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# Before the regime change, the National Museum regularly

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VP
Joined: 25 Nov 2004
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Before the regime change, the National Museum regularly [#permalink]  02 Jan 2005, 22:26
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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.
Director
Joined: 27 Dec 2004
Posts: 908
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Kudos [?]: 19 [0], given: 0

I got 7/13 and 3/5. None of the answer choice so i must be doing something wrong.
Senior Manager
Joined: 19 Nov 2004
Posts: 284
Location: Germany
Followers: 1

Kudos [?]: 13 [0], given: 0

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
VP
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
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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
Director
Joined: 27 Dec 2004
Posts: 908
Followers: 1

Kudos [?]: 19 [0], given: 0

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.
Manager
Joined: 29 Jul 2004
Posts: 61
Followers: 1

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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.
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