RBI Office Attendant 2026 Reasoning Super Mock

RBI Office Attendant 2026 Reasoning Super Mock focuses on the types of questions candidates usually face in the exam and how to approach them with clarity. This mock session introduces common reasoning topics such as inequalities, syllogisms, coding-decoding, and puzzle-based questions in a simple and structured way. It helps students understand question patterns, practise logical thinking, and get comfortable with the level of difficulty expected in the RBI Attendant exam.

Regular practice through such sessions helps reduce confusion during the exam and improves accuracy. It also builds confidence to attempt reasoning questions within the given time limit.

Questions Based on Inequalities

Inequalities problems require determining relationships between elements based on given statements.

Problem 1

• Statement: Z > V ≥ Y = T > P; Q < Y

• Conclusions:

1. Z > Q

2. V > P

• Analysis:

1. To connect Z and Q, follow the path Z > V ≥ Y > Q. This simplifies to Z > Q. The conclusion is correct.

2. To connect V and P, follow V ≥ Y = T > P. This simplifies to V > P. The conclusion is correct.

Problem 2: The 'Either/Or' Case

• Statement: A = R ≥ S = T

• Conclusions:

1. A > T

2. S = R

• Analysis:

• From the statement, we derive A ≥ T.

• Conclusion 1 states A > T.

• Conclusion 2 states S = R. Since A = R and S = T, this implies A = T.

• When the derived relationship is A ≥ T, and the individual conclusions are A > T and A = T, they form an 'Either/Or' complementary pair. This means one of them must be true.

• Therefore, Either conclusion 1 or 2 follows.

Problem 3

• Statement: V < E = N > J

• Conclusions:

1. V ≥ J

2. E > J

• Analysis:

1. There is no clear relationship between V and J because the inequality signs are conflicting (V < E and E > J). The conclusion V ≥ J is incorrect.

2. From E> N > J, it follows that E > J. The conclusion is correct.

• Therefore, only conclusion 2 follows.

Questions Based on Alphanumeric Series

These problems involve analysing a mixed series of letters, numbers, and symbols.

Problem 1

• Question: In the given series, what is the sum of the numbers between the symbols % and @?

• Series Snippet (contextual): ... % ... 4 ... 2 ... @ ...

• Solution: The numbers found between % and @ are 4 and 2. Their sum is 4 + 2 = 6.

Problem 2

• Question: If all the numbers are dropped from the series, which element will be sixth from the left end?

• Procedure: Mentally remove all numbers from the series. Then, count the elements from the left.

• Solution: After removing numbers, the sixth element from the left is G.

Questions based on Syllogisms

Syllogism questions require identifying the correct conclusions based on the statements provided. Venn diagrams are commonly used to understand and solve these questions more clearly.

Problem 1

• Statements:

1. Some Rayon are Nylon.

2. Some Nylon are Silk.

• Conclusions:

1. All Nylon being Silk is a possibility.

2. No Rayon is Silk is a possibility.

• Analysis:

1. Even if Some Nylon are Silk, it's possible that All Nylon are Silk. This is a valid possibility.

2. Since there's no direct definite relation given between Rayon and Silk, No Rayon is Silk remains a valid possibility.

• Therefore, Both conclusions follow.

Problem 2: Understanding "Not a Possibility"

• Statements:

1. All Wifi are Router.

2. All Internet are Router.

3. Some Router are Connection.

• Conclusions:

1. No Internet is Connection is not a possibility.

2. All Router are Connection is not a possibility.

• Pedagogical Rule: "Not a Possibility"

• The phrase "is not a possibility" means the statement inside is definitely false or impossible. If the statement is actually a possibility, then "not a possibility" is false.

• Analysis:

1. No Internet is Connection: Based on the statements, Internet and Connection could overlap or be separate. Thus, No Internet is Connection is a possibility. Therefore, concluding it is "not a possibility" is false.

2. All routers are connected: The statement says some Routers are connected. This means all routers are connected a possibility, not a certainty. Therefore, concluding it is "not a possibility" is false.

• Therefore, neither conclusion follows.

Problem 3

• Statements:

1. Some Pen is not erasers.

2. Some Rubber is not a ball.

3. No Ball is a Scale.

• Conclusions:

1. Some Ball is not Rubber is a possibility.

2. Some Rubber is not Pen is a possibility.

• Analysis:

1. From Some Rubber is not Ball, it's possible that Some Ball is not Rubber (e.g., if some Ball is Rubber, but not all). This is a valid possibility.

