RBI Office Attendant 2026 Reasoning Super Mock - 2 By Vidhu Sir

RBI Office Attendant 2026 Reasoning Super Mock – 2 covers important reasoning topics like Alphanumeric Series, Chinese Coding-Decoding, Syllogisms, Blood Relations, and Arrangement Puzzles. It explains smart shortcuts such as digit comparison, elimination methods, and logical deduction to solve questions quickly and accurately in competitive exams.

The mock also includes detailed solutions with step-by-step logic, helping aspirants improve speed and accuracy. By practicing structured approaches like Venn diagrams and systematic arrangements, candidates can strengthen analytical thinking and boost their performance in the RBI Office Attendant 2026 exam.

RBI Office Attendant 2026 Reasoning Super Mock 3

RBI Office Attendant 2026 Reasoning Super Mock 2– Video Explanation

For better clarity and complete conceptual understanding, watch the detailed video explanation of RBI Office Attendant 2026 Reasoning Super Mock – 2. The video provides step-by-step solutions, shortcut tricks, and smart solving techniques to help you improve speed and accuracy.

Reasoning Super Mock Questions with Solution

This Super Mock explores essential reasoning topics pertinent to competitive exams like the RBI Office Attendant 2026. It provides a detailed breakdown of problem-solving techniques for various question types, from numerical series and coding to logical deductions and complex arrangements. 

Understanding these concepts and strategies is crucial for enhancing analytical skills and improving exam performance.

Alphanumeric Series

An alphanumeric series is a sequence that includes both letters (alphabets) and numbers (digits) arranged in a specific order or following a particular pattern.

Question 1: Highest Number after Ascending Sort

Problem: Given a set of numbers, if all the digits within each number are arranged in ascending order, which of the resulting numbers will be the highest?

Numbers: 787, 948, 664, 537, 429

Solution & Logic:

To find the highest resulting number after arranging its digits in ascending order, we must identify the original number whose smallest digit is the largest among all given numbers. This smallest digit will become the new first digit.

1. 787 → smallest digit is 7

2. 948 → smallest digit is 4

3. 664 → smallest digit is 4

4. 537 → smallest digit is 3

5. 429 → smallest digit is 2

The number 787 has the largest smallest digit (7). Therefore, it will yield the highest number (778) after its digits are sorted. The correct answer is 787.

Question 2: Divisibility by Two after Digit Interchange

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

Solution & Logic:

A number is divisible by 2 if its last digit is an even number. When the second and last digits are interchanged, the second digit of the original number becomes the new last digit. Hence, we only need to check if the second digit of the original number is even.

• 7**8**7 → 8 is even. (Divisible)

• 9**4**8 → 4 is even. (Divisible)

• 6**6**4 → 6 is even. (Divisible)

• 5**3**7 → 3 is odd. (Not divisible)

• 4**2**9 → 2 is even. (Divisible)

There are four such numbers that will be divisible by 2 after the interchange.

Question 3: Lowest Number after Transformation

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

Solution & Logic:

The key strategy is to avoid performing all operations manually. Focus on the digit that determines the final outcome.

1. After dropping the second digit, only the first and third digits remain.

2. Interchanging these remaining digits means the original last digit will become the new first digit of the transformed number.

3. To find the lowest new number, we need to find the original number with the smallest last digit.

• 78**7** → Last digit is 7

• 94**8** → Last digit is 8

• 66**4** → Last digit is 4

• 53**7** → Last digit is 7

• 42**9** → Last digit is 9

The smallest last digit is 4, which belongs to the number 664. This number will become the lowest after the transformation.

Question 4: Second Highest Digit Sum

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

Solution & Logic:

Calculate the sum of digits for each number:

• 787 → 7 + 8 + 7 = 22 (Highest Sum)

• 948 → 9 + 4 + 8 = 21 (Second Highest Sum)

• 664 → 6 + 6 + 4 = 16

• 537 → 5 + 3 + 7 = 15

• 429 → 4 + 2 + 9 = 15

The highest sum is 22, and the second-highest sum is 21. The number corresponding to the sum of 21 is 948.

