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Alex notices that a certain bacterium splits into 2 separate bacteria

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Bunuel
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Alex notices that a certain bacterium splits into 2 separate bacteria [#permalink] New post   18 Jun 2019, 00:18
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Alex notices that a certain bacterium splits into 2 separate bacteria once every 15 minutes. If there was one bacterium on a slide 3 hours ago, how many bacteria are there on the slide now?

(A) 8192
(B) 4096
(C) 2048
(D) 1180
(E) 256
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Lampard42
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Re: Alex notices that a certain bacterium splits into 2 separate bacteria [#permalink] New post   18 Jun 2019, 01:00
Bunuel wrote:
Alex notices that a certain bacterium splits into 2 separate bacteria once every 15 minutes. If there was one bacterium on a slide 3 hours ago, how many bacteria are there on the slide now?

(A) 8192
(B) 4096
(C) 2048
(D) 1180
(E) 256


It is a geometric progression with a = 1; r=2 and n=3*(60/15)=12. Sum of first 12 terms of a GP is a(1-r^n)/(1-r) = 1(1-2^12)/-1=4095. IMO B?

chetan2u Bunuel could you help me find where i have gone wrong.
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Alex notices that a certain bacterium splits into 2 separate bacteria [#permalink] New post   18 Jun 2019, 02:04
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Arvind42 wrote:
Bunuel wrote:
Alex notices that a certain bacterium splits into 2 separate bacteria once every 15 minutes. If there was one bacterium on a slide 3 hours ago, how many bacteria are there on the slide now?

(A) 8192
(B) 4096
(C) 2048
(D) 1180
(E) 256


It is a geometric progression with a = 1; r=2 and n=3*(60/15)=12. Sum of first 12 terms of a GP is a(1-r^n)/(1-r) = 1(1-2^12)/-1=4095. IMO B?

chetan2u Bunuel could you help me find where i have gone wrong.



Why are you trying to find the sum?..
You have to look for the 13th term which is \(a*r^{n-1}=1*2^{13-1}=2^{12}=4096\)
13 term because the first term is at 0 minutes and then 12 more for 3 hours=1+12=13
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Re: Alex notices that a certain bacterium splits into 2 separate bacteria [#permalink] New post   18 Jun 2019, 02:25
chetan2u wrote:
Arvind42 wrote:
Bunuel wrote:
Alex notices that a certain bacterium splits into 2 separate bacteria once every 15 minutes. If there was one bacterium on a slide 3 hours ago, how many bacteria are there on the slide now?

(A) 8192
(B) 4096
(C) 2048
(D) 1180
(E) 256


It is a geometric progression with a = 1; r=2 and n=3*(60/15)=12. Sum of first 12 terms of a GP is a(1-r^n)/(1-r) = 1(1-2^12)/-1=4095. IMO B?

chetan2u Bunuel could you help me find where i have gone wrong.



Why are you trying to find the sum?..
You have to look for the 13th term which is \(a*r^{n-1}=1*2^{13-1}=2^{12}=4096\)
13 term because the first term is at 0 minutes and then 12 more for 3 hours=1+12=13


chetan2u

But the original one also will still be alive right? Example:
at zero min 1 item
at 15 min 1+2 items
at 30 min (3*2)+3 items
at 45 min (9*2)+9 items
at 1 hr (27*2)+27 items
....
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Re: Alex notices that a certain bacterium splits into 2 separate bacteria [#permalink] New post   18 Jun 2019, 02:30
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Arvind42 wrote:
chetan2u wrote:
Bunuel wrote:
Alex notices that a certain bacterium splits into 2 separate bacteria once every 15 minutes. If there was one bacterium on a slide 3 hours ago, how many bacteria are there on the slide now?

(A) 8192
(B) 4096
(C) 2048
(D) 1180
(E) 256




Why are you trying to find the sum?..
You have to look for the 13th term which is \(a*r^{n-1}=1*2^{13-1}=2^{12}=4096\)
13 term because the first term is at 0 minutes and then 12 more for 3 hours=1+12=13


chetan2u

But the original one also will still be alive right? Example:
at zero min 1 item
at 15 min 1+2 items
at 30 min (3*2)+3 items
at 45 min (9*2)+9 items
at 1 hr (27*2)+27 items
....


You would be correct if it said..
Each bacteria produce 2 more bacteria.
But here that bacteria is split into 2
So initial 1, then 1*2, then these split into 2 each, so 4..and so on
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Re: Alex notices that a certain bacterium splits into 2 separate bacteria [#permalink] New post   18 Jun 2019, 05:21
Bunuel wrote:
Alex notices that a certain bacterium splits into 2 separate bacteria once every 15 minutes. If there was one bacterium on a slide 3 hours ago, how many bacteria are there on the slide now?

(A) 8192
(B) 4096
(C) 2048
(D) 1180
(E) 256


in 1 hr we have 4 slots of 15 mins so bacteria will be 2^4 times its original count
so in 3 hrs it would be 2^3*4 ; 2^12 times its original count
IMO B ; 4096= 2^12
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Re: Alex notices that a certain bacterium splits into 2 separate bacteria [#permalink] New post   23 Jun 2019, 19:31
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Bunuel wrote:
Alex notices that a certain bacterium splits into 2 separate bacteria once every 15 minutes. If there was one bacterium on a slide 3 hours ago, how many bacteria are there on the slide now?

(A) 8192
(B) 4096
(C) 2048
(D) 1180
(E) 256


Since there are four 15-minute intervals every hour, in 3 hours there are 12 15-minute intervals. Since there was 1 bacterium on the slide 3 hours ago, now there are 1 x 2^12 = 4096.

Answer: B
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Re: Alex notices that a certain bacterium splits into 2 separate bacteria [#permalink]
23 Jun 2019, 19:31
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