2. Similarly, from Some Pen is not Rubber, it's possible that Some Rubber is not Pen. This is a valid possibility.

• Therefore, Both conclusions follow.

Questions Based on Coding-Decoding

These problems involve figuring out codes from the given examples. The main approach is the process of elimination.

Problem 1: Code for 'M'

• Logic: Identify the codes for other words in the same statement where 'M' appears. By eliminating common codes found in other statements, the remaining code corresponds to 'M'.

• Solution: After eliminating codes for other common words, the code for 'M' is HA.

Problem 2: Code for 'You and I'

• Logic:

1. Find the code for 'I' (e.g., PA from common occurrences).

2. Find the code for 'You' (e.g., NA).

3. The word 'and' is unique. Its code must be new not assigned to any other word.

• Solution: Select the option that includes PA, NA, and a new, unused code.

Problem 3: Code for 'I have pen'

• Logic:

1. Code for 'I' is identified (e.g., PA).

2. Identify 'have' and 'pen' from a statement, typically by eliminating codes for other known words in that statement.

• Solution: The correct option will contain PA and the unique codes for 'have' and 'pen'.

Questions based on Comparison-Based Puzzle (Ranking)

This involves arranging individuals based on comparative conditions.

• Problem: Arrange 6 people (A, B, C, D, E, F) by the number of toffees they have.

• Conditions:

1. B > D > A

2. _ > F > B (F is not maximum)

3. Second-highest has 23 toffees.

4. A does not have the least.

5. B > E

6. All values are distinct integers.

•  Arrangement Derivation:

• Combining conditions 1, 2, and 5: _ > F > B > D > A. We know B > E.

• Condition 4 (A is not least) implies E must be the least.

• The unknown at the top must be the last person, C.

• Final Arrangement: C > F > B > D > A > E.

• From Condition 3, the second-highest (F) has 23 toffees. So, F = 23.

• Question: Which of the following cannot be the number of toffees with A? (Options: 21, 20, 19, 17, 15).

• Analysis:

• We know F = 23. Therefore, B, D, and A must have fewer than 23 toffees.

• Consider the option 21 for A. If A = 21, then D must be greater than 21 (e.g., 22).

• If D = 22, then B must be greater than 22 (e.g., 23).

• However, B cannot be 23 because F is 23 and F > B. This creates a contradiction.

• Therefore, A cannot have 21 toffees.

Questions based on Other Puzzles

Designation-Based Puzzle

• Hierarchy: MD > Director > VP > Senior Associate > Associate > Senior Analyst > Analyst.

• Conditions & Solution:

1. P is an Associate.

2. Two designations between P and R implies R is MD.

3. Q is just junior to R, so Q is Director.

4. One designation between Q and L implies L is Senior Associate.

5. M is senior to O, and O is not the junior-most. This places M as VP and O as Senior Analyst.

6. The remaining person, N, is placed in the last spot.

• Final Question: What is the designation of N?

• Answer: Analyst.

Linear Arrangement Puzzle

• Setup: 6 people in a straight line, each with a different number of rings.

• Arrangement & Solution:

1. Person with 12 rings has two people to their left.

2. Z sits second to the right of the person with 20 rings.

3. Person with 18 rings is to the right of Z. Y is to the immediate left of the person with 18 rings.

4. Combined information starts forming the sequence.

5. V is to the right of the person with 11 rings, and U is to the right of V.

6. Y has one more ring than Z. If Z has 12 rings, Y has 13 rings.

7. Placement of remaining numbers and people (X, W, 17) completes the arrangement.

• Final Question: How many more rings does X (17) have than V (11)?

• Answer: X has 6 more rings than V (17 - 11 = 6). (Note: This is based on logical derivation; speaker's stated answer was 3).

Box-Based Puzzle

• Setup: 8 boxes stacked from 1 (bottom) to 8 (top).

• Case-Based Solution:

1. Condition: More than four boxes are placed between D and A, with A above D, and D on an odd-numbered floor.

2. This generates possible cases for D and A (e.g., D=1, A=7 or 8; or D=3, A=8).

3. Apply subsequent conditions to each case:

• Two boxes between H and A.

• H is just above F.

• Two boxes between B and F.

• B is just above C.

• C is three boxes below G.

4. Systematically eliminate cases that contradict any condition.

• Final Arrangement (Derived): G (8), A (7), B (6), C (5), H (4), F (3), E (2), D (1).

• Final Question: Which box is immediately above E?

• Answer: Box F.

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