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16 Aug 2017, 01:59
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:03) correct
24%
(02:02)
wrong
based on 161
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Marty has a necklace that contains a total of 36 rhinestones, zirconium, and obsidian “gems.” If the ratio of obsidian to zirconium is 2:5, then which of the following could not be the number of rhinestones in Marty’s necklace?
(A) 8
(B) 12
(C) 15
(D) 22
(E) 29
(A) 8
(B) 12
(C) 15
(D) 22
(E) 29
16 Aug 2017, 03:43
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It's a B. Rest all combinations are possible.
If it's 12 , then 36-12= 24, which is not possible
Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app
If it's 12 , then 36-12= 24, which is not possible
Sent from my Lenovo TAB S8-50LC using GMAT Club Forum mobile app
16 Aug 2017, 04:22
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r+z+o= 36
o:z= 2:5=> no of obsidian gem=o= 2k
No. of zirconium gems=z= 5k
total no of o & z= 7k
r+7k=36
k= \(\frac{(36-r)}{7}\)
r=8 => k= \(\frac{28}{7}\)=4
r= 15=> k=\(\frac{24}{7}\)=3
r=22=> k= \(\frac{14}{7}\)=2
r=29=> k=\(\frac{7}{7}\)=1
r=12=> k= not an integer.
Hence \(r \neq{12}\)
o:z= 2:5=> no of obsidian gem=o= 2k
No. of zirconium gems=z= 5k
total no of o & z= 7k
r+7k=36
k= \(\frac{(36-r)}{7}\)
r=8 => k= \(\frac{28}{7}\)=4
r= 15=> k=\(\frac{24}{7}\)=3
r=22=> k= \(\frac{14}{7}\)=2
r=29=> k=\(\frac{7}{7}\)=1
r=12=> k= not an integer.
Hence \(r \neq{12}\)
