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Wes works at a science lab that conducts experiments on bacteria. The

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Wes works at a science lab that conducts experiments on bacteria. The  [#permalink]

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New post 06 Oct 2014, 09:17
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Tough and Tricky questions: Word Problems.



Wes works at a science lab that conducts experiments on bacteria. The population of the bacteria multiplies at a constant rate, and his job is to notate the population of a certain group of bacteria each hour. At 1 p.m. on a certain day, he noted that the population was 2,000 and then he left the lab. He returned in time to take a reading at 4 p.m., by which point the population had grown to 250,000. Now he has to fill in the missing data for 2 p.m. and 3 p.m. What was the population at 3 p.m.?

A. 50,000
B. 62,500
C. 65,000
D. 86,666
E. 125,000

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Re: Wes works at a science lab that conducts experiments on bacteria. The  [#permalink]

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New post 06 Oct 2014, 22:23
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1
1 AM ................. 2000

Let the rate of growth = x

2 AM ....................... 2000x

3 AM ......................... \(2000 x^2\)

4 AM ......................... \(2000 x^3\)

Given that \(2000 x^3 = 250000\)

x = 5

at 3 AM \(= 2000 * 5^2 = 50000\)

Answer = A
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Re: Wes works at a science lab that conducts experiments on bacteria. The  [#permalink]

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New post 06 Oct 2014, 12:07
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Bunuel wrote:

Tough and Tricky questions: Word Problems.



Wes works at a science lab that conducts experiments on bacteria. The population of the bacteria multiplies at a constant rate, and his job is to notate the population of a certain group of bacteria each hour. At 1 p.m. on a certain day, he noted that the population was 2,000 and then he left the lab. He returned in time to take a reading at 4 p.m., by which point the population had grown to 250,000. Now he has to fill in the missing data for 2 p.m. and 3 p.m. What was the population at 3 p.m.?

A. 50,000
B. 62,500
C. 65,000
D. 86,666
E. 125,000


let the rate be x, then population of the bacteria after each hour can be given as 2000, 2000x, 2000(x^2), 2000(x^3)

now population at 4pm =250,000
thus we have 2000(x^3) = 250,000
thus x=5

therefore population at 3pm = 2000(25) = 50,000
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Re: Wes works at a science lab that conducts experiments on bacteria. The  [#permalink]

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New post 17 Mar 2016, 16:09
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Here is a handwritten solution to the question. You could either use the algebraic equation, or try the answers to see which one works.
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Wes works at a science lab that conducts experiments on bacteria. The  [#permalink]

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New post 20 Jul 2017, 12:35
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Bunuel wrote:

Tough and Tricky questions: Word Problems.



Wes works at a science lab that conducts experiments on bacteria. The population of the bacteria multiplies at a constant rate, and his job is to notate the population of a certain group of bacteria each hour. At 1 p.m. on a certain day, he noted that the population was 2,000 and then he left the lab. He returned in time to take a reading at 4 p.m., by which point the population had grown to 250,000. Now he has to fill in the missing data for 2 p.m. and 3 p.m. What was the population at 3 p.m.?

A. 50,000
B. 62,500
C. 65,000
D. 86,666
E. 125,000

I worked from the answer choices because finding the constant rate looked quick for four of the answers. It was quick.

If we have
1 p.m. -- 2,000
2 p.m. -- ???
3 p.m. -- ???
4 p.m. -- 250,000

Then because the answer choices are for 3 p.m., divide 4 p.m. amount by an answer choice to get the multiplication factor.

Use that factor: divide the 3 p.m. amount by the factor to get the 2 p.m. amount, then once more (2 p.m. amt / factor = 1 p.m. amt) to see whether the answer matches 2,000.

1. I started with (C) where 3 p.m. amount is 125,000

\(\frac{250,000}{125,000}\) to get multiplication factor = 2

3:00 --> 2:00 is \(\frac{125,000}{2}\) = 62,500 at 2:00.

Stop. Much too big. The next step, \(\frac{65,000}{2}\) is not going to equal 2,000 at 1 p.m.

2. Try A, where 3 p.m. amount = 50,000

\(\frac{250,000}{50,000}\) = multiplication factor 5

Hour between 2:00 and 3:00 is then \(\frac{50,000}{5}\) = 10,000 at 2:00

2:00 -> 1:00 is \(\frac{10,000}{5}\) = 2,000 at 1 p.m. Correct

Answer A
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Re: Wes works at a science lab that conducts experiments on bacteria. The  [#permalink]

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New post 11 Apr 2018, 11:22
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Ouch this hurts...

Read the question twice..thrice and every time saw 1000 instead of 2000 :sad:
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New post 10 May 2018, 04:56
Bunuel Why can't we use the ARITHMETIC PROGRESSION FORMULA as the bacteria is multiplying at a constant rate?
I tried it but I got 86666 as the answer. Any help would be appreciated.

Thanks in advance. :-)
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New post 10 May 2018, 23:28
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Wes works at a science lab that conducts experiments on bacteria. The  [#permalink]

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New post 15 May 2018, 20:04
1
simple approach this problem:

First, we take out the last 3 zeros for simplicity's sake.

hr 1 = 2
hr 2 = unknown
hr 3 = unknown
hr 4 = 250

Since we know that it is growing at a constant rate, we could assume that there is a patter between the four numbers. Next we look at 250 and see its factors.
250 = 2 x 5 x 5 x 5 <-- notice that there is a 2 here and 3 five's. If the number is growing at a constant rate, then we could plug in the only difference here - the five's

hr 1 = 2
hr 2 = 2 x 5
hr 3 = 2 x 5 x 5 = 50,000 (don't forget the zeros we took out in the beginning)
hr 4 = 2 x 5 x 5 x 5

Answer is A
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Re: Wes works at a science lab that conducts experiments on bacteria. The  [#permalink]

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New post 10 Jun 2019, 00:40
1 AM ................. 2000

Let the rate of growth = x

2 AM ....................... 2000x^1

3 AM ......................... 2000x^2

4 AM ......................... 2000x^3

Given that 2000x^3=250000


at 3 AM =50000
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Re: Wes works at a science lab that conducts experiments on bacteria. The   [#permalink] 10 Jun 2019, 00:40
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