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.
Population - OG-2nd Edition
This topic has expert replies
- vk_vinayak
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I am showing the calculation you left off: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.
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)
I will (Learn. Recognize. Apply)
- niketdoshi123
- Master | Next Rank: 500 Posts
- Posts: 210
- Joined: Thu Mar 08, 2012 11:24 pm
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- Followed by:3 members
The best way to solve this question is to simply multiply '2' 6 times to 10^4.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.
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