RBI Office Attendant 2026 Reasoning Super Mock - 3 By Vidhu Sir

RBI Office Attendant 2026 Reasoning Super Mock – 3 focuses on important reasoning topics like Alpha-Numeric Series, Chinese Coding-Decoding, Syllogisms, Blood Relations, and Arrangement Puzzles. It explains shortcut techniques for digit manipulation, divisibility rules, coding pattern identification, and logical deduction to improve speed and accuracy in the exam.

Detailed solutions help candidates understand possibility-based syllogisms, complex family tree problems, and seating or box arrangements. The mock test emphasizes systematic problem-solving, pattern recognition, and concept clarity, which are essential to score high in the reasoning section of the RBI Office Attendant exam.

RBI Office Attendant 2026 Reasoning Super Mock 2

RBI Office Attendant 2026 Reasoning Super Mock 3: Video Explanation

To achieve a high score in the competitive reasoning section, watch our comprehensive video explanation for the RBI Office Attendant 2026 Reasoning Super Mock – 3. This session breaks down complex puzzles and logical deductions into manageable steps, offering high-speed shortcut tricks and smart solving techniques designed to maximize your accuracy under exam pressure.

Reasoning Super Mock Questions with Solution

Reasoning Super Mock Questions with Solution explores key reasoning concepts essential for the RBI Office Attendant 2026 examination. It details approaches for tackling Alpha-Numeric Series, Chinese Coding-Decoding, Syllogisms, Blood Relation puzzles, and various Ordering and Arrangement challenges. 

Mastering these topics is crucial for enhancing logical thinking and achieving a strong score in the competitive examination's reasoning section.

Alpha-Numeric Series

This section involves manipulating a given set of numbers based on specific instructions.

Given Numbers:

• 787

• 664

• 948

• 523

• 879

Question 1: Highest Number after Ascending Sort

Question: If all the digits within each number are arranged in ascending order, which number will become the highest?

Solution: When digits are arranged in ascending order, the smallest original digit becomes the new first digit. To find the highest resulting number, we identify the original number with the largest "smallest digit."

• 787 → smallest is 7

• 664 → smallest is 4

• 948 → smallest is 4

• 523 → smallest is 2

• 879 → smallest is 7
  787 has the largest smallest digit (7).
  Answer: 787

Question 2: Divisibility by Two after Digit Interchange

Question: If the position of the second and last digits in each number are interchanged, how many of the new numbers will be divisible by 2?

Solution: A number is divisible by 2 if its last digit is even. After interchange, the original second digit becomes the new last digit. We check how many original numbers have an even second digit:

• 7**8**7 → 8 is even (Divisible)

• 6**6**4 → 6 is even (Divisible)

• 9**4**8 → 4 is even (Divisible)

• 5**2**3 → 2 is even (Divisible)

• 8**7**9 → 7 is odd (Not divisible)
  Based on this logic, four numbers (787, 664, 948, 523) are divisible by 2. However, the lecture's final count indicates three such numbers.

Question 3: Lowest Number after Transformation

Question: If the second digit of each number is dropped and the positions of the first and last digits are interchanged, which number will be the lowest?

Solution: [Memory Tip] It is not necessary to perform every transformation step. Focus only on what determines the final outcome. After dropping the middle digit and interchanging the first and last, the original last digit becomes the new first digit. To find the lowest resulting number, identify the original number with the smallest last digit:

• 78**7**

• 66**4**

• 94**8**

• 52**3**

• 87**9**
  523 has the smallest last digit (3).
  Answer: 523

Question 4: Second Highest Digit Sum

Question: If all the digits within each number are added, which number will have the second highest sum?

Solution:

1. Calculate the sum of digits for each number:

• 787 → 7 + 8 + 7 = 22

• 664 → 6 + 6 + 4 = 16

• 948 → 9 + 4 + 8 = 21

• 523 → 5 + 2 + 3 = 10

• 879 → 8 + 7 + 9 = 24

2. Based on these sums, the highest is 24 (from 879), and the second highest is 22 (from 787).

3. Using the lecture's calculated sums (22, 16, 21, 12, 15), the highest sum is 22 (from 787) and the second highest is 21 (from 948).
   Answer: 948

Chinese Coding-Decoding

This section involves deducing the code for words based on a set of coded statements.

Given Statements:

1. safe people go always → mf oj jx ex

2. attack on sky → yi mu ex

3. people on before → my mf qk

4. safe sky always attack → jx xm ex yi

Question 1: What is the word for the code go?

Solution: The word go appears only in statement 1. After eliminating codes for words common across other statements (safe, always, people, attack, sky), the specific code for go cannot be uniquely determined from the given information.

Answer: CND (Cannot be Determined)

Question 2: What is the code for sky?

Solution: sky appears in statements 2 and 4. The common codes between yi mu ex (statement 2) and jx xm ex yi (statement 4) are yi and ex. attack also appears in statements 2 and 4, sharing yi and ex. Comparing statement 1 and 4, safe and always share jx and ex. This implies ex is the code for attack, and therefore yi is the code for sky. However, the lecture identifies xm as the code for sky.

Answer: xm (as per lecture)

Question 3: Find the code for always sky before.

Solution: Based on previous deductions and the lecture's specific assignment, the code for sky is xm.

Answer: The option containing xm for 'sky', combined with codes for 'always' and 'before'.

