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30 Sep 2018, 10:06
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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82% (01:21) correct
18%
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A, B and C are positive integers. Is A > C?
1) A, B and C are distinct factors of 8.
2) B = 4C
1) A, B and C are distinct factors of 8.
2) B = 4C
30 Sep 2018, 11:36
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From statement 1)
We know A, B, and C are distinct factors of 8
So A,B, and C could be 1,2,4 respectively A < C, however if we make it 4,2,1 then A>C
Insufficient.
For 2)
B=4C we have no information on A so insufficient
Now combine
If C = 1 then B = 4 and A could be 2 or 8
A>C yes.
If C = 4 then B = 8 and A could be 1 or 2
A > C no.
Insufficient.
Answer choice E
Posted from my mobile device
We know A, B, and C are distinct factors of 8
So A,B, and C could be 1,2,4 respectively A < C, however if we make it 4,2,1 then A>C
Insufficient.
For 2)
B=4C we have no information on A so insufficient
Now combine
If C = 1 then B = 4 and A could be 2 or 8
A>C yes.
If C = 4 then B = 8 and A could be 1 or 2
A > C no.
Insufficient.
Answer choice E
Posted from my mobile device
30 Sep 2018, 12:44
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CAMANISHPARMAR
A, B and C are positive integers. Is A > C?
1) A, B and C are distinct factors of 8.
2) B = 4C
\(A,B,C\,\,\, \ge 1\,\,\,\,{\rm{ints}}\)1) A, B and C are distinct factors of 8.
2) B = 4C
\({\rm{A}}\,\,\mathop > \limits^? \,\,C\)
Let´s BIFURCATE (1+2) at once, to guarantee the correct answer is (E), indeed.
\(\left( {1 + 2} \right)\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {A,B,C} \right) = \left( {2,4,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,\,{\rm{Take}}\,\,\left( {A,B,C} \right) = \left( {1,8,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\, \hfill \cr} \right.\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
