Question:

P, Q, R, S, T, A and B are consecutive integers (not necessarily in that order), such that the smallest of these is greater than 60 and the greatest is less than 70. It is known that:
(I) A and B both are prime numbers.
(II) T is a multiple of 9.
(III) Both the digits of P are same.
(IV) The average of R and S is 63 and the difference between R and S is 2.
What is the sum of A and Q if A is smaller than B?

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In logic puzzles with multiple constraints, identify the most powerful clues first. The clue about the two prime numbers was key to defining the exact set of integers, making the other clues much easier to apply.
Updated On: Sep 23, 2025
  • 126
  • 128
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  • 122
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The Correct Option is A

Solution and Explanation

Step 1: Determine the set of seven consecutive integers.

- The integers are between 60 and 70 (exclusive). The possible range is from 61 to 69. 
- Clue (I) states that A and B are both prime numbers. The only prime numbers between 61 and 69 are 61 and 67. 
- For a set of seven consecutive integers to contain both 61 and 67, the set must be \{61, 62, 63, 64, 65, 66, 67\}
- From Clue (I) and the condition "A is smaller than B", we can definitively say A = 61 and B = 67.

Step 2: Use the remaining clues to identify the other numbers.

- Clue (IV): The average of R and S is 63, so \(R+S = 126\). Their difference is 2, so \(R-S=2\). Solving these equations gives R = 64 and S = 62. Both are in our set. 
- Clue (III): Both digits of P are the same. In our set, the only number with identical digits is 66. So, P = 66
- Clue (II): T is a multiple of 9. In our set, the only multiple of 9 is 63. So, T = 63.

Step 3: Identify the remaining number, Q.

- The numbers we have identified are A=61, B=67, R=64, S=62, P=66, and T=63. 
- The only integer left in the set \{61, 62, 63, 64, 65, 66, 67\} is 65. 
- Therefore, Q = 65.

Step 4: Calculate the required sum.

- The question asks for the sum of A and Q. 
\[ \text{Sum} = A + Q = 61 + 65 = 126 \]

Final Answer:

\[ \boxed{126} \]

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