Let P, Q, R, S and T be five statements such that: 1. If P is true, then both Q and S are true. 2. If R and S are true, then T is false. Which of the following can be concluded? 1. If T is true, then at least one of P and R must be false. 2. If Q is true, then P is true. Select the correct answer using the code given below:

Correct Option The problem provides two initial statements: 1. If P is true, then both Q and S are true (symbolically: P → (Q ∧ S)). 2. If R and S are true, then T is false (symbolically: (R ∧ S) → ¬T). Let's evaluate Conclusion 1: "If T is true, then at least one of P and R must be false." - From statement 2: (R ∧ S) → ¬T. - The contrapositive of this statement is T → ¬(R ∧ S). - By De Morgan's laws, ¬(R ∧ S) is equivalent to (¬R ∨ ¬S). - Therefore, if T is true, it implies (¬R ∨ ¬S), meaning either R is false or S is false (or both). - Now, we consider the implications of (¬R ∨ ¬S) on P and R: - If S is false (¬S is true): From statement 1 (P → (Q ∧ S)), if P were true, S would necessarily be true. Since S is false, P cannot be true. Thus, P must be false (¬P). If P is false, then the condition "at least one of P and R must be false" (¬P ∨ ¬R) is satisfied. - If R is false (¬R is true): In this case, the condition "at least one of P and R must be false" (¬P ∨ ¬R) is directly satisfied because R is false. - Since both possibilities arising from T being true (¬R ∨ ¬S) lead to "at least one of P and R must be false," Conclusion 1 is valid. Incorrect Options Let's evaluate Conclusion 2: "If Q is true, then P is true." - Statement 1 is P → (Q ∧ S), which logically implies P → Q. - Conclusion 2 proposes Q → P. This is the converse of P → Q. - The converse of a conditional statement is not necessarily true. The truth of P → Q does not guarantee the truth of Q → P. - Consider a scenario where P is false, Q is true, and S is true. In this case, statement 1 (If P is true, then both Q and S are true) holds true because the premise 'P is true' is false, making the entire implication true regardless of Q and S. However, Conclusion 2 (If Q is true, then P is true) would be false in this scenario, as Q is true but P is false. - Therefore, we cannot definitively conclude that if Q is true, then P is true. Conclusion 2 is not valid. Based on the analysis, only Conclusion 1 is valid. Thus, option 1 is the correct answer.

IAS 2023 — CSAT Reasoning

2023mcqmedium

Let P, Q, R, S and T be five statements such that:

If P is true, then both Q and S are true.
If R and S are true, then T is false.

Which of the following can be concluded?

If T is true, then at least one of P and R must be false.
If Q is true, then P is true.

Select the correct answer using the code given below:

Official previous-year question

Verified 2 Aug 2026.

Options

A1 only
B2 only
CBoth 1 and 2
DNeither 1 nor 2

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Solution

Correct Option

The problem provides two initial statements:

If P is true, then both Q and S are true (symbolically: P → (Q ∧ S)).
If R and S are true, then T is false (symbolically: (R ∧ S) → ¬T).

Let's evaluate Conclusion 1: "If T is true, then at least one of P and R must be false."

From statement 2: (R ∧ S) → ¬T.
The contrapositive of this statement is T → ¬(R ∧ S).
By De Morgan's laws, ¬(R ∧ S) is equivalent to (¬R ∨ ¬S).
Therefore, if T is true, it implies (¬R ∨ ¬S), meaning either R is false or S is false (or both).
Now, we consider the implications of (¬R ∨ ¬S) on P and R:
- If S is false (¬S is true): From statement 1 (P → (Q ∧ S)), if P were true, S would necessarily be true. Since S is false, P cannot be true. Thus, P must be false (¬P). If P is false, then the condition "at least one of P and R must be false" (¬P ∨ ¬R) is satisfied.
- If R is false (¬R is true): In this case, the condition "at least one of P and R must be false" (¬P ∨ ¬R) is directly satisfied because R is false.
Since both possibilities arising from T being true (¬R ∨ ¬S) lead to "at least one of P and R must be false," Conclusion 1 is valid.

