A virus population doubles every 3 minutes and reaches a size of K after 1 hour; at what time before the 1 hour mark was the population equal to K divided by 4?

Introduction / Context:
This question involves exponential growth, a common concept in population dynamics, compound interest, and many other areas. The virus population doubles at regular intervals, specifically every 3 minutes. We are told that after 1 hour the population is K and asked when it was K/4. Problems like this test understanding of exponential sequences and backward reasoning through the growth process.

Given Data / Assumptions:
- The virus population doubles once every 3 minutes.
- The population after 1 hour, that is after 60 minutes, is K.
- The growth process is consistent and follows exact doubling with no delays.
- We need the time at which the population was one quarter of K, that is K/4.

Concept / Approach:
Doubling every fixed interval creates a geometric sequence. Going forward in time multiplies the population by 2 each step, while going backward divides by 2. If we know the population at a later time, we can find earlier values by repeatedly halving. To find when the population was K/4, we identify how many doubling steps separate K and K/4. Since K/4 is two halvings before K (because K divided by 2 is K/2 and dividing again by 2 gives K/4), we move back two doubling intervals from 60 minutes. Each interval is 3 minutes, so two intervals is 6 minutes.

Step-by-Step Solution:
Step 1: Understand that each doubling interval is 3 minutes long. Step 2: At 60 minutes, the population is K. Step 3: One doubling period earlier, at 60 - 3 = 57 minutes, the population must have been half of K, that is K/2. Step 4: Another doubling period earlier, at 57 - 3 = 54 minutes, the population must have been half of K/2, that is K/4. Step 5: Therefore, the population was K/4 at 54 minutes from the start. Step 6: Among the given options, 54 minutes matches this calculation exactly.

Verification / Alternative check:
We can check by moving forward in time from 54 minutes. At 54 minutes, suppose the population is K/4. After one doubling interval of 3 minutes, at 57 minutes, the population doubles to 2 * (K/4) = K/2. After another 3 minutes, at 60 minutes, it doubles again to 2 * (K/2) = K. This sequence confirms that starting with K/4 at 54 minutes reaches K at 60 minutes, exactly as stated in the problem. No other option produces this correct chain of doublings.

Why Other Options Are Wrong:
Option 29 minutes: This time is far earlier than needed. From 29 minutes to 60 minutes there are many doubling intervals, so the growth from that time would far exceed K/4 to K in only two steps.
Option 25 minutes: Similar to 29 minutes, this time is too early and would involve too many doublings to reach K at 60 minutes.
Option 57 minutes: At 57 minutes the population is K/2, not K/4, because it is just one halving step from K rather than two.
Option 45 minutes: This is three doubling periods (9 minutes) earlier than 54 minutes, so the population there would be K/32 when we work backward, not K/4.

Common Pitfalls:
One common mistake is to divide the total time proportionally instead of by doubling intervals. Some learners incorrectly think K/2 must occur at 30 minutes and K/4 at 15 minutes, but that assumes linear growth rather than exponential growth. Another error is miscounting how many halving steps are required to go from K to K/4. Remember that exponential processes are controlled by the number of doubling intervals, not by the raw time fraction.

Final Answer:
The virus population was K/4 at 54 minutes.