If x and y are positive numbers such that x + y = 1, which of the following could be...

1. Translate the problem requirements: We need to find which values from the list \(\{80, 140, 199\}\) are possible for the expression \(100\mathrm{x} + 200\mathrm{y}\), given that x and y are positive numbers and \(\mathrm{x} + \mathrm{y} = 1\)
2. Express the target expression using the constraint: Use the constraint \(\mathrm{x} + \mathrm{y} = 1\) to eliminate one variable and rewrite \(100\mathrm{x} + 200\mathrm{y}\) in terms of a single variable
3. Determine the range of possible values: Since x and y must be positive, find the minimum and maximum possible values for our expression
4. Test each given value against the feasible range: Check which of the three given values (80, 140, 199) fall within our determined range

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we're working with here. We have two positive numbers, x and y, that must add up to 1. Think of this like splitting a dollar - if x represents one portion and y represents another portion, together they make the whole dollar (\(\mathrm{x} + \mathrm{y} = 1\)).

We want to find out which of these three values could result from the expression \(100\mathrm{x} + 200\mathrm{y}\):

• I. 80
• II. 140
• III. 199

The key constraint is that both x and y must be positive numbers - neither can be zero or negative.
Process Skill: TRANSLATE

2. Express the target expression using the constraint

Since we know \(\mathrm{x} + \mathrm{y} = 1\), we can say that \(\mathrm{y} = 1 - \mathrm{x}\). This is helpful because now we can write everything in terms of just one variable.

Let's substitute this into our expression:
\(100\mathrm{x} + 200\mathrm{y} = 100\mathrm{x} + 200(1 - \mathrm{x})\)
\(= 100\mathrm{x} + 200 - 200\mathrm{x}\)
\(= 200 - 100\mathrm{x}\)

So our expression simplifies to \(200 - 100\mathrm{x}\). This makes it much easier to work with!
Process Skill: SIMPLIFY

3. Determine the range of possible values

Now we need to figure out what values are possible for \(200 - 100\mathrm{x}\), keeping in mind our constraints.

Since x and y are both positive and \(\mathrm{x} + \mathrm{y} = 1\):

• x must be greater than 0 (positive)
• x must be less than 1 (because if x = 1, then y = 0, which violates the "y is positive" requirement)

So we have: \(0 < \mathrm{x} < 1\)

Now let's see what happens to \(200 - 100\mathrm{x}\) in this range:

• When x approaches 0: \(200 - 100(0)\) approaches 200
• When x approaches 1: \(200 - 100(1)\) approaches 100

Since our expression \(200 - 100\mathrm{x}\) decreases as x increases, the possible values for \(100\mathrm{x} + 200\mathrm{y}\) must be between 100 and 200 (not including the endpoints since x cannot actually equal 0 or 1).
Process Skill: APPLY CONSTRAINTS

4. Test each given value against the feasible range

Now let's check each option against our range of \((100, 200)\):

Option I: 80
Since \(80 < 100\), this value is outside our possible range. This would require \(\mathrm{x} > 1\), which violates our constraint that \(\mathrm{x} + \mathrm{y} = 1\) with both being positive.

Option II: 140
Since \(100 < 140 < 200\), this value is within our possible range.
To verify: if \(200 - 100\mathrm{x} = 140\), then \(100\mathrm{x} = 60\), so \(\mathrm{x} = 0.6\) and \(\mathrm{y} = 0.4\). Both are positive ✓

Option III: 199
Since \(100 < 199 < 200\), this value is within our possible range.
To verify: if \(200 - 100\mathrm{x} = 199\), then \(100\mathrm{x} = 1\), so \(\mathrm{x} = 0.01\) and \(\mathrm{y} = 0.99\). Both are positive ✓

Therefore, options II and III are possible, but option I is not.

4. Final Answer

The values that could be achieved are 140 and 199 (options II and III only).

