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M13-09

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M13-09 [#permalink]

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A certain colony of bacteria doubles every morning while every evening 1000 bacteria die. If by the end of the third day the number of bacteria was 0, what was the original size of the colony by the end of the first day?

A. 825
B. 850
C. 875
D. 910
E. 950
[Reveal] Spoiler: OA

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New post 16 Sep 2014, 00:48
Official Solution:

A certain colony of bacteria doubles every morning while every evening 1000 bacteria die. If by the end of the third day the number of bacteria was 0, what was the original size of the colony by the end of the first day?

A. 825
B. 850
C. 875
D. 910
E. 950


Let \(x\) denote the original size of the colony. Compose the following equation:

\(2*(2*(2*x - 1000) - 1000) - 1000 = 0\), which simplifies to \(8x = 7000\).

From here, \(x = 875\).


Answer: C
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Re: M13-09 [#permalink]

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I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I don't understand this question.
Per my understanding,
Start : 825
Day 1 morning : 1650
Day 1 evening : 650
Day 2 morning : 1300
Day 2 evening : 300
Day 3 morning 600
Day 3 evening : everything disapeared.

What is wrong ?

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Re: M13-09 [#permalink]

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New post 20 Aug 2015, 09:38
Barnal wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I don't understand this question.
Per my understanding,
Start : 825
Day 1 morning : 1650
Day 1 evening : 650
Day 2 morning : 1300
Day 2 evening : 300
Day 3 morning 600
Day 3 evening : everything disapeared.

What is wrong ?

_____
Edited the question.
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Re: M13-09 [#permalink]

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New post 14 Feb 2016, 16:15
based on the wording, why wouldn't it be A/B?

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Re: M13-09 [#permalink]

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New post 28 Jun 2016, 17:22
intogamer wrote:
based on the wording, why wouldn't it be A/B?


Hope it helps:

Day one-
morning: 875*2 =1750
Evening: 1750- 1000= 750

Day two-
morning: 750*2= 1500
Evening: 1500-1000= 500

Day three-
morning: 500*2=1000
Evening: 1000-1000= 0 (all bacteria succumb) :evil:

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Re: M13-09 [#permalink]

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New post 20 Jul 2016, 08:01
well I tried with working backwards from Option approach, and started with option B and it worked, don't know why only option C is correct, maybe the question needs to be rephrased somehow.

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Re: M13-09 [#permalink]

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New post 20 Jul 2016, 08:51
shashanksagar wrote:
well I tried with working backwards from Option approach, and started with option B and it worked, don't know why only option C is correct, maybe the question needs to be rephrased somehow.

What should happen is that in the night of the third day, when 1000 die, you don't get exactly zero bacteria but a negative number

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Re: M13-09 [#permalink]

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New post 20 Jul 2016, 09:07
Number of Bacteria cant be negative.. when I'll wake up in the morning i'll have 0 bacteria on my plate , so its the same with 825 and 850. :)

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Re: M13-09 [#permalink]

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New post 20 Jul 2016, 11:20
shashanksagar wrote:
Number of Bacteria cant be negative.. when I'll wake up in the morning i'll have 0 bacteria on my plate , so its the same with 825 and 850. :)

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Unfortunatelly for the bacteria no, it can't be negative,
but the question explicitly says the number of bacteria must be zero that's why 875 is the best choice, in my humble opinion, since algebraically 875 yields exactly zero.
I'm not sure this question could appear on an official GMAT since it does leave room to interpretation since bacteria cannot be inferior to zero, although it works for algebra, it wouldn't in the real world.

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New post 29 Sep 2016, 07:39
Good question. Thanks for posting !

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Re: M13-09 [#permalink]

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New post 03 Dec 2016, 05:38
why are we starting the first day morning with 2X. ?
if we assuming the day 1 of bacteria then why are we starting the doubling in the morning?

How I solved it was -
2∗(2∗(x−1000)−1000)−1000=0

i am conceptually missing out on something

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Re: M13-09 [#permalink]

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New post 12 Sep 2017, 21:30
Hello gmatdemolisher1234,

I think this is how this question Pans out. Hope this clear things up a bit

  • Original= X
  • (Original +1) Morning = 2X
  • (Original+1)Evening = 2X-1000
  • (Original +2)Morning = 2*(2X-1000) = 4X-2000
  • (Original +2)Evening = 4X-2000 - 1000 = 4X - 3000
  • (Original +3) Morning = 2*(4X - 3000) = 8X - 6000
  • (Original +3) Evening = 8X - 6000 - 1000 = 8X - 7000

The stem says - 8X - 7000 = 0. From this we get X = 875. The Answer is B

Aside Bunuel:

Though I understood this question, I have a small question. Maybe I am over thinking, can you please help clarify?

If we consider Original = X, why dont we consider that 1000 Bacteria die on the original day's evening. Meaning, at the end of original day why is the bacteria count still X and not (X-1000). I am not able to understand from the stem what stops us from thinking so.

Hope I am able to communicate what I intend to say :-)

Thanks in advance!

gmatdemolisher1234 wrote:
why are we starting the first day morning with 2X. ?
if we assuming the day 1 of bacteria then why are we starting the doubling in the morning?

