AMOUNT OF BACTERIA PRESENT Time Amount 1:00 P.M. 10.0 grams 4:00 P.M. x grams 7:00 P.M. 14.4 grams Data for...

1. Translate the problem requirements: We need to find the amount of bacteria at 4:00 P.M. The bacteria increases by the same fraction (percentage) during each 3-hour period. This means if the growth factor is r, then: \(10.0 \times \mathrm{r} = \mathrm{x}\) (amount at 4:00), and \(\mathrm{x} \times \mathrm{r} = 14.4\) (amount at 7:00).
2. Set up the growth relationship: Since the same growth factor applies to both periods, we can express the 7:00 P.M. amount in terms of the 1:00 P.M. amount using two consecutive growth periods.
3. Calculate the growth factor: Use the fact that 10.0 grams grows to 14.4 grams over two identical periods to find the growth factor per period.
4. Apply the growth factor: Multiply the 1:00 P.M. amount by the growth factor to get the 4:00 P.M. amount.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what's happening here. We have bacteria growing over time, and the key insight is that it grows by the same fraction (or percentage) during each 3-hour period.

Think of it like this: if your savings account earns the same interest rate each month, then whatever multiplier applies in month 1 also applies in month 2. Here, whatever growth multiplier applies from 1:00 to 4:00 P.M. is the same multiplier that applies from 4:00 to 7:00 P.M.

So if we call this growth multiplier 'r', then:
• Starting amount (1:00 P.M.): 10.0 grams
• After first growth period (4:00 P.M.): \(10.0 \times \mathrm{r} = \mathrm{x}\) grams
• After second growth period (7:00 P.M.): \(\mathrm{x} \times \mathrm{r} = 14.4\) grams

Process Skill: TRANSLATE - Converting the phrase 'same fraction increase' into a consistent growth multiplier

2. Set up the growth relationship

Since the same growth factor applies twice, we can connect the starting amount directly to the ending amount.

Imagine this like a two-step journey: 10.0 grams → (multiply by r) → x grams → (multiply by r again) → 14.4 grams

This means: \(10.0 \times \mathrm{r} \times \mathrm{r} = 14.4\)
Or more simply: \(10.0 \times \mathrm{r}^2 = 14.4\)

This gives us a way to find r without needing to know x first!

3. Calculate the growth factor

From our equation: \(10.0 \times \mathrm{r}^2 = 14.4\)

Dividing both sides by 10.0:
\(\mathrm{r}^2 = 14.4 ÷ 10.0 = 1.44\)

To find r, we take the square root:
\(\mathrm{r} = \sqrt{1.44} = 1.2\)

This means the bacteria amount gets multiplied by 1.2 each period, which represents a 20% increase each time (since 1.2 = 1 + 0.2).

4. Apply the growth factor

Now that we know the growth factor is 1.2, we can find the amount at 4:00 P.M.:

Amount at 4:00 P.M. = Amount at 1:00 P.M. × growth factor
\(\mathrm{x} = 10.0 \times 1.2 = 12.0\) grams

Let's verify: Does 12.0 grams grow to 14.4 grams with the same growth factor?
\(12.0 \times 1.2 = 14.4\) ✓

Perfect! Our answer checks out.

Final Answer

The amount of bacteria present at 4:00 P.M. was 12.0 grams.

The answer is A. 12.0

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "same fraction increase" as linear growth
Students often think this means the bacteria increases by the same amount each period (e.g., +2.4 grams each time) rather than by the same multiplier. This leads to setting up: \(10.0 + \mathrm{k} = \mathrm{x}\) and \(\mathrm{x} + \mathrm{k} = 14.4\), which gives incorrect results.

2. Setting up addition instead of multiplication relationships
Even when students recognize it's not linear, they might think "same fraction" means adding the same fraction each time: 10.0 + (fraction of 10.0) = x. They fail to recognize that percentage growth compounds multiplicatively.

3. Trying to solve for x directly without using the growth factor
Students may attempt to use the constraint that growth from 1:00-4:00 equals growth from 4:00-7:00 as: \((\mathrm{x} - 10.0) = (14.4 - \mathrm{x})\), missing that "same fraction" refers to the multiplier, not the absolute difference.

Errors while executing the approach

1. Arithmetic errors when calculating the square root
Students may incorrectly calculate \(\sqrt{1.44}\). Common mistakes include thinking \(\sqrt{1.44} = 1.4\) or making decimal errors, leading to wrong growth factors and final answers.

2. Forgetting to divide by the initial amount
When setting up \(10.0 \times \mathrm{r}^2 = 14.4\), students might forget to divide both sides by 10.0, leading to \(\mathrm{r}^2 = 14.4\) instead of \(\mathrm{r}^2 = 1.44\), which gives completely wrong values.

3. Using the wrong relationship in the equation setup
Students might write the equation as \(\mathrm{r}^2 = 10.0/14.4\) instead of \(\mathrm{r}^2 = 14.4/10.0\), confusing which value should be in the numerator versus denominator.

Errors while selecting the answer

No likely faltering points
Once students correctly calculate x = 12.0, the answer choice A is straightforward to select, and verification (\(12.0 \times 1.2 = 14.4\)) is relatively simple to check.