The population of bacteria in Colony A increases by 200% every three

I think its E.

Let the population be equal in n days.

Then, 243*3^(n/3)=9*9^(n/5).

200% increase means it's becoming 3x every 3 days and 800% means it's becoming 9x every 5 days.

So, for colony A, the population first increases to 3x, then next to 3*3x=9x, and so on. Thus we can say it's increasing by a factor of 3^n. But this is happening every 3 days, so the factor is 3^(3/n). Multiplying that by our original population gives us the LHS. The same way we get RHS.

Solving, we get n=45.­

­The population of bacteria in Colony A increases by 200% every three days (means become 3times), while the population of bacteria in Colony B increases by 800% every five days (means become 9times).
Colony A starts with 243 bacteria and Colony B starts with 9 bacteria.
Assuming n are the days after which these colonies have the same population.
243*3^(n/3) = 9*9^(n/5)
(3^5) * (3^(n/3)) = (3^2) * (3^(2n/5))
5 + (n/3) = 2 + (2n/5)
simplifying
75 + 5n = 30 + 6n
n = 45
Answer E
 

For this one can use the compound interest formula of \(A = P(1+\frac{r}{100})^n\)

\(A\) being the final amount.
\(P\) the initial amount.
\(r\) the rate (or percentage).
\(n\) the time it takes.

COLONY A:
\(A\): Not needed here, as one will make Colony A's equation equal to Colony B's. 

\(P\): 243

\(r\): As Colony A, increases by 200%, when plugging 200 in for r, one notices that it is the same as multiplying by 3. In other words, \((1+\frac{r}{100})\) becomes \((3)\).

\(n\): One is solving for this value. However, one cannot use simply n. As the increase occurs every 3 days, one needs to make it \(\frac{n}{3}\)

Completed Equation for Colony A: \(243(3)^{\frac{n}{3}}\)


COLONY B:
\(P\): 9

\(r\): Doing the same here as with Colony A, one will see that plugging in 800 for r makes \((1+\frac{r}{100})\) equal to \((9)\).

\(n\): Doing the same as for Colony A, given that it increases every 5 days n becomes \(\frac{n}{5}\)

Completed Equation for Colony A: \(9(9)^{\frac{n}{5}}\)


SOLVING:

 \(243(3)^{\frac{n}{3}} = 9(9)^{\frac{n}{5}}\)

\(\frac{243}{9} = \frac{9^{\frac{n}{5}}}{3^{\frac{n}{3}}}\)

\(27 = \frac{3^{\frac{2n}{5}}}{3^{\frac{n}{3}}}\)

\(3^3 = 3^{\frac{2n}{5}-\frac{n}{3}}\)

DROP THE BASES

\(3 = \frac{2n}{5}-\frac{n}{3}\)

\(3 = \frac{n}{15}\)

\(n = 45\)

ANSWER E



 ­

­What is the explnation behid dividing the n by 3 or n by 5 ?

Because it’s given in the question the 9x or 3x multiplication happens every 5 or 3 days respectively.
9x is basically 1+800/100 = 900/100 = 9
Similarly 200% too.

Harsha
GMAT Tutor | gmatanchor.com