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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
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Bunuel wrote:
ikokurin wrote:
It takes 30 days to fill a laboratory dish with bacteria. If the size of the bacteria doubles each day, how long did it take for the bacteria to fill one half of dish?

A) 10 days
B) 15 days
C) 24 days
D) 29 days
E) 29.5 days

It is a fairly simple problem, but I am struggling to express what happens here algebraically. Please help on algebra here.


Since it takes 30 days to fill the dish and the population doubles each day, then the dish will be half full after 29 days: 1 day later (so after 30 days) the population will double again and the dish will be full.

Answer: D.

Algebraic approach:

Say initial population occupies 1/n of the disch.
Given: \(\frac{1}{n}*2^{30}=1\)
Question: if \(\frac{1}{n}*2^{x}=\frac{1}{2}\), then \(x=?\)

\(\frac{1}{n}*2^{x}=\frac{1}{2}\) --> \(\frac{1}{n}*2^{x}*2=1\) --> \(\frac{1}{n}*2^{x+1}=1\). Since we know that \(\frac{1}{n}*2^{30}=1\), then \(2^{x+1}=2^{30}\) --> \(x+1=30\) --> \(x=29\).

Hope it helps.


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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
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If we understand the question statement properly this question can be solved in under 30 seconds
If the bacteria in the dish double every day and in 30 days the dish is full,just 1 day before it the dish will be exactly half - full
on 29th day the dish would be half full
Correct answer - D
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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
Let x be the number of bacteria.

Then first day it will be 2x=1st day
2^2=2nd day
2^3=3rd day
....
....
.....

2^30=30th day-Full
These are all powers of 2(double the next day and half the previous day).So on the 29th day,number of bacteria would be just half.
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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
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I found this formula to be easy to apply.

Final population growth = S * P ^ (t/l)
S = starting population
P = progression (doubles = 2, triples = 3 etc.)
t/l = total amount of iterations
t = time
I = intervals

In our question, let the final population be x, in which case ---- x = 1 * 2 ^ (30 days/1 day)

Question, 1/2 * x = 1 * 2^ ( t days / 1 day), we know that x = 2^30...substitute and we get 1/2 * 2^30 = 2 ^ t, from here we can calculate t which is = 29 days
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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
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kalita wrote:
It takes 30 days to fill a laboratory dish with bacteria. If the size of the bacteria doubles each day, how long did it take for the bacteria to fill one half of dish?

A) 10 days
B) 15 days
C) 24 days
D) 29 days
E) 29.5 days


Since the size of the bacteria doubles each day and it takes 30 days to fill the laboratory dish with bacteria, it must be true that it took 29 days to fill half of the dish with bacteria (since the following day, the 30th day, the size of the bacteria will double and fill the entire dish).

Answer: D
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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
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kalita wrote:
It takes 30 days to fill a laboratory dish with bacteria. If the size of the bacteria doubles each day, how long did it take for the bacteria to fill one half of dish?

A) 10 days
B) 15 days
C) 24 days
D) 29 days
E) 29.5 days


KEY INFO: The size of the bacteria doubles each day
So, we can write: (size of bacteria on Day 29)(2) = (size of bacteria on Day 30)

GIVEN: On Day 30, the dish is at 100% capacity
So, we can write: (size of bacteria on Day 29)(2) = (100% capacity)
Divide both sides by 2 to get: (size of bacteria on Day 29) = (100% capacity)/2
Simplify: (size of bacteria on Day 29) = 50% capacity

So, on Day 29, the dish was HALF (aka 50%) full.

Answer: D

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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
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It's fairly simple in an obvious way. If we were to solve it using equations, we can use geometric progression as well:

let the size on the first day = X => x, 2x, 2^2x, 2^3x ...

The nth term of a G.P is [a.r^(n-1)] => on the 30th day we have: X*2^29. Half of which is X*2^28

This is nothing but the 29 term in the G.P. => 29 days.

It's more calculation but can be useful if you don't know where to start.

Ans C
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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
Hi VeritasKarishma - Veritas blog posts are very insightfull. Just wondering, does veritas have a blog post on this topic or problems like this ? While i understand the solution, unless i read some theory on this topic -- i am pretty sure, i will forget it during D - day
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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
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jabhatta@umail.iu.edu wrote:
Hi VeritasKarishma - Veritas blog posts are very insightfull. Just wondering, does veritas have a blog post on this topic or problems like this ? While i understand the solution, unless i read some theory on this topic -- i am pretty sure, i will forget it during D - day


It is a puzzle kind of question. You need to recognise this - if it doubles every day, the previous day it would be half of today.

There are some posts on some popular puzzles here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/catego ... /?s=puzzle

Also, if you do not recognize this, you can solve it using geometric progression.

Say you start with an amount a and at the end of Day1, you have 2a. At the end of day 2, you have 4a and so on... So your GP starts with 2a.

\(t_n = t_{30} = 2a*2^{29} = 1\) (since it takes 30 days to fill up 1 dish)
This is the end of the 30th day.

\(a = 1/2^{30}\\
\)

When the dish is half full, say x days have passed.
\(t_x = 1/2 = 2a*2^{x-1}\)

\(1/2 = 1/2^{30} * 2^x\)

\(2^{29} = 2^x\)

x = 29 days

Geometric Prog is discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... gressions/
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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
BPHASDEU wrote:
I found this formula to be easy to apply.

Final population growth = S * P ^ (t/l)
S = starting population
P = progression (doubles = 2, triples = 3 etc.)
t/l = total amount of iterations
t = time
I = intervals

In our question, let the final population be x, in which case ---- x = 1 * 2 ^ (30 days/1 day)

Question, 1/2 * x = 1 * 2^ ( t days / 1 day), we know that x = 2^30...substitute and we get 1/2 * 2^30 = 2 ^ t, from here we can calculate t which is = 29 days


In the following line:
In our question, let the final population be x, in which case ---- x = 1 * 2 ^ (30 days/1 day)

does 1 stand for initial value? But we don't know that. So are we assuming that the initial value is 1 and final value is X?
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Re: It takes 30 days to fill a laboratory dish with bacteria. If the size [#permalink]
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