Chinese Coding-Decoding

This section involves decoding a fictional language using a process of elimination by comparing statements and their corresponding codes.

Question 1: Word for Code 'go'

Based on the context, the exact word for 'go' often remains ambiguous with another code like 'mf' if they appear together uniquely in a pair. In such cases, if no other information allows for unique assignment, the answer is "Cannot be determined."

Question 2: Code for 'Sky'

To find the code for 'Sky', one must identify lines containing 'Sky' and then eliminate codes for other words present in that line that appear in other statements. The unique code remaining will be for 'Sky'. In this specific problem, xm is the code for 'Sky'.

Question 3: Code for 'always sky before'

A strategic shortcut can be used here. Once the code for a key word (e.g., 'sky' = xm) is found, immediately scan the answer options. If only one option contains this known code (xm), it is highly likely the correct answer, saving time needed to decode other words like 'always' and 'before'.

Question 4: Code for 'People safe'

This requires a more thorough elimination process.

1. Identify common words across different statements and their codes. For instance, ex for 'attack', my for 'on', and qk for 'before' are usually found first.

2. Once these common codes are eliminated from the relevant statements, the remaining unique codes can be paired with the remaining unique words.

3. The code for 'safe' is identified as 'jx'.

4. By further comparing and eliminating, the combined code for 'people safe' is determined to be jx mf.

Syllogisms

Syllogisms test logical deduction from given statements. Venn diagrams are a common and effective method for visualizing relationships.

Question 1

Statements:

• No Apple is Cashew.

• No Cashew is Grape.

• All Pineapple are Grape.

Diagrammatic Representation:

(Apple) ← No → (Cashew) ← No → (Grape [Pineapple inside Grape])

Conclusions:

1. No Apple being Grape is a possibility: Correct. There is no direct connection or restriction between 'Apple' and 'Grape'. Therefore, a "no" relationship between them is a valid possibility.

2. Some Pineapple being Cashew: Incorrect. Since all Pineapples are Grapes, and 'No Cashew is Grape', it definitively follows that 'No Pineapple is Cashew'. A possibility contradicting a definite truth is false.

Result: Only Conclusion 1 follows.

Question 2

Statements:

• All Circle are Round.

• Only a few Shapes are Round.

• No Square is a Circle.

Diagrammatic Representation:

(Shape) ← Some & Some Not → (Round [Circle inside Round]) ← No → (Square)

Conclusions:

1. Some Shape are not Round: Correct. The statement "Only a few Shapes are Round" inherently implies that "Some Shapes are Round" AND "Some Shapes are not Round". This conclusion is definite and true.

2. Some Round are Square is a possibility: Correct. The "No" relationship is specifically between 'Square' and 'Circle'. The remaining portion of 'Round' (outside of 'Circle') has no defined relationship with 'Square'. Therefore, an overlap between 'Some Round' and 'Square' is a valid possibility.

Result: Both conclusions follow.

Question 3

Statements:

• Only a few Doctor is Nurse.

• All Nurse is Patient.

• Some Patient is Bed.

Diagrammatic Representation:

(Doctor) ← Some & Some Not → (Nurse) → All are (Patient) ← Some → (Bed)

Conclusions:

1. All Doctor can be Bed: Correct. There is no direct or indirect negative relationship (like "No Doctor is Bed") established between 'Doctor' and 'Bed'. Thus, it is possible for the entire 'Doctor' category to be contained within 'Bed'.

2. Some Doctor is not Nurse is a possibility: Incorrect. The statement "Only a few Doctor is Nurse" definitively means that "Some Doctor is not Nurse" is 100% true. A possibility of a statement that is already definitely true is considered false in syllogisms because it is certain, not just possible.

Result: Only Conclusion 1 follows.

Blood Relations

Blood relations problems require careful deduction to build a family tree based on given clues. This problem is considered a higher-level question.

Problem Details:

• 9 members in a family.

• 3 married couples.

• Each couple has exactly one child.

• Z and Q do not have siblings.

• P has a male sibling.

Key Deductions & Family Tree Construction:

1. "T is the daughter of Z and P" establishes the first couple: Z-P (Z is male, P is female) and their child T (female).

