A certain firm has 28 lawyers in three states. These lawyers have license either in only one state or all...

1. Translate the problem requirements: We have 28 lawyers total who are licensed in either exactly one state OR all three states (no other combinations allowed). We know 10 have license in X, 11 in Y, and 13 in Z. We need to find how many have license only in Z.
2. Set up variables for the two licensing categories: Define variables for lawyers with single-state licenses and lawyers with all-three-state licenses, since these are the only two possibilities.
3. Create equations using the licensing constraints: Use the fact that the total count for each state equals single-state lawyers plus all-three-state lawyers to build a system of equations.
4. Solve for the number of all-three-state lawyers: Use the total lawyer count and state-specific counts to determine how many lawyers have licenses in all three states.
5. Calculate lawyers with only Z license: Subtract the all-three-state lawyers from the total Z-licensed lawyers to find those with only Z license.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English:

• We have 28 lawyers total working for this firm
• Each lawyer has a very specific licensing pattern: they are licensed in either exactly one state OR all three states - no other combinations are allowed
• When we count by state: 10 lawyers are licensed in state X, 11 in state Y, and 13 in state Z
• We need to find how many lawyers have a license only in state Z

The key insight here is the constraint: lawyers can only be licensed in one state OR all three states. This means we won't have lawyers licensed in just two states.

Process Skill: TRANSLATE - Converting the licensing constraint into mathematical understanding

2. Set up variables for the two licensing categories

Since we only have two types of lawyers, let's use simple variables:

• Let x = number of lawyers licensed only in state X
• Let y = number of lawyers licensed only in state Y
• Let z = number of lawyers licensed only in state Z
• Let a = number of lawyers licensed in all three states

These four groups account for all 28 lawyers, so: \(\mathrm{x + y + z + a = 28}\)

3. Create equations using the licensing constraints

Now let's think about what contributes to each state's total count:

• State X has 10 lawyers total: those with only X license + those with all three = \(\mathrm{x + a = 10}\)
• State Y has 11 lawyers total: those with only Y license + those with all three = \(\mathrm{y + a = 11}\)
• State Z has 13 lawyers total: those with only Z license + those with all three = \(\mathrm{z + a = 13}\)

So our system of equations is:

• \(\mathrm{x + a = 10}\)
• \(\mathrm{y + a = 11}\)
• \(\mathrm{z + a = 13}\)
• \(\mathrm{x + y + z + a = 28}\)

4. Solve for the number of all-three-state lawyers

Let's add up the first three equations:

\(\mathrm{(x + a) + (y + a) + (z + a) = 10 + 11 + 13}\)

\(\mathrm{x + y + z + 3a = 34}\)

But we also know that \(\mathrm{x + y + z + a = 28}\) from the total count.

Substituting this relationship:

\(\mathrm{(x + y + z + a) + 2a = 34}\)

\(\mathrm{28 + 2a = 34}\)

\(\mathrm{2a = 6}\)

\(\mathrm{a = 3}\)

So 3 lawyers are licensed in all three states.

Process Skill: MANIPULATE - Using algebraic substitution to solve the system

5. Calculate lawyers with only Z license

Now we can find the number of lawyers with only a Z license:

From \(\mathrm{z + a = 13}\) and \(\mathrm{a = 3}\):

\(\mathrm{z + 3 = 13}\)

\(\mathrm{z = 10}\)

Let's verify our answer makes sense:

• Lawyers with only X license: \(\mathrm{x = 10 - 3 = 7}\)
• Lawyers with only Y license: \(\mathrm{y = 11 - 3 = 8}\)
• Lawyers with only Z license: \(\mathrm{z = 13 - 3 = 10}\)
• Lawyers with all three licenses: \(\mathrm{a = 3}\)
• Total: \(\mathrm{7 + 8 + 10 + 3 = 28}\) ✓

4. Final Answer

The number of lawyers who have a license only in state Z is 10.

This matches answer choice C.

Common Faltering Points

Errors while devising the approach

• Misinterpreting the licensing constraint: Students often miss or misunderstand the critical constraint that lawyers can ONLY be licensed in exactly one state OR all three states. They may attempt to solve this as a standard Venn diagram problem where lawyers could have any combination of licenses (like just X and Y, or just Y and Z), leading to an overly complex approach with many more variables.
• Confusing the state totals with unique counts: Students may incorrectly think that the numbers 10, 11, and 13 represent lawyers who are licensed ONLY in each respective state, rather than understanding these are the total counts of lawyers licensed in each state (including those licensed in all three).
• Setting up incorrect variables: Due to misunderstanding the constraint, students might create variables for impossible scenarios (like lawyers licensed in exactly two states) or fail to recognize that only four distinct groups of lawyers exist in this problem.

Errors while executing the approach

• Arithmetic errors in the substitution step: When solving the system of equations, students commonly make mistakes in the algebraic manipulation, particularly when adding the three state equations \(\mathrm{(x + a) + (y + a) + (z + a) = 34}\) and then substituting the total constraint. The step involving \(\mathrm{28 + 2a = 34}\) is prone to calculation errors.
• Incorrect equation setup: Students may write the wrong equations for each state's total count. For example, they might write \(\mathrm{x = 10}\) instead of \(\mathrm{x + a = 10}\), forgetting that the all-three-state lawyers (a) contribute to each state's count.

Errors while selecting the answer

• Selecting the wrong calculated value: After correctly calculating all values \(\mathrm{(a = 3, x = 7, y = 8, z = 10)}\), students might select the value of 'a' (3) instead of 'z' (10), or confuse which variable represents what they're looking for. They might see that 'a = 3' and incorrectly choose answer choice A.