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A strain of bacteria multiplies such that the ratio of its p

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A strain of bacteria multiplies such that the ratio of its population in any two consecutive minutes is constant. If the bacteria grows from a a population of 5 million to 40 million over the course on an hour, by what factor does the population increase every 10 minutes?

\(\sqrt{2}\)

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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 07 Oct 2014, 01:40
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taleesh wrote:
Can somebody please show the understandable solution to this question, I appreciate the above mention solutions by Maths prodigies( which isn't me) but i need a simple solution to this question that can be implemented in the time constraint of Gmat.

If geometric progression is applied here - what is the common ratio?
pleaseeeeeeeee help - EXPERTSSSSSSS


A strain of bacteria multiplies such that the ratio of its population in any two consecutive minutes is constant. If the bacteria grows from a a population of 5 million to 40 million over the course on an hour, by what factor does the population increase every 10 minutes?

In 60 minutes the population increased 8 times. So, if the growth rate per minute is r, then \(r^{60}=8=2^3\). The question asks what factor the population will grow by in 10 minutes. The growth rate in 10 minutes is \(r^{10}\). take 6th root from \(r^{60}=2^3\) --> \(r^{10}=\sqrt[6]{(2^3)}=\sqrt{2}\)

Hope it's clear.
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 19 Feb 2010, 23:54
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Currency wrote:
A strain of bacteria multiplies such that the ratio of its population in any two consecutive minutes is constant. If the bacteria grows from a a population of 5 million to 40 million over the course on an hour, by what factor does the population increase every 10 minutes?

\sqrt{2}


These linear and geometric sequences ar rattling me. Can anyone take a run at explaining the answer here?

Thanks

c


Interval = 10 minuets
1 hour = 60 min/10 min = 6 intervals

5 mil (1+x)^6 = 40 mil where (1 + x) is the rate of change (constant ratio).
(1+x)^6 = 8
(1+x)^6 = (sqrt 2)^6
1 + x = sqrt 2

1 + x is the rate of change (constant ratio). So its sqrt 2.
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A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post Updated on: 20 Feb 2010, 10:11
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Consider the values of bacteria as a sequence.
Since ratio of consecutive terms is constant , this is geometric progression.

with 1st term as a =5 , last term as \(ar^{60} = 40\)

=>\(r^{60}= 8\) => \(r^{20}\) = 2
=> \(r^{10}\) = \(\sqrt{2}\)

we need to find the factor after 10 mins
which is \(r^{10}\)
= \(\sqrt{2}\)
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Originally posted by gurpreetsingh on 20 Feb 2010, 00:46.
Last edited by gurpreetsingh on 20 Feb 2010, 10:11, edited 1 time in total.
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 20 Feb 2010, 07:42
gurpreetsingh wrote:
with 1st term as a =5 , last term as

=> =>
=> =

we need to find the factor after 10 mins
which is
=


can you break it down further? Can you put the formula used and define each term? Cutting right to plugging in is moving too fast for me on this questions.

Also where did the 0 come from in your answer, and why did it just disappear when it looked like it should have made the whole side of that equation zero.

Thanks
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 20 Feb 2010, 10:18
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There is no zero, its actually r raised to the power 60, 20 and 10...its not coming in alignment.
( Mods pls correct it if I m doing something wrong)

geometric progression has terms as : a , ar , \(ar^2\) ..... \(ar^(n-1)\)
In Gp series, ratio of consecutive terms is constant.

Since its given that ratio of population of consecutive terms is constant, that means we can consider the values of population as terms of Gp series... with initial term a = 5

after 1 min it will be ar
after 2 min it will be ar^2
.................................
after 10 min it will be ar^10
.................................
after 60 min it will be ar^60 = 40
now just use the value of "a' here and find r^10 which is the population after 10 mins.
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 23 Mar 2010, 21:54
TheSituation wrote:
A strain of bacteria multiplies such that the ratio of its population in any two consecutive minutes is constant. If the bacteria grows from a a population of 5 million to 40 million over the course on an hour, by what factor does the population increase every 10 minutes?

\sqrt{2}


These linear and geometric sequences ar rattling me. Can anyone take a run at explaining the answer here?

Thanks

c



Since the series you know is a geometric one we have to solve like this...

let the series be 5, a1, a2, a3, a4, a5 , 40

where a1 is population after 10 minutes
a2 after 20 and so on....
40 is the population at 60th minute.

