2^e greater than or less than 1? 05 Mar 2019, 05:16
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A
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54% (01:28) correct 46% (01:40) wrong based on 119 sessions

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If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0
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A
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If e is a nonzero integer, is 1/2^e greater than or less than 1? 05 Mar 2019, 05:24
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If e is less than 0 the value of 1/2^e can be greater than 1
1. E>-1
So e>=1
1/2^e greater than 1--> no
So statement 1 is giving us the answer to the question
2. E^2>0
E can be either negative or positive
So statement 2 is not sufficient to ans the question

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If e is a nonzero integer, is 1/2^e greater than or less than 1? 05 Mar 2019, 10:27
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Bunuel
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0
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#1
-e<1
means e=-ve integer
so
1/2^e will always be >1
sufficient
#2
e^2 > 0
e can be +ve or -ve so 1/2^e can be > 1 or <1 ; not sufficient
IMO A
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If e is a nonzero integer, is 1/2^e greater than or less than 1? Updated on: 05 Mar 2019, 10:58
Originally posted by ocelot22 on 05 Mar 2019, 10:46.
Last edited by ocelot22 on 05 Mar 2019, 10:58, edited 1 time in total.
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Bunuel
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0
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(1) −e < 1
(2) e^2 > 0[/quote]


Ok (1/2) ^ e, where e does not equal 0 and e is an integer will be greater than 1 if e <0 (a fraction to a negative exponent gives us 1 divided by some number less than 1, and thus we get a number greater than 1). So the question simplifies to is e <0 subject to the constraint e is an integer >0 Yes/No?

(1) -e <1 gives us e >-1 |e is an integer not equal to 0 --> This means that e = 1,2,3, thus e>0, thus we can mark sufficient.

(2) e^2 >0. Therefore either e>0 or e <0 | e is an integer not equal to 0. First consider e >0. Then e = 1,2,3... thus we get a YES answer. Now lets consider e <0 | e is an integer not equal to 0. Then e = -1,-2,-3...., thus e <0 giving us a NO answer. Thus (2) is NS
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If e is a nonzero integer, is 1/2^e greater than or less than 1? 05 Mar 2019, 10:50
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ocelot22

Please see question : e is a nonzero integer
in #1 you have considered e as a fraction....

ocelot22
Bunuel
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0


Ok (1/2) ^ e, where e does not equal 0 and e is an integer will be greater than 1 if e <0 (a fraction to a negative exponent gives us 1 divided by some number less than 1, and thus we get a number greater than 1). So the question simplifies to is e <0 Yes/No?

(1) -e <1 gives us e >-1 Lete =-1/2. Then (1/2)^(-1/2) = 1/((1/2)^1/2) >1 so we get a YES. Let e =2, then (1/2)^2 = 1/4 and we get a NO. Since we get a Yes and a No, we can mark NS

(2) e^2 >0. Therefore either e>0 or e <0. Let e =-1, then we get a YES answer (since (1/2)^-1 = 2). Let e = 1. then we get (1/2)^1 = 1/2 and we get a No answer NS

(1) and (2) e>-1 and either e>0 or e<0. First lets consider e >-1 subject to e>0. Suppose e = 1. then we get a No answer (since 1/2^1 = 1/2). Now lets consider e >-1 subject to e<0. This gives -1<e<0. Let e = -1/2. Then We get a Yes answer (since 1/2^(-1/2) = 1/((1/2)^1/2) > 0). Since we can get a Yes answer from one case and a no answer from the other case, statements (1) and (2) combined are Not Sufficient.

The answer is E.
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If e is a nonzero integer, is 1/2^e greater than or less than 1? 05 Mar 2019, 10:56
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ocelot22

Please see question : e is a nonzero integer
in #1 you have considered e as a fraction....

ocelot22
Bunuel
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0


Ok (1/2) ^ e, where e does not equal 0 and e is an integer will be greater than 1 if e <0 (a fraction to a negative exponent gives us 1 divided by some number less than 1, and thus we get a number greater than 1). So the question simplifies to is e <0 subject to the constraint e is an integer >0 Yes/No?

(1) -e <1 gives us e >-1 |e is an integer not equal to 0 --> This means that e = 1,2,3, thus e>0, thus we can mark sufficient.

(2) e^2 >0. Therefore either e>0 or e <0 | e is an integer not equal to 0. First consider e >0. Then e = 1,2,3... thus we get a YES answer. Now lets consider e <0 | e is an integer not equal to 0. Then e = -1,-2,-3...., thus e <0 giving us a NO answer. Thus (2) is NS



The answer is A.
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If e is a nonzero integer, is 1/2^e greater than or less than 1? 13 Dec 2019, 01:25
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Archit3110
Bunuel
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0

#1
-e<1
means e=-ve integer
so
1/2^e will always be >1
sufficient
#2
e^2 > 0
e can be +ve or -ve so 1/2^e can be > 1 or <1 ; not sufficient
IMO A
Show more

At the end of #1 you say means e=-ve integer.
Shouldn't it be the other way around (e=+ve integer)?
But the answer remains same for either of them. Just wanted to clarify..
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