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An old woman sits on a park bench and begins dropping rice on the grou

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An old woman sits on a park bench and begins dropping rice on the grou  [#permalink]

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New post 24 Apr 2017, 04:32
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Question Stats:

65% (01:44) correct 35% (01:58) wrong based on 208 sessions

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An old woman sits on a park bench and begins dropping rice on the ground at a constant rate of 7 grains per second. Initially, there is no rice on the ground and no pigeons are present, but after the woman has been dropping rice for 10 seconds, a pigeon shows up and begins eating rice at a constant rate of 3 grains per second. If the woman continues dropping rice at the same rate, if one new pigeon shows up every 10 seconds, if no pigeons leave, and if all pigeons eat at the same constant rate as long as there is rice on the ground, how many pigeons will be present the next time there is no rice on the ground?

A. 3
B. 4
C. 5
D. 6
E. 7

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An old woman sits on a park bench and begins dropping rice on the grou  [#permalink]

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New post 24 Apr 2017, 08:53
5
3
To start: 70 grains

Next 10:140 grains
First Bird: 30
Rice Left: 110

Next 10: 180 grains
Two Birds: 60
Rice Left: 120

Next 10: 190 grains
Three Birds: 90
Rice Left: 100

Next 10: 170
Four Birds: 120
Rice Left: 50

Next 10: 120
FIVE BIRDS: 150
Rice Left: ZERO

Answer Choice "C"
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An old woman sits on a park bench and begins dropping rice on the grou  [#permalink]

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New post 24 Apr 2017, 06:23
Grains being dropped = 7 grains/second
Pigeon's rate of eating grains = 3 grains/second

The woman drops 70 grains in the first 10 seconds.

After 10 seconds, as a pigeon joins & starts to eat the rice grains on the road, there is a net addition of 40 grains

From the 10-20 second period, since there are 2 pigeons, there is a net addition of 10 grains (70 - 2*30). Total: 120

From the 20-30 second period, since there are 3 pigeons, there is a net loss of 20 grains (70 - 3*30). Total: 100

From the 30-40 second period, since there are 4 pigeons, there is a net loss of 50 grains (70 - 4*30). Total: 50

Therefore, somewhere between 40 and 50 seconds, after the 5th pigeon joins, there is a net loss s.t there is no grain on the ground.(Option C)
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Re: An old woman sits on a park bench and begins dropping rice on the grou  [#permalink]

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New post 25 Apr 2017, 20:24
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Bunuel wrote:
An old woman sits on a park bench and begins dropping rice on the ground at a constant rate of 7 grains per second. Initially, there is no rice on the ground and no pigeons are present, but after the woman has been dropping rice for 10 seconds, a pigeon shows up and begins eating rice at a constant rate of 3 grains per second. If the woman continues dropping rice at the same rate, if one new pigeon shows up every 10 seconds, if no pigeons leave, and if all pigeons eat at the same constant rate as long as there is rice on the ground, how many pigeons will be present the next time there is no rice on the ground?

A. 3
B. 4
C. 5
D. 6
E. 7

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Re: An old woman sits on a park bench and begins dropping rice on the grou  [#permalink]

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New post 25 Apr 2017, 21:08
1
First 10s: +70 (= 7x10) -> Total grain: 70
Next 10s: +40 (=(7-3)x10) -> Total grain: 70+40=110
Next 10s: +10 (=7-3-3)x10) -> ...


As we can see we have the following sequence +70,+40,+10,-20,-50,-80.
If we add the sequence we can see that in the 6th period we became negative.

Given that in the first period we had 0 pigeons it means that in the 6th period we have 5 pigeons.
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Re: An old woman sits on a park bench and begins dropping rice on the grou  [#permalink]

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New post 09 May 2017, 14:11
70+7n=(10+20+....+n)3
solving, n=140/30= max 5
ans:5
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Re: An old woman sits on a park bench and begins dropping rice on the grou  [#permalink]

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New post 03 May 2018, 10:27
ishaan.wares wrote:
70+7n=(10+20+....+n)3
solving, n=140/30= max 5
ans:5



please explain .
Re: An old woman sits on a park bench and begins dropping rice on the grou &nbs [#permalink] 03 May 2018, 10:27
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