Kay began a certain game with x chips. On each of the next two plays, she lost one more than...

1. Translate the problem requirements: Kay starts with \(\mathrm{x}\) chips. In each play, she loses "one more than half" of what she has at the start of that play. After two plays, she has 5 chips left. We need to find which interval contains \(\mathrm{x}\).
2. Set up the loss pattern for each play: For any number of chips \(\mathrm{n}\) at the start of a play, she loses \(\mathrm{(n/2 + 1)}\) chips, leaving her with \(\mathrm{n - (n/2 + 1) = n/2 - 1}\) chips.
3. Work backwards from the final result: Since she ends with 5 chips after two plays, we can determine how many chips she had before the second play, then before the first play.
4. Calculate the starting amount and match to intervals: Use our backwards calculation to find \(\mathrm{x}\), then identify which given interval contains this value.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in plain English first. Kay starts with some number of chips (we'll call this \(\mathrm{x}\)). She plays the game twice, and each time she loses chips according to a specific rule. After both plays, she has exactly 5 chips left.

The loss rule is: "she lost one more than half the number of chips she had at the beginning of that play." Let's understand this with a simple example. If she starts a play with 10 chips, then half would be 5, and "one more than half" would be 6. So she loses 6 chips.

Mathematically, if she has \(\mathrm{n}\) chips at the start of any play:

• She loses: \(\mathrm{(n/2) + 1}\) chips

• She keeps: \(\mathrm{n - [(n/2) + 1] = n/2 - 1}\) chips

Process Skill: TRANSLATE - Converting the problem's language into clear mathematical understanding

2. Set up the loss pattern for each play

Now that we understand the rule, let's see what happens after each play.

If Kay starts any play with \(\mathrm{n}\) chips:

• Chips lost = \(\mathrm{n/2 + 1}\)

• Chips remaining = \(\mathrm{n - (n/2 + 1) = n/2 - 1}\)

So after any play, she keeps exactly "half minus one" of what she started that play with. This pattern will apply to both plays.

3. Work backwards from the final result

Since we know she ends with 5 chips, let's work backwards to find how many chips she had before each play.

Before the second play:

Let's say she had \(\mathrm{y}\) chips before the second play. After the second play, she has 5 chips.

Using our pattern: \(\mathrm{y/2 - 1 = 5}\)

Solving: \(\mathrm{y/2 = 6}\), so \(\mathrm{y = 12}\)

So Kay had 12 chips before her second play.

Before the first play (this is \(\mathrm{x}\)):

Kay started with \(\mathrm{x}\) chips. After the first play, she had 12 chips.

Using our pattern: \(\mathrm{x/2 - 1 = 12}\)

Solving: \(\mathrm{x/2 = 13}\), so \(\mathrm{x = 26}\)

Process Skill: INFER - Working backwards from the end result to find the starting value

4. Calculate the starting amount and match to intervals

We found that \(\mathrm{x = 26}\).

Let's verify this makes sense:

• Kay starts with 26 chips

• After first play: \(\mathrm{26/2 - 1 = 13 - 1 = 12}\) chips ✓

• After second play: \(\mathrm{12/2 - 1 = 6 - 1 = 5}\) chips ✓

Now we need to find which interval contains \(\mathrm{x = 26}\):

1. \(7 \leq \mathrm{x} \leq 12\) (No, 26 is too large)
2. \(13 \leq \mathrm{x} \leq 18\) (No, 26 is too large)
3. \(19 \leq \mathrm{x} \leq 24\) (No, 26 is too large)
4. \(25 \leq \mathrm{x} \leq 30\) (Yes! 26 falls in this range)
5. \(31 \leq \mathrm{x} \leq 35\) (No, 26 is too small)

Final Answer

The answer is D. Kay started with 26 chips, which falls in the interval \(25 \leq \mathrm{x} \leq 30\).

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the loss rule: Students often misread "one more than half" as simply "half" or confuse it with "half of one more." For example, if Kay has 10 chips, students might think she loses 5 chips (half) instead of the correct 6 chips (one more than half). This fundamental misunderstanding will lead to completely wrong calculations throughout the problem.

2. Setting up forward calculation instead of backward: Many students try to work forward from \(\mathrm{x}\) to 5, setting up complex equations like \(\mathrm{x}\) → (some expression) → 5. However, working backwards from 5 is much cleaner and less error-prone. Students who insist on forward calculation often get bogged down in complicated algebraic manipulations.

3. Misunderstanding what "chips remaining" means: Some students confuse the chips remaining after each play with the total chips lost. They might think that if she starts with \(\mathrm{n}\) chips and loses \(\mathrm{(n/2 + 1)}\), then she has \(\mathrm{(n/2 + 1)}\) chips left, when actually she has \(\mathrm{n - (n/2 + 1) = n/2 - 1}\) chips remaining.

Errors while executing the approach

1. Arithmetic errors in the backward calculation: When working backwards, students often make simple arithmetic mistakes. For example, from \(\mathrm{y/2 - 1 = 5}\), they might incorrectly solve to get \(\mathrm{y/2 = 4}\) (instead of 6) or make errors when doubling to find \(\mathrm{y = 8}\) (instead of 12). These small errors compound through both steps.

2. Forgetting to apply the loss rule consistently: Students might correctly apply the "\(\mathrm{n/2 - 1}\)" pattern for one play but then use a different formula for the second play, or they might switch between using "chips lost" and "chips remaining" formulas inconsistently.

3. Sign errors in algebraic manipulation: When simplifying \(\mathrm{n - (n/2 + 1)}\), students commonly make sign errors, getting expressions like \(\mathrm{n/2 + 1}\) instead of \(\mathrm{n/2 - 1}\). This happens because they lose track of the negative sign when distributing.

Errors while selecting the answer

1. Not checking which interval contains the calculated value: After correctly calculating \(\mathrm{x = 26}\), some students hastily pick an answer without carefully checking which interval actually contains 26. They might pick option C (\(19 \leq \mathrm{x} \leq 24\)) because 26 is "close to" the range, not realizing 26 exceeds the upper bound.

2. Misreading interval notation: Students sometimes confuse the inequality symbols or boundaries. For example, they might think 26 fits in the range "\(19 \leq \mathrm{x} \leq 24\)" because they misread it as "\(19 \leq \mathrm{x} \leq 29\)" or don't understand that \(26 > 24\) means it's outside this interval.