IAS 2025 — CSAT Reasoning

2025mcqmedium

A question is given followed by two Statements I and II. Consider the Question and the Statements and mark the correct option.

Question: Let P, Q, R, S be distinct non-zero digits. If PP × PQ = RRSS, where P ≤ 3 and Q ≤ 4, then what is Q equal to?

Statement I: R = 1.
Statement II: S = 2.

Which one of the following is correct in respect of the above Question and the Statements?

Official previous-year question

Verified 2 Aug 2026.

Options

AThe Question can be answered by using one of the Statements alone, but cannot be answered using the other statement alone.
BThe Question can be answered by using either Statement alone.
CThe Question can be answered by using both the Statements together, but cannot be answered using either Statement alone.
DThe Question can be answered even without using any of the Statements.

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Solution

Correct Option (4)

The problem requires determining the value of Q from the equation PP × PQ = RRSS, given that P, Q, R, S are distinct non-zero digits, with P ≤ 3 and Q ≤ 4.

The given equation can be translated into an algebraic form:

PP represents 11P.
PQ represents 10P + Q.
RRSS represents 1100R + 11S.

Substituting these into the equation yields: 11P × (10P + Q) = 1100R + 11S.

Dividing both sides of the equation by 11 simplifies it to: P(10P + Q) = 100R + S.

Now, we apply the given constraints:

P is a non-zero digit and P ≤ 3, so P can be 1, 2, or 3.
R is a non-zero digit, meaning 100R + S must be at least 100 (specifically, at least 101, as S is also a non-zero digit).

Let's test the possible values for P:

If P = 1: The left side becomes 1 × (10 × 1 + Q) = 10 + Q. Since Q ≤ 4, the maximum value is 10 + 4 = 14. This is less than 100, so P cannot be 1.
If P = 2: The left side becomes 2 × (10 × 2 + Q) = 2(20 + Q). Since Q ≤ 4, the maximum value is 2(20 + 4) = 48. This is also less than 100, so P cannot be 2.
Therefore, P must be 3.

Substitute P = 3 into the simplified equation: 3(10 × 3 + Q) = 100R + S, which simplifies to 3(30 + Q) = 100R + S.

Next, consider the constraints for Q:

Q is a non-zero digit, Q ≤ 4, and Q must be distinct from P (which is 3).
Thus, Q can be 1, 2, or 4.

Let's test these possible values for Q:

If Q = 1: 3(30 + 1) = 3(31) = 93. This is less than 100, so Q cannot be 1.
If Q = 2: 3(30 + 2) = 3(32) = 96. This is less than 100, so Q cannot be 2.
If Q = 4: 3(30 + 4) = 3(34) = 102. This value fits the form 100R + S, where R = 1 and S = 2.

The unique set of distinct non-zero digits that satisfies all conditions is P=3, Q=4, R=1, S=2. These values also satisfy P ≤ 3 and Q ≤ 4.

Since Q = 4 is uniquely determined using only the information provided in the question, neither Statement I (R=1) nor Statement II (S=2) is necessary to answer the question.

Incorrect Options:

Options 1, 2, and 3 are incorrect because the value of Q can be definitively established through a systematic analysis of the given equation and its constraints alone. The information provided in Statement I (R=1) and Statement II (S=2) is consistent with the derived solution but not required for its determination.

Your note

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Correct Option (4)

The problem requires determining the value of Q from the equation PP × PQ = RRSS, given that P, Q, R, S are distinct non-zero digits, with P ≤ 3 and Q ≤ 4.

The given equation can be translated into an algebraic form:

PP represents 11P.
PQ represents 10P + Q.
RRSS represents 1100R + 11S.

Substituting these into the equation yields: 11P × (10P + Q) = 1100R + 11S.

Dividing both sides of the equation by 11 simplifies it to: P(10P + Q) = 100R + S.

Now, we apply the given constraints:

P is a non-zero digit and P ≤ 3, so P can be 1, 2, or 3.
R is a non-zero digit, meaning 100R + S must be at least 100 (specifically, at least 101, as S is also a non-zero digit).

Let's test the possible values for P:

If P = 1: The left side becomes 1 × (10 × 1 + Q) = 10 + Q. Since Q ≤ 4, the maximum value is 10 + 4 = 14. This is less than 100, so P cannot be 1.
If P = 2: The left side becomes 2 × (10 × 2 + Q) = 2(20 + Q). Since Q ≤ 4, the maximum value is 2(20 + 4) = 48. This is also less than 100, so P cannot be 2.
Therefore, P must be 3.

Substitute P = 3 into the simplified equation: 3(10 × 3 + Q) = 100R + S, which simplifies to 3(30 + Q) = 100R + S.

Next, consider the constraints for Q:

Q is a non-zero digit, Q ≤ 4, and Q must be distinct from P (which is 3).
Thus, Q can be 1, 2, or 4.

Let's test these possible values for Q:

If Q = 1: 3(30 + 1) = 3(31) = 93. This is less than 100, so Q cannot be 1.
If Q = 2: 3(30 + 2) = 3(32) = 96. This is less than 100, so Q cannot be 2.
If Q = 4: 3(30 + 4) = 3(34) = 102. This value fits the form 100R + S, where R = 1 and S = 2.

The unique set of distinct non-zero digits that satisfies all conditions is P=3, Q=4, R=1, S=2. These values also satisfy P ≤ 3 and Q ≤ 4.

Since Q = 4 is uniquely determined using only the information provided in the question, neither Statement I (R=1) nor Statement II (S=2) is necessary to answer the question.

IAS 2025 — CSAT Reasoning

Options

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Solution

Correct Option (4)

Incorrect Options:

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IAS 2025 — CSAT Reasoning

Options

Did you get this right?

Solution

Correct Option (4)

Incorrect Options:

Your note

IAS Reasoning in other years

More IAS Reasoning questions

Other CSAT topics for IAS

Comprehension

Maths

Keep practicing Reasoning