It is currently 18 Nov 2017, 20:42

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Before the regime change, the National Museum regularly

Author Message
VP
Joined: 25 Nov 2004
Posts: 1481

Kudos [?]: 130 [0], given: 0

Before the regime change, the National Museum regularly [#permalink]

### Show Tags

02 Jan 2005, 23:26
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Before the regime change, the National Museum regularly displayed only one third of the artifacts in its collection; the other two thirds were kept in storage. The National Museumâ€™s curator estimates that one seventh to one fifth of the artifacts in the museumâ€™s collection disappeared during the regime change. If the National Museum now wishes to display the same number of artifacts as it did before the regime change, and if none of the missing artifacts are recovered, then the museum will keep in storage:

A. 4/5 to 6/7 of the artifacts in its collection.

B. 9/14 to 13/20 of the artifacts in its collection.

C. 3/5 to 5/7 of the artifacts in its collection.

D. 7/12 to 11/18 of the artifacts in its collection..

E. 7/18 to 5/12 of the artifacts in its collection.

Kudos [?]: 130 [0], given: 0

Director
Joined: 27 Dec 2004
Posts: 894

Kudos [?]: 54 [0], given: 0

### Show Tags

03 Jan 2005, 08:14
I got 7/13 and 3/5. None of the answer choice so i must be doing something wrong.

Kudos [?]: 54 [0], given: 0

Senior Manager
Joined: 19 Nov 2004
Posts: 281

Kudos [?]: 47 [0], given: 0

Location: Germany

### Show Tags

03 Jan 2005, 09:03
can be worked with algebraic expression, but a quicker way would be to work with hypothetical (yet consistent) numbers.

A quick look would reveal that 2100 can be used as the original number of total artifacts (before the regime change). Hence, the number of artifacts displayed = 700 and that kept in store = 1400.

During the regime change, 300 to 420 artifcats went missing.

Therefore, the remaining number of artifactsis between

(2100 - 300 = 1800) and (2100 - 420 = 1680)

In order to display the same number of artifacts as before the regime change (700), the fraction that needs to be kept in storage = (1800 - 700 = 1100) to (1680 - 700 = 980).

=> the fraction of artifacts in storage = (1100/1800 = 11/18) to (980/1680 = 7/12), hence the correct answer is D

Kudos [?]: 47 [0], given: 0

VP
Joined: 30 Sep 2004
Posts: 1480

Kudos [?]: 425 [0], given: 0

Location: Germany

### Show Tags

03 Jan 2005, 09:24
D it is ?

lets assume 105 artifacts.

1. 1/3 of 105 are displayed = 35
2. 2/3 of 105 are storaged = 70
3. 1/7 to 1/5 disappaered => 15 to 21
4. 105 - 15 = 90 - 35 = 55 => x * 90 = 55 => x = 55/90 = 11/18
5. 105 - 21 = 84 -35 = 49 => x * 84 = 49 => 49 /84 = 7/12

Kudos [?]: 425 [0], given: 0

Director
Joined: 27 Dec 2004
Posts: 894

Kudos [?]: 54 [0], given: 0

### Show Tags

03 Jan 2005, 09:27
Thanks for the explanation. I had used plugging of number approach however i didn't plug in number that was a multiple of 7 (don't ask me why ). Instead i plugged in 30 thus the reason why i was off.

Kudos [?]: 54 [0], given: 0

Manager
Joined: 29 Jul 2004
Posts: 61

Kudos [?]: [0], given: 0

### Show Tags

03 Jan 2005, 10:59
christoph wrote:
D it is ?

lets assume 105 artifacts.

1. 1/3 of 105 are displayed = 35
2. 2/3 of 105 are storaged = 70
3. 1/7 to 1/5 disappaered => 15 to 21
4. 105 - 15 = 90 - 35 = 55 => x * 90 = 55 => x = 55/90 = 11/18
5. 105 - 21 = 84 -35 = 49 => x * 84 = 49 => 49 /84 = 7/12

I also used 105 also because 7*5*3=105. I said that 6/7 to 4/5 are left, then subtracted 35 from the respective numbers. Other than that I did it the same way as you.

Kudos [?]: [0], given: 0

03 Jan 2005, 10:59
Display posts from previous: Sort by