Of the 3{,}600 families in Township X, 1/2 subscribe to the Beacon, 1/4 subscribe to the Herald, and 1/5 subscribe...

1. Translate the problem requirements: Convert the fractions into actual family counts and clarify that we need families subscribing to neither paper (outside both subscription sets)
2. Calculate individual subscription counts: Find how many families subscribe to each paper separately using the given fractions
3. Apply overlap principle: Use the families subscribing to both papers to find the total number subscribing to at least one paper
4. Find the complement: Subtract families with at least one subscription from total families to get those with no subscriptions

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have and what we need to find.

We have 3,600 families total in Township X. Think of this as the entire population we're working with.

Now let's convert those fractions into actual numbers of families:

• Families subscribing to the Beacon: \(\frac{1}{2} \times 3600 = 1800\) families
• Families subscribing to the Herald: \(\frac{1}{4} \times 3600 = 900\) families
• Families subscribing to BOTH papers: \(\frac{1}{5} \times 3600 = 720\) families

What we're looking for is families that subscribe to NEITHER paper - basically, families that are completely outside both subscription groups.

Process Skill: TRANSLATE - Converting fractions to concrete numbers makes the problem much more manageable

2. Calculate individual subscription counts

We already calculated the key numbers in step 1, but let's organize them clearly:

• Total families: 3,600
• Beacon subscribers: 1,800 families
• Herald subscribers: 900 families
• Both papers subscribers: 720 families

These are our building blocks for the next step.

3. Apply overlap principle

Here's where we need to think carefully about what "at least one subscription" means.

If we just added Beacon subscribers (1,800) + Herald subscribers (900), we'd get 2,700. But wait - this counts the 720 families who subscribe to BOTH papers twice!

Think of it this way: if someone subscribes to both papers, they appear in both the Beacon count AND the Herald count. So when we add those counts together, we're double-counting those 720 families.

To find families with at least one subscription, we need to subtract out that double-counting:
Families with at least one subscription = \(1800 + 900 - 720 = 1980\) families

Process Skill: VISUALIZE - Imagining overlapping circles helps us see why we need to subtract the overlap

4. Find the complement

Now for the final step. If 1,980 families subscribe to at least one paper, then the remaining families subscribe to neither paper.

Families subscribing to neither paper = Total families - Families with at least one subscription
= \(3600 - 1980 = 1620\) families

Let's double-check this makes sense: \(1620 + 1980 = 3600\) ✓

Final Answer

1,620 families do not subscribe to either the Beacon or the Herald.

This matches answer choice A.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting "1/5 subscribe to both papers"
Students often misread this as "1/5 of the Beacon subscribers also subscribe to Herald" or "1/5 of the Herald subscribers also subscribe to Beacon" instead of understanding it correctly as "1/5 of ALL 3,600 families subscribe to both papers." This fundamental misunderstanding leads to calculating the overlap as a fraction of one group rather than the total population.

Faltering Point 2: Confusing "neither" with "only one"
Students sometimes misinterpret the question asking for families that "do not subscribe to either" paper. They might think this means families that subscribe to only one paper (but not both), rather than families that subscribe to zero papers. This conceptual error leads to solving for the wrong quantity entirely.

Errors while executing the approach

Faltering Point 1: Forgetting to subtract the overlap when finding "at least one"
Even when students understand they need to find families with at least one subscription, they often simply add Beacon subscribers (1,800) + Herald subscribers (900) = 2,700, forgetting that this double-counts the 720 families who subscribe to both papers. This is the most common arithmetic error in set problems.

Faltering Point 2: Subtracting the overlap twice
Some students recognize they need to account for overlap but make the error of subtracting it from both individual counts before adding them together. They might calculate: \((1800 - 720) + (900 - 720) = 1080 + 180 = 1260\), which incorrectly removes the overlap entirely instead of just avoiding double-counting.

Errors while selecting the answer

Faltering Point 1: Selecting the "at least one" value instead of the "neither" value
Students correctly calculate that 1,980 families subscribe to at least one paper, but then select this number as their final answer instead of recognizing they need to subtract this from 3,600 to find families subscribing to neither paper. They might see 1,980 is close to answer choice B (1,200) and select it without completing the final step.