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08 Jun 2024, 14:23
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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
08 Jun 2024, 14:23
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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
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
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
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
14 Apr 2025, 01:09
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Can also solve through working out a few scenarios.
Population of A increases 200% every 3 days, means it becomes 3 times every 3 days.Initially it's 243 which is 3^4, after every 3 days, the power of 3 will increase by 1. For ex, on 3rd day, population of A will be 3^5, on 6th day, it will be 3^6, on 9th day, it will be 3^7 and on 12th day it will be 3^8.
For Colony B, we have a 800% increase, means it becomes 9 times every 5 days. So on day 1 it is 3^2, on day 5 it will be 3^2x9 => 3^4 (so every 5th day the power increases 2 times)
On 10th day it will be 3^6 and on 15th day it will be 3^8
From these, we can eliminate A, B and C. Now for 27 days, we only need to check for Colony B.
on 20th day, it comes 3^10, 25th day it becomes 3^11 so on 27th day it's still 3^11.
For colony A, 15th day will give ys 3^9, 18th day will give us 3^10 and 21st day will give us 3^11 and 24th day will give us 3^12.
On 24th day, population of A > population of B on 27th day so we can eliminate 27 as well
Hence, the only possible answer is (E) 45
Population of A increases 200% every 3 days, means it becomes 3 times every 3 days.Initially it's 243 which is 3^4, after every 3 days, the power of 3 will increase by 1. For ex, on 3rd day, population of A will be 3^5, on 6th day, it will be 3^6, on 9th day, it will be 3^7 and on 12th day it will be 3^8.
For Colony B, we have a 800% increase, means it becomes 9 times every 5 days. So on day 1 it is 3^2, on day 5 it will be 3^2x9 => 3^4 (so every 5th day the power increases 2 times)
On 10th day it will be 3^6 and on 15th day it will be 3^8
From these, we can eliminate A, B and C. Now for 27 days, we only need to check for Colony B.
on 20th day, it comes 3^10, 25th day it becomes 3^11 so on 27th day it's still 3^11.
For colony A, 15th day will give ys 3^9, 18th day will give us 3^10 and 21st day will give us 3^11 and 24th day will give us 3^12.
On 24th day, population of A > population of B on 27th day so we can eliminate 27 as well
Hence, the only possible answer is (E) 45
