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M03 #01

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Re: M03 #01 [#permalink] New post 19 Feb 2012, 10:27
statement 1: test a decimal between 1 & 2 (ie. 1.5) you'll still get a positive
statement 2: 1.5 still works
eliminate d
eliminate c
E
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Re: M03 #01 [#permalink] New post 17 Jan 2013, 12:41
I go for ans A, because the solution set for stat 1 is x>3 which ans the que. Stat 2 is insufficient.
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Re: M03 #01 [#permalink] New post 30 Jan 2014, 03:44
Ans is E.

Consider 1st statement;
(x-3)(x-2)(x-1)>0
This is true in 4 cases:
Case 1: (x-3), (x-2), (x-1) all are > 0, which is possible if x>3 {x-3 > 0}
Case 2: (x-3) > 0 and (x-2) and (x-1) < 0 which is not possible if x>3
Case 3: (x-3) and (x-2)< 0 and (x-1) > 0 which is possible if 1<x<2
Case 4: (x-3) and (x-1)< 0 and (x-2) > 0 which is not possible if x>2 [since x-1 can not be negative for x>2]
This gives two possible answers from Case 1 (x>3) & case 3 (1<x<2). Therefore, this statement is NOT SUFFICIENT

Consider 2nd statement;
x>1
That does not tell us if x>3 or x<3, x could be 1.5, 2, 2.5 etc...Hence, this is NOT SUFFICIENT

Combining 1st and 2nd statement does not give us exact value of x. Hence, the answer should be E
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Re: M03 #01 [#permalink] New post 20 Apr 2014, 05:48
1) Both X=4 and X=1.5 satisfy (X-3)(X-2)(X-1)>0 -> insufficient
2) Clearly insufficient
Combine 2 stats: still cannot, using the same examples as in 1)

-> Choose E
Re: M03 #01   [#permalink] 20 Apr 2014, 05:48
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