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Rate/Work Question

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Manager
Status: Darden '12 Alumni
Affiliations: USMC
Joined: 15 Jun 2009
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Schools: Darden '12
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Rate/Work Question [#permalink]  29 Aug 2009, 17:07
00:00

Difficulty:

5% (low)

Question Stats:

100% (05:14) correct 0% (00:00) wrong based on 1 sessions
One more. Thanks guys.

Annie and Jessie go on a jog together. Annie begins to jog at a rate of X feet per second, but Jessie has to tie her shoe. After tying her shoe, Jessie begins to jog, but jogs 5 feet per second faster than Annie to catch up. After both jogging 1800 feet, both Annie and Jessie are side by side. How long did Jessie tie her shoe for?

Any help would be appreciated.
Manager
Joined: 30 May 2009
Posts: 221
Followers: 3

Kudos [?]: 43 [1] , given: 0

Re: Rate/Work Question [#permalink]  29 Aug 2009, 17:38
1
KUDOS
Are u sure u have the right correction....I feel some information is missing.
Manager
Joined: 19 Jul 2009
Posts: 54
Schools: McIntire, Fuqua, Villanova, Vandy, WashingtonU
Followers: 0

Kudos [?]: 10 [1] , given: 4

Re: Rate/Work Question [#permalink]  29 Aug 2009, 18:27
1
KUDOS
Jessie
(1) jogs 5 feet/sec faster
(2) ran 1,800 feet before catching up

1800 / 5 = 360 seconds = 6 minutes. It took Jessie 6 minutes to catch up after tying her shoes. We need more info
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Manager
Joined: 25 Aug 2009
Posts: 180
Followers: 1

Kudos [?]: 53 [1] , given: 12

Re: Rate/Work Question [#permalink]  29 Aug 2009, 19:30
1
KUDOS
Time to tie up shoes = Time taken by Annie - Time taken by Jessie
= 1800/X - 1800/(X+5)

We need X to solve this problem.
Manager
Status: Darden '12 Alumni
Affiliations: USMC
Joined: 15 Jun 2009
Posts: 183
Schools: Darden '12
Followers: 1

Kudos [?]: 26 [0], given: 22

Re: Rate/Work Question [#permalink]  29 Aug 2009, 23:38
Sorry guys. Guess my memory isn't as good as I thought it was. Let me rethink this question, and I'll come back if I remember.

Thanks anyway.
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
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Kudos [?]: 115 [0], given: 10

Re: Rate/Work Question [#permalink]  01 May 2011, 23:40
1800/ A = 1800/ A+5 +t
Time taken by Ann has to be given here.
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Re: Rate/Work Question   [#permalink] 01 May 2011, 23:40
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