GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Jul 2018, 12:38

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Marty has a necklace that contains a total of 36 rhinestones, zirconiu

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47037
Marty has a necklace that contains a total of 36 rhinestones, zirconiu [#permalink]

### Show Tags

16 Aug 2017, 01:59
00:00

Difficulty:

35% (medium)

Question Stats:

73% (01:46) correct 27% (01:34) wrong based on 67 sessions

### HideShow timer Statistics

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

_________________
Manager
Joined: 02 Nov 2015
Posts: 171
GMAT 1: 640 Q49 V29
Re: Marty has a necklace that contains a total of 36 rhinestones, zirconiu [#permalink]

### Show Tags

16 Aug 2017, 03:43
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
Intern
Joined: 22 Apr 2015
Posts: 28
WE: Business Development (Internet and New Media)
Re: Marty has a necklace that contains a total of 36 rhinestones, zirconiu [#permalink]

### Show Tags

16 Aug 2017, 04:22
2
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 $$r \neq{12}$$
Intern
Joined: 04 Aug 2017
Posts: 10
Re: Marty has a necklace that contains a total of 36 rhinestones, zirconiu [#permalink]

### Show Tags

02 Sep 2017, 13:05
1
Another way to look at the problem:
We are told that the ratio between obsidian and zirconium "gems" is 2x:5x and that the total amount of gems are 36.
We can plug in numbers to find out how many rhinestones would be possible for us to have given the limitations set by the problem.
2(1)+5(1)+r=36 -> r=29
2(2)+5(2)+r=36 -> r=22
2(3)+5(3)+r=36 -> r=15
2(4)+5(4)+r=36 -> r=8
2(5)+5(5)+r=36 -> r=1

As you can see from this list that out of the options given to us, 12 is the only number of rhinestones that isn't possible.
Re: Marty has a necklace that contains a total of 36 rhinestones, zirconiu   [#permalink] 02 Sep 2017, 13:05
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.