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Is b<0?

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Is b<0?  [#permalink]

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New post 27 Jan 2015, 07:44
4
12
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A
B
C
D
E

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Question Stats:

43% (01:19) correct 57% (01:10) wrong based on 382 sessions

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Re: Is b<0?  [#permalink]

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New post 02 Feb 2015, 02:50
Bunuel wrote:
Is b<0?

(1) Integer b is a multiple of 13
(2) b^3 < 13

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

E. The key to this problem is recognizing that b could equal 0. Statement 1 should clearly not be sufficient, as there are infinitely many negative multiples of 13 (including -13, -26, -130, etc., all providing the answer "yes") but also infinitely many positive values (13, 26, 130, etc., each providing the answer "no"). And statement 2 should also pretty clearly not be sufficient, as any negative number would satisfy the statement, but so would an integer like 1 or 2.

Taken together, you might think that the statements are sufficient, as statement 1 rules out b = 1 and b = 2. But what about 0? 0 is a multiple of 13 and when cubed it's less than 13, but it's also not negative. All other solutions are negative multiples of 13 (-13, -26, -130, etc.) all providing the answer "yes" but 0 provides the answer "no", meaning that the statements together are not sufficient.
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Re: Is b<0?  [#permalink]

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New post 27 Jan 2015, 08:46
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Quote:
(1) Integer b is a multiple of 13
(2) b^3 < 13

(1) -> 13*i=b. but is b<0? sometimes yes, sometimes no.
(2) b could be 1 or -100, b<0? sometimes yes, sometimes no.
(1/2) -> (13*i)^3<13 ->for negative multiples of 13 this is ok.
but wait, whats about 0. if se set b=0 then we got 0<13 and 13*i=0, but b is still not <0. sometimes yes, sometimes no.

it should be E
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Re: Is b<0?  [#permalink]

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New post 27 Jan 2015, 14:11
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Bunuel wrote:
Is b<0?

(1) Integer b is a multiple of 13
(2) b^3 < 13

Kudos for a correct solution.


1) 0 is a multiple of 13. Not sufficient

2) b< 13^1/3 which is a tad more than 2 Not sufficient

1+2) 0 is still a valid option. Not sufficient

Answer E.
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Re: Is b<0?  [#permalink]

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New post 27 Jan 2015, 15:08
Is b<0?

(1) Integer b is a multiple of 13
(2) b^3 < 13

(1) not sufficient, b could be 26 or -26 so could be greater or less than 0
(2) b could be 1,2 or any negative number so also not sufficient

Taking both statements together will also not work, b=0 makes both statements true, and b=-26 also makes both statements true, so taking both statements together does not show that b<0

Answer is E
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Re: Is b<0?  [#permalink]

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New post 28 Jan 2015, 03:57
Is b<0?

(1) Integer b is a multiple of 13
(2) b^3 < 13
Statement 1. If b=13 then answer is No but if b=-13 then answer is Yes. Not sufficient
Statement 2. B could be equal to 2 or -2. Hence not sufficient.
Both statements together. If b=(-13) then answer is Yes. But if b=0 then Answer is No. Hence answer E
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Re: Is b<0?  [#permalink]

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New post 28 Jan 2015, 07:09
Bunuel wrote:
Is b<0?

(1) Integer b is a multiple of 13
(2) b^3 < 13

Kudos for a correct solution.



E:

from 1: b can be 13, 26, 39 ....or -13, -26, -39 and 0 so NSF
from 2: considering only this statement b lies in the interval (-infinity,3) considering b as an integer --> NSF

1+ 2
values that b can take are ......-39,-26,-13 and 0. so b can be negative or 0 NSF

E
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Re: Is b<0?  [#permalink]

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New post 02 Jun 2016, 03:18
Bunuel wrote:
Bunuel wrote:
Is b<0?

(1) Integer b is a multiple of 13
(2) b^3 < 13

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

E. The key to this problem is recognizing that b could equal 0. Statement 1 should clearly not be sufficient, as there are infinitely many negative multiples of 13 (including -13, -26, -130, etc., all providing the answer "yes") but also infinitely many positive values (13, 26, 130, etc., each providing the answer "no"). And statement 2 should also pretty clearly not be sufficient, as any negative number would satisfy the statement, but so would an integer like 1 or 2.

Taken together, you might think that the statements are sufficient, as statement 1 rules out b = 1 and b = 2. But what about 0? 0 is a multiple of 13 and when cubed it's less than 13, but it's also not negative. All other solutions are negative multiples of 13 (-13, -26, -130, etc.) all providing the answer "yes" but 0 provides the answer "no", meaning that the statements together are not sufficient.


If the problem was worded differently (in this case: is b>0?) would the answer have been C?
Using both 1 and 2 stems, we know that b is either 0 or negative. If b is 0, then the answer to the question is 'no'. If b is negative, the answer to the question still in 'no'.
Please help.

Best,
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Re: Is b<0?  [#permalink]

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New post 04 Jun 2016, 10:24
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tae808 wrote:
Bunuel wrote:
Bunuel wrote:
Is b<0?

(1) Integer b is a multiple of 13
(2) b^3 < 13

Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

E. The key to this problem is recognizing that b could equal 0. Statement 1 should clearly not be sufficient, as there are infinitely many negative multiples of 13 (including -13, -26, -130, etc., all providing the answer "yes") but also infinitely many positive values (13, 26, 130, etc., each providing the answer "no"). And statement 2 should also pretty clearly not be sufficient, as any negative number would satisfy the statement, but so would an integer like 1 or 2.

Taken together, you might think that the statements are sufficient, as statement 1 rules out b = 1 and b = 2. But what about 0? 0 is a multiple of 13 and when cubed it's less than 13, but it's also not negative. All other solutions are negative multiples of 13 (-13, -26, -130, etc.) all providing the answer "yes" but 0 provides the answer "no", meaning that the statements together are not sufficient.


If the problem was worded differently (in this case: is b>0?) would the answer have been C?
Using both 1 and 2 stems, we know that b is either 0 or negative. If b is 0, then the answer to the question is 'no'. If b is negative, the answer to the question still in 'no'.
Please help.

Best,
Tae

Yes you are correct. If the question was "is b>0" then taking both statements together would mean that statement 1 rules out 1 and 2 and we have only 0 and negative multiples of 13. In both cases answer would be "no"
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Re: Is b<0?  [#permalink]

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New post 07 Jun 2017, 10:14
IMO E
Both are insufficient as 0 is a multiple of 13 hence both the statements fail

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Re: Is b<0?  [#permalink]

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New post 08 Jun 2017, 09:19
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Bunuel wrote:
Is b<0?

(1) Integer b is a multiple of 13
(2) b³ < 13



Target question: Is b less than 0?

Statement 1: Integer b is a multiple of 13
There are several values of b that satisfy statement 1. Here are two:
Case a: b = -13, in which case x IS less than 0
Case b: b = 0, in which case x is NOT less than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: b³ < 13
There are several values of b that satisfy statement 2. Here are two:
Case a: b = -13, in which case x IS less than 0
Case b: b = 0, in which case x is NOT less than 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: b = -13, in which case x IS less than 0
Case b: b = 0, in which case x is NOT less than 0
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer:

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Re: Is b<0?   [#permalink] 05 Oct 2018, 17:54
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