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24 Feb 2025, 10:02
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B.
A: No, because it's possible some of the Puerta supporters support Quintana, if not all. We don't have any info on whether or not P supporters also support Ramirez or not. It's possible all segments support Ramirez.
B. Yes. Since there is at least 1 segment that does support both Q and R, then Mashika's condition doesn't hold true, which means not all P supporters support Q... but some can still. This further means that there is at least 1 segment of P supporters that does not support Q.
C, D, E All No. Can't reasonably infer these. Have no info. It may be possible but that's not enough.
A: No, because it's possible some of the Puerta supporters support Quintana, if not all. We don't have any info on whether or not P supporters also support Ramirez or not. It's possible all segments support Ramirez.
B. Yes. Since there is at least 1 segment that does support both Q and R, then Mashika's condition doesn't hold true, which means not all P supporters support Q... but some can still. This further means that there is at least 1 segment of P supporters that does not support Q.
C, D, E All No. Can't reasonably infer these. Have no info. It may be possible but that's not enough.
gmatt1476
24 Feb 2025, 13:33
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Mashika -
if Ms. Puerta’s supporters also support Mr. Quintana, then no segment that supports Mr. Quintana can also support Mr. Ramirez.
Salim-
This directly contradicts Mashika’s conditional statement if the premise about Ms. Puerta’s supporters also supporting Mr. Quintana were true.
Inference -
Since Salim’s statement is factual (based on conclusive polling data), Mashika’s conditional claim must be false.
The only way Mashika’s statement is false is if Ms. Puerta’s supporters do not significantly support Mr. Quintana.
Thus, it must be true that at least one segment supports Ms. Puerta but not Mr. Quintana.
Answer Choices -
(A) At least one segment of the electorate provides significant support neither to Mr. Quintana nor to Mr. Ramirez.
Not necessarily true. The argument doesn’t say anything about segments that support neither candidate.
(B) At least one segment of the electorate provides significant support to Ms. Puerta but not to Mr. Quintana.
Correct. This is what we inferred: for Mashika’s conditional claim to be false, there must be at least one segment that supports Ms. Puerta but does not support Mr. Quintana.
(C) Each segment of the electorate provides significant support to Ms. Puerta.
Too strong. The passage only says some segments support Ms. Puerta, not all.
(D) Each segment of the electorate provides significant support to Mr. Quintana.
Too strong. The passage never claims that every segment supports Mr. Quintana.
(E) Each segment of the electorate provides significant support to Mr. Ramirez.
Too strong. The passage only mentions one segment supporting both Quintana and Ramirez, not all.
if Ms. Puerta’s supporters also support Mr. Quintana, then no segment that supports Mr. Quintana can also support Mr. Ramirez.
Salim-
This directly contradicts Mashika’s conditional statement if the premise about Ms. Puerta’s supporters also supporting Mr. Quintana were true.
Inference -
Since Salim’s statement is factual (based on conclusive polling data), Mashika’s conditional claim must be false.
The only way Mashika’s statement is false is if Ms. Puerta’s supporters do not significantly support Mr. Quintana.
Thus, it must be true that at least one segment supports Ms. Puerta but not Mr. Quintana.
Answer Choices -
(A) At least one segment of the electorate provides significant support neither to Mr. Quintana nor to Mr. Ramirez.
Not necessarily true. The argument doesn’t say anything about segments that support neither candidate.
(B) At least one segment of the electorate provides significant support to Ms. Puerta but not to Mr. Quintana.
Correct. This is what we inferred: for Mashika’s conditional claim to be false, there must be at least one segment that supports Ms. Puerta but does not support Mr. Quintana.
(C) Each segment of the electorate provides significant support to Ms. Puerta.
Too strong. The passage only says some segments support Ms. Puerta, not all.
(D) Each segment of the electorate provides significant support to Mr. Quintana.
Too strong. The passage never claims that every segment supports Mr. Quintana.
(E) Each segment of the electorate provides significant support to Mr. Ramirez.
Too strong. The passage only mentions one segment supporting both Quintana and Ramirez, not all.
25 Jul 2026, 08:52
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The Toolkit You Should Memorize
For these questions, always identify:
Facts vs. conditionals — Don't treat an "if" statement as a fact.
Trigger → Result — Mark the antecedent and consequent.
Universal vs. existential — Words like all, no, each versus some, at least one.
Contradiction pairs:
All A are B ↔ Some A are not B
No A are B ↔ Some A are B
Modus Tollens:
If A → B
Not B
Therefore, not A.
Negating universals:
Not(all A are B) = Some A are not B.
Detailed form
Inference CR - One-Page Cheat Sheet
A conditional states only a relationship, not that the trigger actually occurred.
Form: If A -> B
Never assume A happened just because the conditional exists.
Trigger (Antecedent) -> Result (Consequent)
Always identify the trigger first.
Only if -> A -> B
Only -> B -> A
Unless A -> Not A -> B (equivalently, Not B -> A)
Existential quantifiers require only one example: Some, At least one, Exists, Several, Many, Often, Usually.
Remember: Some means >= 1.
