Question

A population of a colony of bacteria increases by 20 percent every 3 minutes. If at 9:00am the colony had a
population of 144,000, what was the population of the colony at 8:54am?

Answer 1

Ok, well.. Simplest way I see to do it:
we know that 6 minutes passed between 8:54am to 9:00am
Let's find out what the population was 3 minutes earlier at 8:57am.
We know it grew up to 144,000 by those 3 minutes. This increase is 20% of what it was at 8:57am.. so:
\[
120\% = 144,000 \\
1\% = \frac{144,000}{120} = 1,200 \\
100\% = 1\% \cdot 100 = 1,200 \cdot 100 = 120,000
\]
So at 8:57am we had population of 120,000. Now same idea 3 minutes earlier:
\[
120\% = 120,000 \\
1\% = \frac{120,000}{120} = 1,000 \\
100\% = 1\% \cdot 100 = 1,000 \cdot 100 = 100,000
\]
If my calculations are right, the population should be 100,000 at 8:54am

Answer 2

Pitamar is right, if you calculate it as a check, you start at

100,000 + 20% = 120,000 -> 8:57

then another 3 minutes

120,000 + 20% = 120 000 +24 000 = 144 000

Answer 3

Ye.. well, that's the 'simple' way doing it, but if they would ask you what was it at 8:03am it's not practical.
So we could define a function for this f(t), where 't' is the number of minutes passed from 9:00am
\[
f(t) = 144,000 \cdot \bigg( \frac{6}{5} \bigg)^{\Large \frac{t}{3}}
\]
By the given data the result is known to be true only if 't' is divisible by 3.
Then to calculate 8:54 we do:
\[
8:54am - 9:00am = -6_{minutes} \\
f(-6_{minutes}) = 144,000 \cdot \bigg( \frac{6}{5} \bigg)^{\Large \frac{-6}{3}} = \\
= 144,000 \cdot \bigg( \frac{6}{5} \bigg)^{-2} = 144,000 \cdot \bigg( \frac{5}{6} \bigg)^{2} = \\
= 144,000 \cdot \frac{25}{36} = \frac{3,600,000}{36} = 100,000
\]

Answer 4

thanks pitamar , this formula was just what i was looking for. I did solve it using simple way as u did before.

Answer 5

actually i did it a bit more simply. If x = original quantity , we know 20 % change occured twice, so x*(1.2)*(1.2) = 144000. So solving for x we get x = 100000

Answer 6

Ye.. I guess I tend to do it over-detailed.. Well glad i could help