A batch of fresh grapes is fully dried to make raisins by 2 sequential processes. The first drying process is...

1. Translate the problem requirements: We need to find k such that k kilograms of fresh grapes becomes exactly 1 kilogram of raisins after two sequential weight reduction processes - first losing \(65\%\) weight, then losing \(45\%\) of the remaining weight.
2. Track weight through sequential reductions: Calculate what fraction of original weight remains after each drying process by determining what percentage survives each reduction.
3. Establish the relationship equation: Set up an equation where k kilograms multiplied by the total fraction remaining equals 1 kilogram of final raisins.
4. Solve for k using the weight survival fractions: Isolate k by dividing 1 by the product of survival fractions to find the initial weight needed.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what happens in plain English. We begin with some amount of fresh grapes (k kilograms) and we want to end up with exactly 1 kilogram of raisins after two drying processes.

Think of it like this: imagine you have a bag of fresh grapes. You put them outside to dry, and they lose a lot of their water weight. Then you put these partially dried grapes into a special drying machine, and they lose even more weight to become raisins.

The problem tells us:
• First drying: grapes lose \(65\%\) of their weight
• Second drying: the partially dried grapes lose \(45\%\) of their remaining weight
• We want exactly 1 kg of raisins at the end
• We need to find how many kg of fresh grapes (k) we should start with

Process Skill: TRANSLATE - Converting the percentage language into a clear understanding of what survives each process

2. Track weight through sequential reductions

Instead of thinking about what we lose, let's think about what survives each drying process. This makes the math much clearer.

If something loses \(65\%\) of its weight, then \(35\%\) of the original weight remains (since \(100\% - 65\% = 35\%\)).
If something loses \(45\%\) of its weight, then \(55\%\) of that weight remains (since \(100\% - 45\% = 55\%\)).

Let's trace through with a concrete example first:
Suppose we start with 10 kg of fresh grapes:
• After first drying: \(10 \text{ kg} \times 0.35 = 3.5 \text{ kg}\) of partially dried grapes
• After second drying: \(3.5 \text{ kg} \times 0.55 = 1.925 \text{ kg}\) of raisins

So our general pattern is:
Fresh grapes → (keep \(35\%\)) → Partially dried grapes → (keep \(55\%\)) → Raisins

The total fraction that survives both processes = \(0.35 \times 0.55\)

3. Establish the relationship equation

Now we can set up our equation in plain English:

Starting weight × Fraction that survives = Final weight

We know:
• Starting weight = k kilograms (what we're solving for)
• Fraction that survives = \(0.35 \times 0.55\)
• Final weight = 1 kilogram

So our equation becomes:
\(\mathrm{k} \times (0.35 \times 0.55) = 1\)

This equation captures the entire drying process: k kilograms of fresh grapes, after losing weight through both drying steps, results in exactly 1 kilogram of raisins.

4. Solve for k using the weight survival fractions

To find k, we need to isolate it on one side of the equation:

\(\mathrm{k} \times (0.35 \times 0.55) = 1\)

Divide both sides by \((0.35 \times 0.55)\):
\(\mathrm{k} = 1 \div (0.35 \times 0.55)\)

This can be written as:
\(\mathrm{k} = \frac{1}{(0.35)(0.55)}\)

Let's verify this makes sense: if we need more than 1 kg of fresh grapes to end up with 1 kg of raisins, then k should be greater than 1. Since \((0.35)(0.55) = 0.1925\), we get \(\mathrm{k} = \frac{1}{0.1925} \approx 5.19 \text{ kg}\), which is indeed greater than 1 kg. This confirms our logic is correct.

Final Answer

Looking at the answer choices, our expression \(\mathrm{k} = \frac{1}{(0.35)(0.55)}\) matches exactly with choice A.

The answer is A: \(\frac{1}{(0.35)(0.55)}\)

This tells us that to get exactly 1 kilogram of raisins, we need to start with 1 divided by the product of the survival fractions from both drying processes.

