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# If a ≠ 0, is b < a < c ?

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Manager
Joined: 25 Dec 2018
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If a ≠ 0, is b < a < c ?  [#permalink]

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05 Mar 2019, 10:25
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Difficulty:

55% (hard)

Question Stats:

54% (01:28) correct 46% (01:43) wrong based on 61 sessions

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If a ≠ 0, is b < a < c ?

I. b < |a| < c
II. |b| < |a| <|c|
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Joined: 24 Jun 2008
Posts: 1392
Re: If a ≠ 0, is b < a < c ?  [#permalink]

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05 Mar 2019, 12:13
1
Using both statements, the answer can easily be yes if we just let b=1, a=2 and c=3, say.

But we can also get a no answer, if we just make a negative, and set b and c carefully. Say c = 10, then we might have a = -7 and b = -3. In this case, a is the smallest of the three unknowns, and both statements are true.

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Re: If a ≠ 0, is b < a < c ?  [#permalink]

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05 Mar 2019, 12:54
2
mangamma wrote:
If a ≠ 0, is b < a < c ?

I. b < |a| < c
II. |b| < |a| <|c|

(1) b < |a| <c This gives two possibilities 1) b <a< c when a >0 which gives us a Yes answer, and -c < a < -b. when a <0. Suppose a =-1, -b = -1/2 and -c =-2, satisfying -c<a< -b Then b =1/2, c =2, and a<b<c, which gives us a NO answer. Since we get a Yes and a No, we can mark NS

(2) |b|<|a|<|c|. Let |c| = 3, |a| =2, |b| = 1. Suppose c =3, a = 2, b =1. Then b<a<c, and we get a YES answer. suppose c=-3, a=2, b=1, then c<b<a and we get a YES answer. Since we get a Yes and a No, we can mark NS

(1) and (2) ( b<a<c or -c<a<-b) AND |b|<|a|<|c| Suppose b<a<c such that |b|<|a|<|C| let b =1 a =2 c =3, Both constraints are satisfied and we get a YES answer. Suppose -c<a<-b such that |b|<|a|<|c| Let b =2, a = -3, c =4, then we have satisfied -c <a<-b (-4<-3<-2) as well as |b| <|a|<|c| (2<3<4), but a<b<c (-3<2<4) giving us a NO answer. Since we get both a Yes and NO, we can mark NS

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If a ≠ 0, is b < a < c ?  [#permalink]

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06 Mar 2019, 17:28
1
mangamma wrote:
If a ≠ 0, is b < a < c ?

I. b < |a| < c
II. |b| < |a| <|c|

Statements combined:

Case 1: b=1, a=2, c=3
Statement 1 --> 1 < |2| < 3
Statement 2 --> |1| < |2| < |3|
In this case, b < a < c, so the answer to the question stem is YES.

Case 2: b=-1, a=-2, c=3
Statement 1 --> -1 < |-2| < 3
Statement 2 --> |-1| < |-2| < |3|
In this case, b > a, so the answer to the question stem is NO.

Since the answer is YES in Case 1 but NO in Case 2, the two statements combined are INSUFFICIENT.

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Re: If a ≠ 0, is b < a < c ?  [#permalink]

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07 Mar 2019, 18:26
1
1
mangamma wrote:
If a ≠ 0, is b < a < c ?

I. b < |a| < c
II. |b| < |a| <|c|

$$?\,\,\,:\,\,\,b < a < c\,\,\,\,\left( {a \ne 0} \right)$$

$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,0,2} \right)\,\,\,\,\,\,\left[ {0 < \left| 1 \right| < 2\,\,\,{\rm{and}}\,\,\left| 0 \right| < \left| 1 \right| < \left| 2 \right|\,} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( { - 1,0,2} \right)\,\,\,\,\,\,\left[ {0 < \left| { - 1} \right| < 2\,\,\,{\rm{and}}\,\,\left| 0 \right| < \left| { - 1} \right| < \left| 2 \right|} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\,\,\,\,\,$$

We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: If a ≠ 0, is b < a < c ?   [#permalink] 07 Mar 2019, 18:26
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