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Bunuel
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I think this is a high-quality question and I agree with explanation.
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Hi Bunuel,
Can we do reverse calculation and arrive at the answer?

Bunuel
Official Solution:

If a bacterial colony doubles every day at 9 a.m. and loses 320 bacteria every day at 9 p.m., what was the population size of the colony at 8 a.m. on Day 3 if exactly 320 bacteria were lost at 9 p.m. on Day 5, resulting in extinction of the colony?

A. 160
B. 240
C. 280
D. 480
E. 560


Suppose the size of the colony at 8 a.m. on Day 3 was \(x\) bacteria;

At 9 a.m. on Day 3 it would become \(2x\) and at 9 p.m. of the same day it would become \(2x - 320\);

At 9 a.m. on Day 4 it would become \(2(2x - 320)\) and at 9 p.m. of the same day it would become \(2(2x - 320) - 320\);

At 9 a.m. on Day 5 it would become \(2*(2(2x - 320) - 320)\) and at 9 p.m. of the same day it would become \(2*(2(2x - 320) - 320) - 320\);

Since at that point the colony was extinct, then \(2*(2(2x - 320) - 320) - 320=0\).

Solving gives \(x = 280\)


Answer: C
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Hi Bunuel,
Can we do reverse calculation and arrive at the answer?

Bunuel
Official Solution:

If a bacterial colony doubles every day at 9 a.m. and loses 320 bacteria every day at 9 p.m., what was the population size of the colony at 8 a.m. on Day 3 if exactly 320 bacteria were lost at 9 p.m. on Day 5, resulting in extinction of the colony?

A. 160
B. 240
C. 280
D. 480
E. 560


Suppose the size of the colony at 8 a.m. on Day 3 was \(x\) bacteria;

At 9 a.m. on Day 3 it would become \(2x\) and at 9 p.m. of the same day it would become \(2x - 320\);

At 9 a.m. on Day 4 it would become \(2(2x - 320)\) and at 9 p.m. of the same day it would become \(2(2x - 320) - 320\);

At 9 a.m. on Day 5 it would become \(2*(2(2x - 320) - 320)\) and at 9 p.m. of the same day it would become \(2*(2(2x - 320) - 320) - 320\);

Since at that point the colony was extinct, then \(2*(2(2x - 320) - 320) - 320=0\).

Solving gives \(x = 280\)


Answer: C
We are told the bacterial colony doubles every day at 9 a.m. and loses 320 bacteria every day at 9 p.m. On Day 5 at 9 p.m., exactly 320 bacteria are lost and the colony becomes extinct. We are asked to find the size of the colony at 8 a.m. on Day 3.

We work backward from the extinction point:

9 p.m. Day 5: colony is extinct, so population = 0
Before losing 320: population must have been 320

9 a.m. Day 5: population = 320
Before doubling at 9 a.m.: population = 160

9 p.m. Day 4: population = 160
Before losing 320: population = 480

9 a.m. Day 4: population = 480
Before doubling at 9 a.m.: population = 240

9 p.m. Day 3: population = 240
Before losing 320: population = 560

9 a.m. Day 3: population = 560
Before doubling at 9 a.m.: population = 280

So, the population at 8 a.m. on Day 3 was 280.

Answer: C.
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