Common Faltering Points

Errors while devising the approach

1. Confusing weight loss percentages with weight remaining percentages

Students often struggle with the key conceptual shift from "loses \(65\%\)" to "keeps \(35\%\)." They might try to multiply by 0.65 instead of 0.35, thinking that a \(65\%\) weight loss means multiplying by \(65\%\). This fundamental misunderstanding of percentages leads to using the wrong fractions throughout the problem.

2. Misinterpreting the sequential nature of the processes

Students may treat the two percentage reductions as independent rather than sequential. For example, they might think the total weight loss is simply \(65\% + 45\% = 110\%\), rather than understanding that the second \(45\%\) reduction applies to the already-reduced weight from the first process.

3. Setting up the equation backwards

Students might set up the relationship as "1 kg of fresh grapes produces k kg of raisins" instead of "k kg of fresh grapes produces 1 kg of raisins." This leads to an equation like \(1 \times (0.35)(0.55) = \mathrm{k}\) instead of \(\mathrm{k} \times (0.35)(0.55) = 1\).

Errors while executing the approach

1. Arithmetic errors in percentage calculations

Students may correctly understand that \(65\%\) loss means \(35\%\) remains, but then make calculation errors like writing \(100\% - 65\% = 45\%\) instead of \(35\%\). Similarly, they might calculate \(100\% - 45\% = 65\%\) instead of \(55\%\).

2. Incorrect algebraic manipulation when solving for k

When isolating k from the equation \(\mathrm{k} \times (0.35)(0.55) = 1\), students might incorrectly multiply both sides by \((0.35)(0.55)\) instead of dividing, leading to \(\mathrm{k} = (0.35)(0.55)\) instead of \(\mathrm{k} = \frac{1}{(0.35)(0.55)}\).

Errors while selecting the answer

1. Matching incorrect decimal values to answer choices

Students who made errors in the approach phase might arrive at expressions involving 0.65 and 0.45 (the loss percentages) and incorrectly select answer choice B: \(\frac{1}{(0.45)(0.65)}\), thinking these are the correct values to use since they appear in the problem statement.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a smart starting value

Let's work backwards from the desired final result. Since we want exactly 1 kilogram of raisins at the end, let's trace through what happens to a convenient starting weight.

Let's choose \(\mathrm{k} = 100 \text{ kg}\) of fresh grapes as our smart number (choosing 100 makes percentage calculations very clean).

Step 2: Apply the first drying process

Fresh grapes lose \(65\%\) of their weight in open air drying.

Weight remaining = \(100\% - 65\% = 35\%\) of original

Partially dried grapes = \(100 \times 0.35 = 35 \text{ kg}\)

Step 3: Apply the second drying process

Partially dried grapes lose \(45\%\) of their weight in the dehydrator.

Weight remaining = \(100\% - 45\% = 55\%\) of partially dried weight

Final raisins = \(35 \times 0.55 = 19.25 \text{ kg}\)

Step 4: Find the scaling factor

We started with 100 kg and got 19.25 kg of raisins.

This means: 100 kg fresh grapes → 19.25 kg raisins

To get exactly 1 kg of raisins, we need: k kg fresh grapes → 1 kg raisins

Step 5: Set up the proportion

100 kg fresh : 19.25 kg raisins = k kg fresh : 1 kg raisins

\(\mathrm{k} = \frac{100}{19.25} = \frac{100}{35 \times 0.55} = \frac{100}{0.35 \times 0.55} = \frac{1}{0.35 \times 0.55}\)

Step 6: Verify the pattern

Notice that \(19.25 = 100 \times 0.35 \times 0.55\), so the general formula is:

\(\mathrm{k} = \frac{1}{0.35 \times 0.55}\)

This matches answer choice A: \(\frac{1}{(0.35)(0.55)}\)

Why this smart number works: Using 100 kg as our starting weight makes percentage calculations intuitive and allows us to clearly see the pattern that emerges. The specific value doesn't matter - any starting weight would give us the same ratio, but 100 makes the arithmetic clean and the relationship transparent.