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01 Nov 2018, 03:04
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
77% (00:47) correct
23%
(00:59)
wrong
based on 216
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If a carton containing a dozen mirrors is dropped, which of the following cannot be the ratio of broken mirrors to unbroken mirrors?
(A) 2:1
(B) 3:1
(C) 3:2
(D) 1:1
(E) 7:5
(A) 2:1
(B) 3:1
(C) 3:2
(D) 1:1
(E) 7:5
01 Nov 2018, 03:49
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Because we have a dozen mirrors, we would have to find a ratio that corresponds with 12.
A) 2:1 --> 2+1 = 3 and 12 can be divided by 3
B) 3:1 --> 3+1 = 4 and 12 can be divided by 4
C) 3:2 --> 3+2 = 5 and 12 can not be divided by 5 --> answer C cannot be the ratio of broken mirrors to unbroken mirrors.
D) 1:1 --> 1+1 = 2 and 12 can be divided by 2
E) 7:5 ---> 7+5 = 12 and 12 can be divided by 12.
A) 2:1 --> 2+1 = 3 and 12 can be divided by 3
B) 3:1 --> 3+1 = 4 and 12 can be divided by 4
C) 3:2 --> 3+2 = 5 and 12 can not be divided by 5 --> answer C cannot be the ratio of broken mirrors to unbroken mirrors.
D) 1:1 --> 1+1 = 2 and 12 can be divided by 2
E) 7:5 ---> 7+5 = 12 and 12 can be divided by 12.
01 Nov 2018, 04:02
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Bunuel
\(12\,\,{\rm{mirrors}}\,\,\,\left\{ \matrix{\\
{\rm{broken}}\,\,{\rm{:}}\,\,b \hfill \cr \\
{\rm{unbroken}}\,\,{\rm{:}}\,\,u \hfill \cr} \right.\)
\(?:\,\,\,b:u\,\,\,\underline {{\rm{impossible}}}\)
The k technique is very useful here!
\(\left( {\rm{A}} \right)\,\,\,\left\{ \matrix{\\
\,b = 2k \hfill \cr \\
u = k \hfill \cr} \right.\,\,\,\left( {k \ge 1\,\,{\mathop{\rm int}} } \right)\,\,\,\,\,\, \Rightarrow \,\,\,3k = 12\,\,\, \Rightarrow \,\,\,{\rm{viable}}\)
\(\left( {\rm{B}} \right)\,\,\,\left\{ \matrix{\\
\,b = 3k \hfill \cr \\
u = k \hfill \cr} \right.\,\,\,\left( {k \ge 1\,\,{\mathop{\rm int}} } \right)\,\,\,\,\,\, \Rightarrow \,\,\,4k = 12\,\,\, \Rightarrow \,\,\,{\rm{viable}}\)
\(\left( {\rm{C}} \right)\,\,\,\left\{ \matrix{\\
\,b = 3k \hfill \cr \\
u = 2k \hfill \cr} \right.\,\,\,\left( {k = 3k - 2k \ge 1\,\,\underline {{\mathop{\rm int}} } } \right)\,\,\,\,\,\, \Rightarrow \,\,\,5k = 12\,\,\, \Rightarrow \,\,\,\underline {{\rm{not}}} \,\,{\rm{viable}}\)
The correct answer is therefore (C).
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
