RBI Assistant Reasoning Mock Test 2026: 10 Days 10 Mocks Practice Test 1

Preparing for the RBI Assistant 2026 exam requires consistent practice and smart revision, especially in the Reasoning section. To help you build speed and accuracy, this “RBI Assistant Reasoning Mock Test 2026: 10 Days 10 Mocks” series is designed to simulate the real exam experience. 

In this series, Practice Test 1 will help you assess your current level, identify weak areas, and get familiar with the question pattern. By solving these mocks daily, you can strengthen concepts, improve time management, and boost your overall performance step by step.

1. Syllogism

Syllogism questions test logical deduction from given statements. Constructing a clear Venn diagram is key to accurate conclusions.

Question 1:

• Statements:

• Some One are Two.

• All Three are Two.

• No Two is Four.

• Conclusions & Reasoning:

1. Some Three are not Four: This conclusion is correct. Since "No Two is Four" and "All Three are Two", it implies that all of Three are part of Two. Therefore, no part of Three can be Four.

2. Some One are not Four: This conclusion is correct. As "Some One are Two" and "No Two is Four", the part of One that is Two cannot be Four.

• Answer: Both conclusions are correct.

Question 2:

• Statements:

• All Looks are Larks.

• No Lark is a Bus.

• All Buses are Blocks.

• Pedagogical Note: When multiple sets have similar starting letters (e.g., Looks, Larks, Blocks, Buses), pay close attention to avoid confusion.

• Conclusions & Reasoning:

1. No Block is a Lark: This conclusion is incorrect. While "No Lark is a Bus," and "All Buses are Blocks," it is possible that some Blocks are not Buses and might be Larks. We can only definitively state that "Some Blocks are not Larks" (specifically, the portion of Blocks that are Buses cannot be Larks), but "No Block is a Lark" is too strong and cannot be confirmed.

2. No Look is a Bus: This conclusion is correct. Since "All Looks are Larks" and "No Lark is a Bus," it logically follows that no Look can be a Bus.

• Answer: Only the second conclusion follows.

Question 3:

• Statements:

• Some Cities are NCR.

• Some NCR are States.

• All States are Country.

• Conclusions & Reasoning (Possibilities):

1. All NCR can be Countries: This is a possible conclusion, as no direct negative relationship prevents it.

2. Some City can be State: This is a possible conclusion, as no direct negative relationship prevents it.

3. All Cities can be State: This is a possible conclusion, as no direct negative relationship prevents it.

• Answer: All possibilities are correct.

2. Inequality

Inequality questions involve deriving relationships between elements based on given comparisons.

Question 1:

• Conclusions & Reasoning:

1. Relationship between B and J: Analyzing the path between B and J shows B > J.

2. Relationship between L and S: A direct path from L to S confirms the established relationship.

• Answer: Only the second conclusion follows.

Question 2:

• Key Relationship: Observe the presence of U = P.

• Reasoning: If U = P, substitute P for U in the expressions. This allows for a clear comparison between P and A. By combining given inequalities and the U = P relation, specific relationships (e.g., P is greater than or equal to A) can be fulfilled.

• Answer: All conditions for the correct option are fulfilled.

Question 3:

• Conclusions & Reasoning:

1. Relationship between W and 1: A direct equality connection is not possible.

2. Relationship between 4 and D: A direct path from 4 to D establishes the relationship.

• Memory Tip: Either/Or in Inequalities: If you have doubts regarding "Either/Or" conditions in Inequalities, review the specific rules for "Either/Or" cases as they involve distinct logical conditions.

3. Blood Relation

Solving blood relation puzzles requires careful mapping of family members and their relationships, often aided by a family tree diagram.

• Problem Setup: A family of seven members.

• Constraint: "Either both or none of the parents were alive." This implies that for any child in the problem scope, both parents are present in the family tree, or neither is explicitly mentioned.

• Step-by-step Construction:

1. S is U's male sibling.

2. Q is married to S, and S is P's son.

3. U has a child (implying U is married).

4. R is female.

5. T is P's son-in-law (T married to U).

6. V is R's granddaughter (V is child of U/T).

7. Count members and place remaining individuals.

• Question: Who was the mother of V?

• Solution: Based on the constructed family tree, U is the mother of V.

4. Direction and Distance

These problems involve tracking movements and turns to determine final direction or distance.

• Problem Setup: Priyanka travels to a mall following a series of directions and distances.

• Movements:

1. 8 km South.

2. Right turn, 12 km (West).

3. East direction, 9 km.

4. 270° clockwise turn, 4 meters.

• Pedagogical Explanation: Turns:

• A standard left or right turn implies a 90° turn.

• For a 270° clockwise turn:

• From facing East (after moving 9 km East):

• 90° clockwise: South

• 180° clockwise: West

• 270° clockwise: North

• So, after the 270° clockwise turn, Priyanka is facing North.

• Question: Which direction is she facing when she reaches the mall?

• Solution: North.

5. Comparison (Cricket Wickets)

Comparison problems require setting up equations from given relational statements and solving for unknown values.

