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# m13 #9

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A certain colony of bacteria doubles every morning while [#permalink]

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17 Feb 2008, 05:42
A certain colony of bacteria doubles every morning while every evening 1000 bacteria die. What was the original size of the colony if the colony completely disappeared by the end of the third day?

 825
 850
 875
 910
 950
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17 Feb 2008, 05:59
1
KUDOS
C
--
x = original size:
at the end of 1st day: 2x-1000
at the end of 2nd day: 2(2x-1000) -1000 = 4x-3000
at the end of 3rd day: 2(4x-3000)-1000
--
8x-7000=0; x= 875
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16 Jun 2009, 00:50
The first plausible number I found in reasonable time that causes 0 bacteria on Day 3 - evening is 800, but to have exactly 0 bacteria at the end of D3 the starting number for the colony must be 875:
Day 1 - morning: 1750
Day 1 - evening: 750
Day 2 - morning: 1500
Day 2 - evening: 500
Day 3 - morning: 1000
Day 3 - evening: 0
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16 Jun 2009, 01:08
I worked on this problem..backwards, as in i strtd from the end of day 3 and took 0, and worked their onwards.

Is tis right??

or culd it be 875, since 1750 is going to be the number on the morning of Day1.
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17 Jun 2009, 18:22

Day Morning Evening

I x x-1000
II 2x-2000 2x-3000
III 4x-6000 4x-7000

4x-7000=0

x=1750
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17 Jun 2009, 18:30
You can't do it this way because it's compounding on itself.

If you want to start backwards you have to do it like this (which I think was actually already posted in another method).

End of Day 3 = 0.....before losing 1000 = 1000, that morning before it doubled = 500

End of Day 2 = 500...before losing 1000 that night = 1500...that morning before doubled=750

End of Day 1 = 750...before losing 1000 that night = 1750...that morning before doubled = 875.

So 875 is the start, as said previously

My problem with this question though that it appears to asks for a certain number to start with, but in reality, there is a range which would still produce 0 at the end of day 3. if you start with 800 on day 1, you still end up with 0 at the end of day 3.

If you think of it as a range, then 875 > x > 750. If x has a value anywhere in that range, you will get 0 at the end of the 3rd day.

hemantsood wrote:

Day Morning Evening

I x x-1000
II 2x-2000 2x-3000
III 4x-6000 4x-7000

4x-7000=0

x=1750

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02 Jan 2011, 20:20
I think the best way to approach this question would be to backsolve. I picked D, then backsolved. That didnt work, so i checked B. B came out to be negative. Thus, the answer would be C.
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04 Jan 2011, 09:12
Day 0 Day 1 Day 2 Day3

Morning
Evening Must be 1000 before 1000 die and disappear

Solving backwards and filling in the table

Day 0 Day 1 Day 2 Day3

Morning 1750 1500 1000
Evening 875 750 500 Must be 1000 before 1000 die and disappear

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04 Jan 2011, 09:15
Day3
Evening: Must be 1000 before 1000 die and disappear.

Day3
Morning : 1000
Evening : Must be 1000 before 1000 die and disappear

Day2
Morning: 1500
Evening: 500

Day1
Morning: 1750
Evening: 750

Day 0: 1750/2 = 875

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13 Feb 2012, 08:29
kamilshahid wrote:
I think the best way to approach this question would be to backsolve. I picked D, then backsolved. That didnt work, so i checked B. B came out to be negative. Thus, the answer would be C.

Yes, back solving should help us and I did in that way only. But there is no need to pick any answer and in fact that wastes time.

My back solving is like this:

third day night: input & output are 1000,0
third day morning: input & output are 500, 1000

second day night: input & output are 1500,500
second day morning: input & output are 750, 1500

first day night: input & output are 1750,750
first day morning: input & output are 875, 1750

But after seeing the first reply, I liked it very much and suggest to follow it because it is very methodical and probability of error is less.
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13 Feb 2012, 08:30
srp wrote:
C
--
x = original size:
at the end of 1st day: 2x-1000
at the end of 2nd day: 2(2x-1000) -1000 = 4x-3000
at the end of 3rd day: 2(4x-3000)-1000
--
8x-7000=0; x= 875

Thanks, I like this better than my method
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-Aravind Chembeti

Re: Arithmetic   [#permalink] 13 Feb 2012, 08:30
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# m13 #9

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