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

Question

The population of bacteria in Colony A increases by 200% every three days, while the population of bacteria in Colony B increases by 800% every five days. Initially, Colony A starts with 243 bacteria and Colony B starts with 9 bacteria. In how many days will these colonies have the same population?

A. 3
B. 9
C. 15
D. 27
E. 45

A. 3
B. 9
C. 15
D. 27
E. 45

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Bunuel · Accepted Answer

Official Solution:

The population of bacteria in Colony A increases by 200% every three days, while the population of bacteria in Colony B increases by 800% every five days. Initially, Colony A starts with 243 bacteria and Colony B starts with 9 bacteria. In how many days will these colonies have the same population?

A. 3
B. 9
C. 15
D. 27
E. 45

A. 3
B. 9
C. 15
D. 27
E. 45

An increase of 200% triples the original number, and an increase of 800% multiplies the original number by nine. Therefore, the population of Colony A triples every three days, while the population of Colony B increases ninefold every five days. Assuming these colonies reach the same population after  n  days, the population of Colony A will have tripled  \frac{n}{3}  times, and the population of Colony B will have increased ninefold  \frac{n}{5}  times. Consequently, we have the equation:
 
 243*3^{\frac{n}{3}}=9*9^{\frac{n}{5}}

3^5*3^{\frac{n}{3}}=3^2*9^{\frac{n}{5}}

3^3*3^{\frac{n}{3}}=9^{\frac{n}{5}}

3^{3 + \frac{n}{3}}=3^{2*\frac{n}{5}}

3 + \frac{n}{3} = 2*\frac{n}{5}

45 + 5n = 6n

n=45 
 
Therefore, these colonies will have the same population in 45 days.

Answer: E

GMATinsight · Answer

Important thing for GMAT aspirants to know

200% increase = 3 times the original population

So population is becoming 3 times every three days in Colony A
and population is becoming 9 times every five days in Colony B

LCM (3, 5) = 15

So, Population in colony A becomes 3^5 every 15 days
and, Population in colony B becomes 9^3 every 15 days

Let's assume that population becomes same in t cycles of 15 days

Population of A in t cycles of 15 days =  243*3^{5t} 
Population of B in t cycles of 15 days =  9*9^{3t}

Now,  243*3^{5t} = 9*9^{3t} 
i.e.  3^5*3^{5t} = 3^2*3^{2*3t} 
i.e.  3^{5t+5} = 3^{6t+2}

i.e. 5t+5 = 6t+2
i.e. t = 3

i.e. Total days = 15*3 = 45

Answer: Option E

pranjalshah · Answer

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.

Purnank · Answer

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

Nidzo · Answer

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

shanoo.revankar · Answer

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

Adarsh_24 · Answer

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.

HarshavardhanR · Answer

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

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