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Population - OG-2nd Edition

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Master | Next Rank: 500 Posts
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by shanice » Fri Aug 24, 2012 1:00 am
Hi everyone,

I've asked this question before and was given understandable solutions. However, I would like to know whether the below ways works for this type of problem:-

A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^4, what was the number in the population 1 hour later.

A)2(10^4)
B)6(10^4)
C)(2^6)(10^4)
D)(10^6)(10^4)
E)(10^4)^6

The answer is C.

Can I solve it this way?

Time Elapsed Population
Initially(Now) 10^4
10 minutes 10^4 + 10^4 = 10^4(2)
20 minutes (2)10^4 + (2)10^4 = got stuck
30 minutes
40 minutes
50 minutes
60 minutes

For ex: Double 5 = 2(5) = 10 or 5+5=10.

So in this case, I can use 2(10^4) which the answer will lead to (2^6)(10^4) but I would like to try adding it. Is it possible? Can anyone show me how?

I hope you guys understand what I'm trying to say here. I don't how to put it in right words.

Thank you.
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by vk_vinayak » Fri Aug 24, 2012 1:54 am
shanice wrote:Hi everyone,

I've asked this question before and was given understandable solutions. However, I would like to know whether the below ways works for this type of problem:-

A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^4, what was the number in the population 1 hour later.

A)2(10^4)
B)6(10^4)
C)(2^6)(10^4)
D)(10^6)(10^4)
E)(10^4)^6

The answer is C.

Can I solve it this way?

Time Elapsed Population
Initially(Now) 10^4
10 minutes 10^4 + 10^4 = 10^4(2)
20 minutes (2)10^4 + (2)10^4 = got stuck
30 minutes
40 minutes
50 minutes
60 minutes

For ex: Double 5 = 2(5) = 10 or 5+5=10.

So in this case, I can use 2(10^4) which the answer will lead to (2^6)(10^4) but I would like to try adding it. Is it possible? Can anyone show me how?

I hope you guys understand what I'm trying to say here. I don't how to put it in right words.

Thank you.
I am showing the calculation you left off:

Time Elapsed Population
Initially(Now) 10^4
10 minutes 10^4 + 10^4 = 10^4(2)
20 minutes (2)10^4 + (2)10^4 => 2 (10^4 + 10^4) => 2 (2)(10^4) => 2^2 (10^4)
30 minutes 2^2 (10^4) + 2^2 (10^4) => 2^2 (10^4 + 10^4) => 2^2 (2) 10^4 => 2^3 (10^4)
40 minutes
50 minutes
60 minutes

You can see that 2 (10^4) is common in all these, and the power of 2 is increasing with each 10 minutes.

The power of 2:
Initially 0, After 10 minutes => 1, After 20 minutes =>2, After 30 minutes =>3, ... so after 60 minutes, the power of 2 will be 6.

Therefore ans is : 2^6 (10^4)
- VK

I will (Learn. Recognize. Apply)
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by niketdoshi123 » Fri Aug 24, 2012 1:59 am
shanice wrote:Hi everyone,

I've asked this question before and was given understandable solutions. However, I would like to know whether the below ways works for this type of problem:-

A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^4, what was the number in the population 1 hour later.

A)2(10^4)
B)6(10^4)
C)(2^6)(10^4)
D)(10^6)(10^4)
E)(10^4)^6

The answer is C.

Can I solve it this way?

Time Elapsed Population
Initially(Now) 10^4
10 minutes 10^4 + 10^4 = 10^4(2)
20 minutes (2)10^4 + (2)10^4 = got stuck
30 minutes
40 minutes
50 minutes
60 minutes

For ex: Double 5 = 2(5) = 10 or 5+5=10.

So in this case, I can use 2(10^4) which the answer will lead to (2^6)(10^4) but I would like to try adding it. Is it possible? Can anyone show me how?

I hope you guys understand what I'm trying to say here. I don't how to put it in right words.

Thank you.
The best way to solve this question is to simply multiply '2' 6 times to 10^4.
However if you want to solve it by adding


Time Elapsed________________Population
Initially(Now)________________10^4
10 minutes________________10^4 + 10^4 = 10^4(2)
20 minutes________________(2)10^4 + (2)10^4 = (4)*10^4=(2^2)*10^4
30 minutes________________(4)10^4 + (4)10^4 = (8)*10^4=(2^3)*10^4
40 minutes________________(8)10^4 + (8)10^4 = (16)*10^4=(2^4)*10^4
50 minutes________________(16)10^4 + (16)10^4 = (32)*10^4=(2^5)*10^4
60 minutes________________(32)10^4 + (32)10^4 = (64)*10^4=(2^6)*10^4

As you can see, solving this way is pretty complex.Avoid this if you can :-)
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by shanice » Sat Aug 25, 2012 6:24 am
I won't be using addition for this type of problem. I just wanted to know whether it was possible to solve it through this method.

Thank you.
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