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

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New post 19 Feb 2010, 20:44
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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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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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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 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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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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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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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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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
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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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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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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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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.


Bang On!!!!

Thanks a lot... +1 to you... :thanks
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New post 07 Oct 2014, 10:25
Thanks a lot Bunnel and Paresh.
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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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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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Re: A strain of bacteria multiplies such that the ratio of its p  [#permalink]

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New post 29 Nov 2019, 07:56
Hi chetan2u Bunuel generis VeritasKarishma Gladiator59

I have 1 very basic confusion. Since the interval is 60 minutes, shouldn't n be 60 minutes which will make the last term r^59 because of (n-1).

I usually get confused by this. What is the best way to sort this out ? Will appreciate any help with strategy to know the n part. Here as it appears n should be 61, but why and how.

Thanks in advance.
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New post 01 Dec 2019, 23:35
altairahmad wrote:
Hi chetan2u Bunuel generis VeritasKarishma Gladiator59

I have 1 very basic confusion. Since the interval is 60 minutes, shouldn't n be 60 minutes which will make the last term r^59 because of (n-1).

I usually get confused by this. What is the best way to sort this out ? Will appreciate any help with strategy to know the n part. Here as it appears n should be 61, but why and how.

Thanks in advance.


Think about it in terms of beginning of a minute to end of a minute. From beginning of 1st minute to end of 1st minute, from beginning of 2nd minute (which is the same as end of 1st minute) to end of second minute and so on till beginning of 60th minute to end of 60th minute - in 60 minutes the bacteria grew 8 times.
If the ratio of growth every minute is R, at the end of 1st minute, it became R times 5 million. After the end of 2nd minute, it became R^2 times 5 million.
So at the end of 60th minute, it became R^60 times 5 million.

I do understand that GP formula gives
Last term (A60) = First term * R^59
But that is so because R is multiplied for the first time to first term to get the second term. Here, after 2 mins, R has been multiplied to the initial value twice. So it will always depend on context. In time related questions, think in terms of beginning and end of the time period.
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Re: A strain of bacteria multiplies such that the ratio of its p   [#permalink] 01 Dec 2019, 23:35
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