2. "P has a male sibling." Let's call this sibling Y for now.

3. "Z is the brother-in-law of Y." This confirms Y is P's brother.

4. "M is the son of Y." Since each couple has one child, Y must be married. So, Y (male) forms the second couple.

5. "Q is the brother-in-law of X." Since Q has no siblings, his spouse must have a sibling X. This means Q is married to X's sibling.

6. "X is the sister of S." This connects X and S.

Combining these, we deduce the family structure:

• Couple 1: Z (male) and P (female) → Child: T (female)

• P has a brother, Y (male). So, P and Y are siblings.

• Couple 2: Y (male) is married to X (female) → Child: M (male)

• X has a sister, S (female). So, X and S are siblings.

• Couple 3: S (female) is married to Q (male) → Child: F (gender not explicitly given, assume male for completion based on typical problem patterns if not specified for specific questions).

• This arrangement makes Q the husband of X's sister S, thus Q is X's brother-in-law.

• This also means P, Y, and S are siblings, covering the condition "P has a male sibling (Y)".

• The 9 members are Z, P, T, Y, X, M, S, Q, F.

• The 3 couples are (Z-P), (Y-X), (S-Q).

• Each couple has one child: T, M, F.

• Z and Q having no siblings is consistent with their positions as in-laws or spouses of siblings.

Final Family Tree Summary:

• Generation 1 Couples: (Z - P), (Y - X), (S - Q)

• Children (Generation 2): T (child of Z-P), M (child of Y-X), F (child of S-Q)

• Siblings: P, Y, S are siblings. X is also a sister of S.

Question: How is X related to T?

Answer: T's mother is P. P's brother is Y. Y's wife is X. Therefore, X is the wife of T's paternal uncle. X is T's Aunt.

Arrangement & Puzzles

These problems require systematic arrangement of people or objects based on given conditions.

Puzzle 1: Ordering of Purchases

Problem: Seven people (B, F, J, N, R, and two others) buy seven different items (Dumbbell, Pers, Satchel, Laptop, Mirror, and two others) one after another.

Final Arrangement (Sequence of Purchase):

1. R

2. N

3. Laptop (bought by someone)

4. Mirror (bought by someone)

5. Pers (bought by someone)

6. J

7. B (bought Satchel)

8. F

9. Dumbbell (bought by someone)
   Note: This sequence represents the relative order derived from the problem's conditions.

Question: Who purchased the item immediately before B?

Answer: Based on the final sequence, J purchased immediately before B.

Puzzle 2: Box Arrangement

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

Final Arrangement (Top to Bottom):

1. R

2. Q

3. U

4. S

5. P

6. V

7. T

8. W

Solution Logic:

The puzzle is solved by considering multiple cases generated from initial clues, then eliminating inconsistent cases as further conditions are applied. For example, conditions like "One box between P and W," "Three boxes between V and S," and "Q was kept above S" are used to build and validate the arrangement, narrowing down possibilities to a single correct order.

Question: Which box is kept immediately above T?

Answer: Box V is kept immediately above T.

Puzzle 3: Square Table Arrangement

Problem: Eight people (A, B, C, D, E, F, G, H) are seated around a square table. Four sit at the corners (facing inward), and four sit in the middle of the sides (facing outward).

Final Arrangement (Relative positions, e.g., Clockwise from a corner):

• Corners (Facing Inward): H, E, B, F

• Sides (Facing Outward): D, G, A, C

A possible clockwise sequence starting from H (corner) would be:

1. H (Corner, Inward)

2. D (Side, Outward)

3. E (Corner, Inward)

4. G (Side, Outward)

5. B (Corner, Inward)

6. A (Side, Outward)

7. F (Corner, Inward)

8. C (Side, Outward)

Question: Who sits third to the left of F?

Answer: F is at a corner, facing inward. Counting three positions to F's left (clockwise direction from F's perspective), we find A.

Why Is This Reasoning Super Mock Important?

This Reasoning Super Mock is important because reasoning is a scoring section in the RBI Office Attendant exam. Regular practice improves problem-solving speed, accuracy, and time management. It also helps candidates understand common question patterns and apply shortcuts effectively during the actual exam.

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