Question

Choose the ordered pair of statements where the first statement implies the second, and the two statements are logically consistent with the main statement.

Main statement: Only if the teaching standard is destroyed, will examination result be poor.
Statements: 
1. Examination result is poor. 
2. Teaching standard is not destroyed. 
3. Examination result is not poor. 
4. Teaching standard is destroyed. 
Choose the ordered pair in which the first statement implies the second, and both are logically consistent with the main statement.

• A. 2, 3
• B. 2, 4
• C. 1, 3
• D. 1, 2

Hint:

“Only if $Q$ then $P$” translates to $P \to Q$. Always write its contrapositive $\neg Q \to \neg P$—it often unlocks the correct option in implication questions.

Answer 1

The Correct Option is A

Translate the main statement into logic
“Only if teaching standard is destroyed, examination result will be poor” means:
Poor result $\to$ Teaching destroyed.
Let $P =$ “Exam result is poor”, $D =$ “Teaching standard is destroyed”. Then the main statement is:
$P \to D$. Use the contrapositive
From $P \to D$ we get the logically equivalent contrapositive: $\neg D \to \neg P$. So the main statement allows two consistent conditionals:
- If results are poor, teaching must be destroyed. ($P \to D$)
- If teaching is not destroyed, results cannot be poor. ($\neg D \to \neg P$)
Test each ordered pair
(A) 2, 3: First is “Teaching standard is not destroyed” ($\neg D$). Second is “Exam result is not poor” ($\neg P$). From the contrapositive $\neg D \to \neg P$, statement 2 implies statement 3. Both are consistent with the main statement. Valid.
(B) 2, 4: $\neg D$ implies $D$? That would be a contradiction and not supported by the main statement. Invalid.
(C) 1, 3: $P$ implies $\neg P$? This is self-contradictory and does not follow from $P \to D$. Invalid.
(D) 1, 2: From the main statement, $P$ implies $D$, not $\neg D$. So $P \to \neg D$ contradicts the main statement. Invalid.
Conclusion
Only the pair (2, 3) satisfies “first implies second” while remaining consistent with the main statement.

Answer 2

The main statement, "Only if the teaching standard is destroyed, will examination result be poor," tells us one thing for certain: a poor result can never happen while the teaching standard stays intact. Using this single rule, let's test each of the four ordered pairs directly.

1. 2, 3 (Teaching standard is not destroyed, so examination result is not poor): Since a poor result cannot occur while the teaching standard is intact, saying the teaching standard is not destroyed guarantees the result cannot be poor. The first statement forces the second to be true, and neither statement contradicts the main statement.
2. 2, 4 (Teaching standard is not destroyed, so teaching standard is destroyed): These two statements directly contradict each other. Nothing in the main statement lets "not destroyed" lead to "destroyed," so this pair fails immediately.
3. 1, 3 (Examination result is poor, so examination result is not poor): This is a direct contradiction between the two statements themselves, independent of the main statement altogether. It cannot be a valid pair.
4. 1, 2 (Examination result is poor, so teaching standard is not destroyed): The main statement tells us a poor result can only occur when the teaching standard has been destroyed, so a poor result should lead to "teaching standard is destroyed," the opposite of what this pair claims. This pair goes against the main statement.

Only the first pair keeps both statements true together and respects what the main statement rules out.

Therefore, the correct answer is 2, 3.

Answer 3

The main statement, "only if the teaching standard is destroyed, will examination result be poor," fixes a subset relationship: every situation where the result is poor must also be a situation where the teaching standard is destroyed. In other words, the set of "poor result" cases sits entirely inside the set of "teaching destroyed" cases, though the destroyed-teaching set can be bigger and include cases where the result still isn't poor. Let's test each ordered pair against this containment.

1. 2, 3 (Teaching standard is not destroyed, so examination result is not poor): Standing outside the larger set, "teaching destroyed," automatically means standing outside the smaller set contained within it, "poor result." So being outside "destroyed" forces being outside "poor," which is exactly what the second statement says. This pair holds by simple containment.
2. 2, 4 (Teaching standard is not destroyed, so teaching standard is destroyed): This asks a case to be simultaneously outside and inside the very same set, which is impossible on its own terms, and the main statement offers no route from one to the other.
3. 1, 3 (Examination result is poor, so examination result is not poor): This is a case demanded to be both inside and outside the "poor result" set at once, a plain contradiction that doesn't even need the main statement to be ruled out.
4. 1, 2 (Examination result is poor, so teaching standard is not destroyed): Being inside the smaller set, "poor result," guarantees being inside the larger set that contains it, "teaching destroyed," not outside it. So this pair asks for the opposite of what the containment actually guarantees.

Only the containment relationship between "poor result" and "teaching destroyed" supports the first pair, since standing outside the larger set forces standing outside the smaller one it contains.

Therefore, the correct answer is 2, 3.