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If a ≠ 0, is b < a < c ?
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05 Mar 2019, 10:25
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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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Re: If a ≠ 0, is b < a < c ?
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05 Mar 2019, 12:13
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. So the answer is E.
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Re: If a ≠ 0, is b < a < c ?
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05 Mar 2019, 12:54
mangamma wrote: If a ≠ 0, is b < a < c ?
I. b < a < c II. b < a <c IF YOU FIND MY SOLUTION HELPFUL, PLEASE GIVE ME KUDOS (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 The answer is E.



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If a ≠ 0, is b < a < c ?
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06 Mar 2019, 17:28
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 ?
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07 Mar 2019, 18:26
mangamma wrote: If a ≠ 0, is b < a < c ?
I. b < a < c II. b < a <c
\(?\,\,\,:\,\,\,b < a < c\,\,\,\,\left( {a \ne 0} \right)\) \(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,0,2} \right)\,\,\,\,\,\,\left[ {0 < \left 1 \right < 2\,\,\,{\rm{and}}\,\,\left 0 \right < \left 1 \right < \left 2 \right\,} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {  1,0,2} \right)\,\,\,\,\,\,\left[ {0 < \left {  1} \right < 2\,\,\,{\rm{and}}\,\,\left 0 \right < \left {  1} \right < \left 2 \right} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\,\,\,\,\,\) The correct answer is (E). 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 ?
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