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Re: A strain of bacteria multiplies such that the ratio of its population [#permalink]
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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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Re: A strain of bacteria multiplies such that the ratio of its population [#permalink]
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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 population [#permalink]
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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 population [#permalink]
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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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Re: A strain of bacteria multiplies such that the ratio of its population [#permalink]
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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 population [#permalink]
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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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Re: A strain of bacteria multiplies such that the ratio of its population [#permalink]
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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 population [#permalink]
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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 population [#permalink]
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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Re: A strain of bacteria multiplies such that the ratio of its population [#permalink]
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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 population [#permalink]
Given: A strain of bacteria multiplies such that the ratio of its population in any two consecutive minutes is constant.
Asked: 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?

Let the factor by which the population increase every 10 minutes be x.

5 million * xˆ6 = 40 million
xˆ6 = 40/5 = 8 = 2ˆ3
\(x = \sqrt{2}\)

IMO B
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Re: A strain of bacteria multiplies such that the ratio of its population [#permalink]
Bunuel KarishmaB

Every reply seems to have taken r^60 but as the question asks that bacteria grows every 2 min interval thus over a period of one hour there should be 30 such intervals

Thus r^30 = 8 =2^3

Now question asks the ratio of growth for 10 min interval

That should be IMO r^10/r^5 = r^5 (considering 5 2min interval in 10 mins and 10 2min interval in 20 min )

Now r^30 = 2^8 from first equation and r^5 = ?
This gives the same answer = sqrt(2)

But I wanted to know if my line of thought is correct or do I need to review something or am I missing something?

Please assist

Posted from my mobile device
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Re: A strain of bacteria multiplies such that the ratio of its population [#permalink]
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TPW wrote:
Bunuel KarishmaB

Every reply seems to have taken r^60 but as the question asks that bacteria grows every 2 min interval thus over a period of one hour there should be 30 such intervals

Thus r^30 = 8 =2^3

Now question asks the ratio of growth for 10 min interval

That should be IMO r^10/r^5 = r^5 (considering 5 2min interval in 10 mins and 10 2min interval in 20 min )

Now r^30 = 2^8 from first equation and r^5 = ?
This gives the same answer = sqrt(2)

But I wanted to know if my line of thought is correct or do I need to review something or am I missing something?

Please assist

Posted from my mobile device


The statement "A strain of bacteria multiplies such that the ratio of its population in any two consecutive minutes is constant" indicates that the bacteria's population growth occurs every minute, not every two minutes.
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Re: A strain of bacteria multiplies such that the ratio of its population [#permalink]
I tried solving this question by using GP. An= ar^(n-1). So this should be 5*R^59. Can someone help me with whats wrong here? Why have we taken 5r^60= 40?
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