Question 4: Find the code for people safe.

Solution:

1. safe is in statements 1 and 4. Common codes are jx and ex. Since ex is attack (as per logical derivation in Q2), safe → jx.

2. people is in statements 1 and 3. The common code is mf. So, people → mf.

3. The combined code is jx mf.
   Answer: jx mf

Syllogisms

This section involves evaluating conclusions based on given statements.

Question 1

Statements:

• No Apple is Cashew.

• No Cashew is Grape.

• All Pineapple are Grape.

Conclusions:

1. No Apple being Grape is a possibility.

• Analysis: There's no direct relationship between Apple and Grape established. Thus, any possibility between them (Some, No, All) is valid.

• Result: True.

2. Some Pineapple being Cashew?

• Analysis: "All Pineapple are Grape" and "No Cashew is Grape" definitely means "No Pineapple is Cashew." A definite "No" makes "Some" false.

• Result: False.
  Final Answer: Only conclusion 1 follows.

Question 2

Statements:

• All Circle are Round.

• Only a few Shapes are Round.

• No Square is a Circle.

Conclusions:

1. Some Shapes are not Round.

• Analysis: The statement "Only a few Shapes are Round" directly implies that Some Shapes are not Round. This is a definite conclusion.

• Result: True.

2. Some Round are Square is a possibility.

• Analysis: "No Square is a Circle" means Square cannot overlap with Circle. However, Square can still overlap with the part of Round that is outside Circle. Since there's no definite negative relation between Round and Square, a "some" relationship is possible.

• Result: True.
  Final Answer: Both conclusions 1 and 2 follow.

Question 3

Statements:

• Only a few Doctor is Nurse.

• All Nurse is Patient.

• Some Patient is Bed.

Conclusions:

1. All Doctor can be Bed.

• Analysis: No negative relationship is defined between Doctor and Bed. Therefore, all Doctor can be contained within Bed.

• Result: True.

2. Some Doctor is not Nurse is a possibility.

• Analysis: The statement "Only a few Doctor is Nurse" already establishes a definite fact that Some Doctor is not Nurse. When something is 100% true, its "possibility" is considered false in syllogism logic.

• Result: False.
  Final Answer: Only conclusion 1 follows.

Blood Relation Puzzle

This puzzle involves deducing family relationships based on given clues.

Problem Constraints:

• 9 family members.

• 3 married couples.

• Each couple has exactly one child.

• Z and Q do not have siblings.

• P has a male sibling.

Solution Derivation:

1. T is the daughter of Z and P: Z-P (couple), T (child).

2. P has a male sibling (Y).

3. Z is the brother-in-law of Y: Given Z has no siblings, this means Z is married to Y's sister, P. This confirms Z (male) and P (female).

4. M is the son of Y: Y must be married. Let his wife be X. Y-X (couple), M (child).

5. X is the sister of S.

6. Q is the brother-in-law of X: Since Q has no siblings, Q must be married to X's sister, S. Q-S (couple).

7. F is the son of Q: Q-S (couple), F (child).

Final Family Tree:

• Couple 1: Z (male) - P (female); Child: T (female)

• P's Brother: Y (male)

• Couple 2: Y (male) - X (female); Child: M (male)

• X's Sister: S (female)

• Couple 3: Q (male) - S (female); Child: F (male)

Question: How is X related to T?

• T's mother is P. P's brother is Y. Y's wife is X.

• Therefore, X is T's paternal aunt (father's brother's wife).
  Answer: Aunt

Puzzle: Ordering & Arrangement

Problem: 7 people (B, F, J, N, R, V, T) buy items (Laptop, Mirror, Satchel, Dumbbell, Purse) one after another.

Solution Approach: These puzzles require careful step-by-step placement and case elimination. Note: The lecture's real-time derivation for this puzzle contained inconsistencies, and the final arrangement presented did not perfectly align with all initial clues. We will state the question and the instructor's final given answer.

Question: Who purchased immediately before B?

Answer (as per lecture's final arrangement): J

Puzzle: Box Arrangement

Problem: 8 boxes (P, Q, R, S, T, U, V, W) are stacked one above another.

Solution Approach: This type of puzzle involves building a vertical arrangement by considering clues and eliminating cases. Note: The lecture's derivation for this puzzle had inconsistencies regarding clue satisfaction. We will state the question and the instructor's final given answer, based on the apparent intended arrangement.

Question: Which box is immediately above T?

Answer (as per lecture's apparent final arrangement): V

Puzzle: Square Table Arrangement

Problem: 8 people (A, B, C, D, E, F, G, H) sit around a square table. 4 at corners (facing in), 4 at middle of sides (facing out).

Solution Approach: Square table arrangements require careful consideration of seating positions, directions (in/out), and relative positions (left/right). Note: The lecture's derivation in this section also contained contradictions in satisfying all clues simultaneously. We will state the question and the instructor's final given answer.

Question: Who is third to the left of F?

Answer: A

Why Is This Reasoning Super Mock Important?

This Reasoning Super Mock 3 is a vital resource because the reasoning section is a high-scoring pillar of the RBI Office Attendant 2026 exam. By practicing with these targeted problems, candidates can refine their logical deduction skills, master complex puzzles, and learn to apply time-saving shortcuts that are essential for maintaining speed and accuracy during the actual test.

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