The answer is (E) II and III.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint "positive numbers"
Students often overlook that BOTH x and y must be positive, meaning neither can be zero. They might incorrectly allow \(\mathrm{x} = 0\) or \(\mathrm{y} = 0\), which would expand the range to include endpoints (100 and 200) that are actually not achievable.

2. Not recognizing the substitution opportunity
Students might attempt to work with two variables throughout the problem instead of using the constraint \(\mathrm{x} + \mathrm{y} = 1\) to express everything in terms of a single variable. This makes the problem unnecessarily complex and harder to visualize.

3. Failing to identify this as a range problem
Students might try to test each value individually by setting up equations rather than first determining the complete range of possible values for the expression \(100\mathrm{x} + 200\mathrm{y}\).

Errors while executing the approach

1. Incorrect range determination
When finding the range of \(200 - 100\mathrm{x}\), students might incorrectly include the endpoints, thinking the range is \([100, 200]\) instead of the open interval \((100, 200)\). This happens when they forget that x cannot actually equal 0 or 1.

2. Sign errors in algebraic manipulation
During the substitution step: \(100\mathrm{x} + 200(1-\mathrm{x}) = 100\mathrm{x} + 200 - 200\mathrm{x} = 200 - 100\mathrm{x}\), students might make sign errors, particularly when distributing the 200 or combining like terms.

3. Boundary analysis errors
Students might incorrectly analyze what happens as x approaches 0 and 1, potentially getting the direction of the inequality wrong or miscalculating the limiting values of the expression.

Errors while selecting the answer

1. Misreading the answer choices
Students might correctly identify that options II and III work but then select choice (D) "I and III" instead of choice (E) "II and III" due to careless reading of the Roman numerals.

2. Including boundary values incorrectly
If students incorrectly determined that the range includes the endpoints, they might think additional values are possible or reject values that are actually achievable, leading to wrong answer selection.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose strategic values for x and y

Since we need \(\mathrm{x} + \mathrm{y} = 1\) with both x and y positive, let's test specific boundary and middle values to understand the range of \(100\mathrm{x} + 200\mathrm{y}\).

Step 2: Test extreme cases using smart number selection

Case 1: x is very small, y is close to 1
Let \(\mathrm{x} = 0.1\), then \(\mathrm{y} = 0.9\)
\(100\mathrm{x} + 200\mathrm{y} = 100(0.1) + 200(0.9) = 10 + 180 = 190\)

Case 2: x is close to 1, y is very small
Let \(\mathrm{x} = 0.9\), then \(\mathrm{y} = 0.1\)
\(100\mathrm{x} + 200\mathrm{y} = 100(0.9) + 200(0.1) = 90 + 20 = 110\)

Case 3: x and y are equal
Let \(\mathrm{x} = 0.5\), then \(\mathrm{y} = 0.5\)
\(100\mathrm{x} + 200\mathrm{y} = 100(0.5) + 200(0.5) = 50 + 100 = 150\)

Step 3: Test values closer to boundaries

For a value near 200 (testing if 199 is possible):
Let \(\mathrm{x} = 0.01\), then \(\mathrm{y} = 0.99\)
\(100\mathrm{x} + 200\mathrm{y} = 100(0.01) + 200(0.99) = 1 + 198 = 199\) ✓

For a value near 100 (testing if 80 is possible):
We need \(100\mathrm{x} + 200\mathrm{y} = 80\)
Let's try \(\mathrm{x} = 0.8\), then \(\mathrm{y} = 0.2\)
\(100\mathrm{x} + 200\mathrm{y} = 100(0.8) + 200(0.2) = 80 + 40 = 120\) (too high)

Since our minimum from Case 2 was 110, and we need positive values, 80 is impossible.

Step 4: Verify our findings

From our smart number tests:
• The range appears to be between just over 100 and just under 200
• Value I (80): Impossible ✗
• Value II (140): Possible (falls between our calculated range) ✓
• Value III (199): Possible (we found exact values) ✓

Answer: E (II and III only)