Incorrect Options

Let's evaluate Conclusion 2: "If Q is true, then P is true."

Statement 1 is P → (Q ∧ S), which logically implies P → Q.
Conclusion 2 proposes Q → P. This is the converse of P → Q.
The converse of a conditional statement is not necessarily true. The truth of P → Q does not guarantee the truth of Q → P.
Consider a scenario where P is false, Q is true, and S is true. In this case, statement 1 (If P is true, then both Q and S are true) holds true because the premise 'P is true' is false, making the entire implication true regardless of Q and S. However, Conclusion 2 (If Q is true, then P is true) would be false in this scenario, as Q is true but P is false.
Therefore, we cannot definitively conclude that if Q is true, then P is true. Conclusion 2 is not valid.

Based on the analysis, only Conclusion 1 is valid. Thus, option 1 is the correct answer.

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Two identical straight rods are painted in five distinct colours so that each of them gets divided into five equal parts along the length. In one of them, the portions are marked P1, P2, P3, P4 and P5 (not necessarily in that order) whereas in the other, they are marked Q1, Q2, Q3, Q4 and Q5 (not necessarily in that order). When the rods are kept parallel to each other side by side, P1 and Q3 match, P4 matches Q1 or Q2, and Q4 matches P3 or P5. If Q3 and Q5 are adjacent, which of the following is/are possible? 1. Q3 is marked at the middle portion of the straight rod. 2. P2 is marked at one of the extreme portions of the straight rod. Select the answer using the code given below.

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Correct Option

The problem provides two initial statements:

If P is true, then both Q and S are true (symbolically: P → (Q ∧ S)).
If R and S are true, then T is false (symbolically: (R ∧ S) → ¬T).

Let's evaluate Conclusion 1: "If T is true, then at least one of P and R must be false."

From statement 2: (R ∧ S) → ¬T.
The contrapositive of this statement is T → ¬(R ∧ S).
By De Morgan's laws, ¬(R ∧ S) is equivalent to (¬R ∨ ¬S).
Therefore, if T is true, it implies (¬R ∨ ¬S), meaning either R is false or S is false (or both).
Now, we consider the implications of (¬R ∨ ¬S) on P and R:
- If S is false (¬S is true): From statement 1 (P → (Q ∧ S)), if P were true, S would necessarily be true. Since S is false, P cannot be true. Thus, P must be false (¬P). If P is false, then the condition "at least one of P and R must be false" (¬P ∨ ¬R) is satisfied.
- If R is false (¬R is true): In this case, the condition "at least one of P and R must be false" (¬P ∨ ¬R) is directly satisfied because R is false.
Since both possibilities arising from T being true (¬R ∨ ¬S) lead to "at least one of P and R must be false," Conclusion 1 is valid.

Incorrect Options

Let's evaluate Conclusion 2: "If Q is true, then P is true."

Statement 1 is P → (Q ∧ S), which logically implies P → Q.
Conclusion 2 proposes Q → P. This is the converse of P → Q.
The converse of a conditional statement is not necessarily true. The truth of P → Q does not guarantee the truth of Q → P.
Consider a scenario where P is false, Q is true, and S is true. In this case, statement 1 (If P is true, then both Q and S are true) holds true because the premise 'P is true' is false, making the entire implication true regardless of Q and S. However, Conclusion 2 (If Q is true, then P is true) would be false in this scenario, as Q is true but P is false.
Therefore, we cannot definitively conclude that if Q is true, then P is true. Conclusion 2 is not valid.

Based on the analysis, only Conclusion 1 is valid. Thus, option 1 is the correct answer.

IAS 2023 — CSAT Reasoning

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IAS 2023 — CSAT Reasoning

Options

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Correct Option

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IAS Reasoning in other years

More IAS Reasoning questions

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