No A are B -> A and B have no overlap.
Some A are B -> A and B overlap.
Some A are not B -> Part of A lies outside B.
Not all A are B -> Some A are not B.
No A are B contradicts Some A are B.
Every A is B contradicts At least one A is not B.
One counterexample destroys a universal statement.
If A -> B
A
Therefore B.
Modus Tollens:
If A -> B
Not B
Therefore Not A.
Contrapositive:
A -> B is logically equivalent to Not B -> Not A.
Chain Rule:
A -> B
B -> C
Therefore A -> C.
If A -> B
B
Does NOT imply A.
Denying the Antecedent:
If A -> B
Not A
Does NOT imply Not B.
No A are B -> Some A are B.
Some A are B -> No A are B.
Some A are not B -> All A are B.
Every -> Not every (Some do not).
Always -> Not always (Sometimes not).
Evidence shows Not B.
Apply Modus Tollens.
Conclusion: Not A.
Do NOT conclude "No A are B."
Step 2: Identify Trigger -> Result.
Step 3: Identify quantifiers (All, Every, No, Some, At least one, None).
Step 4: Convert statements into set relationships (inside, overlap, outside, disjoint).
Step 5: Check whether later evidence contradicts the conditional result.
Step 6: If yes, apply Modus Tollens.
Step 7: Negate universal statements correctly.
All P are Q.
No Q are R.
But:
Some Q are R.
Therefore:
"No Q are R" is false.
Apply Modus Tollens.
Therefore:
Not(All P are Q).
Final inference:
Some P are not Q.
Golden Rules
Some means >= 1.
One counterexample destroys "All" and "No."
Never assume the trigger of a conditional occurred.
Translate English -> Logic -> Sets before evaluating answer choices.
For elite Inference CR, think in this order:
Facts -> Conditionals -> Quantifiers -> Sets -> Contradictions -> Modus Tollens -> Inference.
For these questions, always identify:
Facts vs. conditionals — Don't treat an "if" statement as a fact.
Trigger → Result — Mark the antecedent and consequent.
Universal vs. existential — Words like all, no, each versus some, at least one.
Contradiction pairs:
All A are B ↔ Some A are not B
No A are B ↔ Some A are B
Modus Tollens:
If A → B
Not B
Therefore, not A.
Negating universals:
Not(all A are B) = Some A are not B.
Detailed form
Inference CR - One-Page Cheat Sheet
- Facts vs Conditionals
A conditional states only a relationship, not that the trigger actually occurred.
Form: If A -> B
Never assume A happened just because the conditional exists.
- Identify the Structure
Trigger (Antecedent) -> Result (Consequent)
Always identify the trigger first.
- English to Logic
Only if -> A -> B
Only -> B -> A
Unless A -> Not A -> B (equivalently, Not B -> A)
- Quantifiers
Existential quantifiers require only one example: Some, At least one, Exists, Several, Many, Often, Usually.
Remember: Some means >= 1.
- Set Translation
No A are B -> A and B have no overlap.
Some A are B -> A and B overlap.
Some A are not B -> Part of A lies outside B.
Not all A are B -> Some A are not B.
- Contradiction Patterns
No A are B contradicts Some A are B.
Every A is B contradicts At least one A is not B.
One counterexample destroys a universal statement.
- Valid Conditional Rules
If A -> B
A
Therefore B.
Modus Tollens:
If A -> B
Not B
Therefore Not A.
Contrapositive:
A -> B is logically equivalent to Not B -> Not A.
Chain Rule:
A -> B
B -> C
Therefore A -> C.
- Invalid Conditional Rules
If A -> B
B
Does NOT imply A.
Denying the Antecedent:
If A -> B
Not A
Does NOT imply Not B.
- Negation Rules
No A are B -> Some A are B.
Some A are B -> No A are B.
Some A are not B -> All A are B.
Every -> Not every (Some do not).
Always -> Not always (Sometimes not).
- Master Conditional Pattern
Evidence shows Not B.
Apply Modus Tollens.
Conclusion: Not A.
- Universal Failure Rule
Do NOT conclude "No A are B."
- 30-Second Attack Plan
Step 2: Identify Trigger -> Result.
Step 3: Identify quantifiers (All, Every, No, Some, At least one, None).
Step 4: Convert statements into set relationships (inside, overlap, outside, disjoint).
Step 5: Check whether later evidence contradicts the conditional result.
Step 6: If yes, apply Modus Tollens.
Step 7: Negate universal statements correctly.
- Most Common Elite Pattern
All P are Q.
No Q are R.
But:
Some Q are R.
Therefore:
"No Q are R" is false.
Apply Modus Tollens.
Therefore:
Not(All P are Q).
Final inference:
Some P are not Q.
Golden Rules
Some means >= 1.
One counterexample destroys "All" and "No."
Never assume the trigger of a conditional occurred.
Translate English -> Logic -> Sets before evaluating answer choices.
For elite Inference CR, think in this order:
Facts -> Conditionals -> Quantifiers -> Sets -> Contradictions -> Modus Tollens -> Inference.
