Every brick house on River Street has a front yard.
Conclusion (so)
Most of the BH's on R St. have two stories.
Evidence
Every BH on R St. has a front yard. Most of the H's on R St. w/ front yards also have two stories.
Evaluation
Since this argument has symbol matches ("brick house" / "river street" / "front yard" / "two stories" all appear twice) and quantifiers ("every / most / most"), this argument is ripe for being turned into an algebraic recipe.
In doing so, we don't really need to worry about the flaw. As long as we re-create this recipe, it'll re-create the flaw.
Every [BH, R] has FY. A and B → Y Most [H, R, FY] have 2S. Most *A, B, Y = Z ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Most [BH, R] have 2S. Most A, B = Z
The *A is because the first sentence is about "brick houses" and the second sentence is about "houses".
That switch from a narrower category "brick houses" to a broader one "houses" is why we can't infer the conclusion.
Goal
We have our recipe, but we can hopefully make quick eliminations by just scanning for things such as - 1 All Premise, 1 Most Premise, 1 Most Conclusion - The All Premise should be a subset of the group mentioned in the Most Premise