• Problem Setup: Five bowlers (A, B, C, D, E) took wickets in IPL.

• Conditions:

1. A took five wickets more than B. (A = B + 5)

2. B took three wickets less than C. (B = C - 3)

3. D took four wickets more than C. (D = C + 4)

4. E took six wickets more than A. (E = A + 6)

5. B and C together took 27 wickets. (B + C = 27)

• Solving Strategy:

1. Substitute B = C - 3 into B + C = 27: (C - 3) + C = 27 => 2C - 3 = 27 => 2C = 30 => C = 15.

2. Calculate B: B = 15 - 3 => B = 12.

3. Calculate A: A = 12 + 5 => A = 17.

4. Calculate D: D = 15 + 4 => D = 19.

5. Calculate E: E = 17 + 6 => E = 23.

• Summary of Wickets: E=23, D=19, A=17, C=15, B=12.

• Question: Which of the given statements is true? (Answer verified by checking calculated values).

• Pedagogical Insight: Some comparison questions might have relationships stated directly in the question that can be verified without full calculation, offering a potential shortcut.

6. Number/Alphanumeric Series

These questions involve analyzing patterns or specific positions within a given string of numbers or characters.

Question 1 (Relative Position):

• Problem: Find the number that is 10th to the left of the number which is 6th from the right end of the string.

• Rule: When directions are opposite (left of right), add the positions. The count is from the stated end (right end).

• Calculation: 6 (from right) + 10 (left of that) = 16th from the right end.

• Solution: Count 16 positions from the right end of the given string to find the number (e.g., 3).

Question 2 (Filtering Prime Numbers):

• Problem: If prime numbers are dropped from the string, count the remaining elements.

• Definition: Prime numbers in this context are 2, 3, 5, 7.

• Strategy: Identify and remove all occurrences of 2, 3, 5, and 7 from the string, then count the remaining elements.

• Solution: After dropping primes, count the rest (e.g., 22 remaining elements).

Question 3 (Filtering Specific Digits):

• Problem: If 0, 1, and 8 are dropped from the string, find the 9th number from the left end.

• Strategy: Count from the left end, skipping any occurrences of 0, 1, or 8.

• Pedagogical Caution: This is a simple counting task. Avoid careless errors by carefully skipping the specified digits.

• Solution: Count the 9th visible number from the left after filtering (e.g., 2).

Question 4 (Counting elements to the left):

• Problem: How many numbers are to the left of the 15th number from the right end?

• Strategy:

1. Locate the 15th number from the right end.

2. Count all numbers appearing to the left of that identified number.

• Solution: (e.g., 20 numbers to the left).

7. Floor Puzzle

Floor puzzles involve arranging people or items on different floors of a building based on given conditions.

• Problem Setup: Eight people living on eight different floors.

• Solving Strategy: Create an 8-floor setup (8 to 1). Start with definite conditions and build cases.

• Key Conditions & Deductions:

1. D lives on an odd-numbered floor below 5 (Floor 1 or 3).

2. Only two persons live between D and Q.

3. The number of persons below Q is one more than above M.

4. Only two persons live between M and E; E is not adjacent to Q.

5. C lives two floors above H (C - _ - H).

6. P does not live two floors above M.

7. G lives above P.

• Question: What is the position of P from the bottom?

• Solution: (e.g., Second position from the bottom).

8. Box Puzzle

Box puzzles require arranging boxes vertically based on a set of conditional statements.

• Problem Setup: Eight boxes arranged vertically, without numbering.

• Solving Strategy: Start with the most restrictive conditions and build the arrangement.

• Key Conditions & Deductions:

1. Four boxes are kept between X and B.

2. Z is kept just above A, which is kept above B (Z - A - ... - B).

3. A is not kept just above B (at least one box between A and B).

4. Z is not kept near X.

5. B is not kept at the bottom.

6. Q is kept just above Y (Q - Y).

7. R is not kept near A.

• Self-Correction during Solving: It's important to double-check conditions and correct mistakes, like accidentally repeating a variable, which can lead to incorrect arrangements.

• Question: How many boxes are kept between R and P?

• Solution: (e.g., Two boxes).

9. Circular Arrangement

Circular arrangement problems involve seating people around a circular table based on their relative positions and preferences.

• Problem Setup: Six people seated around a circular table.

• Exam Context: This type of question might be slightly more challenging for RBI Assistant/Attendant exams due to variables (people and flowers), but the principles of direct placement apply. Do not get stressed; focus on applying conditions directly.

• Key Conditions & Deductions:

1. Akshat sits immediately right of Ishika, who likes Rose.

2. Vanshika sits opposite Akshat and likes Lotus.

3. Ayushi sits third to the right of Ishika.

4. Sumit likes Tulip and sits immediately right of Vanshika.

5. The person immediately right of Akshat likes Lily.

6. Ayushi likes Sunflower.

7. Akshat likes Daisy.

8. Priyanshu is the remaining person.

• Question: Who sits second to the right of the person who likes Tulip?

• Solution: The person who likes Daisy sits second to the right of the person who likes Tulip.