We need to find the ratio of increase which is driven by a formula (40/5)^(1/6)

here 40 is last term
5 is first term
6 comes from 5 intervals + 1

so we get sqrt2...

Hope it helps
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 24 Mar 2010, 03:20
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Series will be 5 5x 5x^2 5x^3----5x^10--------5x^60

X^10 we need to find out.

So
5x^60 = 40
X^60 = 8
((X^10)^2)^3 = 8
X^10 = 2^(1/2)
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 03 Oct 2014, 05:40
Yet to find a more apt explanation to it.

My take on it: p - population

t=0, p = 5 mil

t= 60 minutes, p = 40 mil

let r be the rate of increase.

at t=1, we have pr
t =2, pr^2 (where p is the initial population - 5)

hence, t=60,
40 = 5*r^60
--> r^60 = 8
Since we are look for r at every 10 minutes, we can take the sixth root both sides
r^10 = 2^0.5
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New post 06 Oct 2014, 16:56
Can somebody please show the understandable solution to this question, I appreciate the above mention solutions by Maths prodigies( which isn't me) but i need a simple solution to this question that can be implemented in the time constraint of Gmat.

If geometric progression is applied here - what is the common ratio?
pleaseeeeeeeee help - EXPERTSSSSSSS
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A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 07 Oct 2014, 02:13
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Bacteria.................. Start

5 ................................ 0

5x .............................. 10 Minutes (Say x is the multiplication factor)

\(5x^2\) ........................... 20 Minutes

\(5x^3\) ........................... 30 Minutes

\(5x^4\) ........................... 40 Minutes

\(5x^5\) ........................... 50 Minutes

\(5x^6\) ........................... 60 Minutes

Given that \(5x^6 = 40\)

\(x^6 = 8 = 2^3\)

\(x = 2^{\frac{3}{6}} = \sqrt{2}\)
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 07 Oct 2014, 02:27
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Someone please correct me in the following procedure

5X^6 = 40
X^6 = 8
X^6 = 2^3
X^3 = 2
X = (2)^1/3

What's wrong with the above procedure ?
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New post 07 Oct 2014, 03:05
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Thoughtosphere wrote:
Someone please correct me in the following procedure

5X^6 = 40
X^6 = 8
X^6 = 2^3
X^3 = 2
X = (2)^1/3

What's wrong with the above procedure ?


When taking the cube root from \(x^6 = 2^3\), we get \((x^6)^{(\frac{1}{3})} = (2^3)^{(\frac{1}{3})}\) --> \(x^2 = 2\) --> \(x = \sqrt{2}\).

Hope it helps.
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 07 Oct 2014, 03:14
Bunuel wrote:
Thoughtosphere wrote:
Someone please correct me in the following procedure

5X^6 = 40
X^6 = 8
X^6 = 2^3
X^3 = 2
X = (2)^1/3

What's wrong with the above procedure ?


When taking the cube root from \(x^6 = 2^3\), we get \((x^6)^{(\frac{1}{3})} = (2^3)^{(\frac{1}{3})}\) --> \(x^2 = 2\) --> \(x = \sqrt{2}\).

Hope it helps.


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New post 07 Oct 2014, 10:25
Thanks a lot Bunnel and Paresh.
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A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 17 May 2015, 01:39
This question gives us more information than we need... in 1 hour (60 min) population grows by factor 8... so it doubles every 20 min --> k^20 = 2 --> k^20/2 or k^10 = 2^1/2

Alternative
K^60 = 8 (40 Mio / 5 Mio) --> k^10 = 8^1/6 = 2^1/2
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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 18 May 2015, 04:09
Bunuel wrote:
taleesh wrote:
Can somebody please show the understandable solution to this question, I appreciate the above mention solutions by Maths prodigies( which isn't me) but i need a simple solution to this question that can be implemented in the time constraint of Gmat.

If geometric progression is applied here - what is the common ratio?
pleaseeeeeeeee help - EXPERTSSSSSSS


A strain of bacteria multiplies such that the ratio of its population in any two consecutive minutes is constant. If the bacteria grows from a a population of 5 million to 40 million over the course on an hour, by what factor does the population increase every 10 minutes?

In 60 minutes the population increased 8 times. So, if the growth rate per minute is r, then \(r^{60}=8=2^3\). The question asks what factor the population will grow by in 10 minutes. The growth rate in 10 minutes is \(r^{10}\). take 6th root from \(r^{60}=2^3\) --> \(r^{10}=\sqrt[6]{(2^3)}=\sqrt{2}\)

Hope it's clear.


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