How I solved it was -
2∗(2∗(x−1000)−1000)−1000=0

i am conceptually missing out on something

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Re: M13-09 [#permalink]

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New post 12 Sep 2017, 23:11
susheelh wrote:
Hello gmatdemolisher1234,

I think this is how this question Pans out. Hope this clear things up a bit

  • Original= X
  • (Original +1) Morning = 2X
  • (Original+1)Evening = 2X-1000
  • (Original +2)Morning = 2*(2X-1000) = 4X-2000
  • (Original +2)Evening = 4X-2000 - 1000 = 4X - 3000
  • (Original +3) Morning = 2*(4X - 3000) = 8X - 6000
  • (Original +3) Evening = 8X - 6000 - 1000 = 8X - 7000

The stem says - 8X - 7000 = 0. From this we get X = 875. The Answer is B

Aside Bunuel:

Though I understood this question, I have a small question. Maybe I am over thinking, can you please help clarify?

If we consider Original = X, why dont we consider that 1000 Bacteria die on the original day's evening. Meaning, at the end of original day why is the bacteria count still X and not (X-1000). I am not able to understand from the stem what stops us from thinking so.

Hope I am able to communicate what I intend to say :-)

Thanks in advance!

gmatdemolisher1234 wrote:
why are we starting the first day morning with 2X. ?
if we assuming the day 1 of bacteria then why are we starting the doubling in the morning?

How I solved it was -
2∗(2∗(x−1000)−1000)−1000=0

i am conceptually missing out on something


The question asks about the original size of the colony by the end of the first day.
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Re: M13-09 [#permalink]

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New post 12 Sep 2017, 23:24
Answer is C
First day morning 875*2=1750
First day eve 1750-1000=750

Second day morning 750*2= 1500
Second day eve 1500-1000=500

Third day morning 500*2=1000
Third day eve 1000-1000=0


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Re: M13-09 [#permalink]

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New post 13 Sep 2017, 00:44
Thanks a Ton Bunuel for replying!!

Please bear with me for few more mins. Its just that I want to be sure I understand questions such as this correctly when I see them next.

I am still unable to see where the question asks about - "the original size of the colony by the end of the first day".

I am breaking down the question into bits to explain what I understood from each of the part. Can you please see where I am going wrong?

Quote:
A certain colony of bacteria doubles every morning while every evening 1000 bacteria die.


This is a fact to be used while calculating.

Quote:
If by the end of the third day the number of bacteria was 0


This tells us what is the Needed. Meaning what is needed at the end of third day.

Quote:
what was the original size of the colony?


The Stem says what's the original size of the colony. But does not seem to mention if this original is the Morning or the Evening. Of course if it had said - "what was the original size of the colony at the end of first day" - I would not be having this question.

I do understand I might be simply Over thinking here :(

Thanks in advance for your time!
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susheelh wrote:
Thanks a Ton Bunuel for replying!!

Please bear with me for few more mins. Its just that I want to be sure I understand questions such as this correctly when I see them next.

I am still unable to see where the question asks about - "the original size of the colony by the end of the first day".

I am breaking down the question into bits to explain what I understood from each of the part. Can you please see where I am going wrong?

Quote:
A certain colony of bacteria doubles every morning while every evening 1000 bacteria die.


This is a fact to be used while calculating.

Quote:
If by the end of the third day the number of bacteria was 0


This tells us what is the Needed. Meaning what is needed at the end of third day.

Quote:
what was the original size of the colony?


The Stem says what's the original size of the colony. But does not seem to mention if this original is the Morning or the Evening. Of course if it had said - "what was the original size of the colony at the end of first day" - I would not be having this question.

I do understand I might be simply Over thinking here :(

Thanks in advance for your time!


Edited the question to end this confusion.
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Re: M13-09 [#permalink]

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New post 15 Sep 2017, 05:54
I still find the question unclear.

"original size of the colony by the end of the first day" is ambiguous. Why not just ask for the original size of the colony at the start of day one?

By the end of the first day, the original size of 875 would have become 750, which was not a choice, which got me very confused.

I think this a poorly worded item and needs editing.

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Re: M13-09 [#permalink]

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New post 01 Oct 2017, 00:15
alainca wrote:
I still find the question unclear.

"original size of the colony by the end of the first day" is ambiguous. Why not just ask for the original size of the colony at the start of day one?

By the end of the first day, the original size of 875 would have become 750, which was not a choice, which got me very confused.

I think this a poorly worded item and needs editing.


I Agree. If the question intends for us to solve for the original size of the colony why not write "Original size of the colony".
End of first day made me sure we were implying the count after we had already cycled once.
Let x = original be size at the instant the cycle started (2x-1000) Brings us to the evening of the first day, which we will use for the morning of the second day. and so on for the third.

\([2*(2*(2*x-1000)-1000)-1000]=0\) which brings us to the correct finishing point of \(\frac{7000}{8}=x=875\)


However when we consider x = original size at END of day one, we are assuming one cycle has already passed. Thus we only cycle 2 more times.
\(2x(2x-1000)-1000=0\)
Which gives us \(4x=3000\) which is unsolvable through any of the answer choices. (Max would have to be 750)

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Re: M13-09 [#permalink]

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New post 09 Oct 2017, 14:02
alainca wrote:
I still find the question unclear.

"original size of the colony by the end of the first day" is ambiguous. Why not just ask for the original size of the colony at the start of day one?

By the end of the first day, the original size of 875 would have become 750, which was not a choice, which got me very confused.

I think this a poorly worded item and needs editing.


I agree. End of the first day implies the answer should be 750.

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Re: M13-09   [#permalink] 09 Oct 2017, 14:02

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