BITSAT 2025 Logical Reasoning Questions with Solutions

What is Asked in BITSAT Logical Reasoning?

Candidates preparing for BITSAT 2027 should focus on the following Logical Reasoning topics, as they are frequently asked in the exam and carry significant weightage.

BITSAT 2027 Logical Reasoning Questions

As per the BITSAT exam pattern, the exam will have a total of 4 sections. One such section is Logical Reasoning and English proficiency. The BITSAT logical reasoning section will comprise 20 BITSAT Logical Reasoning questions. Each question will be worth 3 marks, thus making the logical reasoning section a significant component. The BITSAT 2027 logical reasoning questions will test candidates on various aspects of logical reasoning, analytical thinking, and problem-solving skills. Check the BITSAT logical reasoning and verbal reasoning questions below.

Q1. In the following question, a matrix of certain characters is given. These characters follow a certain trend, row-wise or column-wise. Find out this trend and choose the missing character accordingly.

1. 18

2. 24

3. 36

4. 58

Solution:

(7 x 3) = 21 and (9 x 3) = 27

and (4 x 9) = 36 and (2 x 9) = 18

Therefore (9 x 6) = 54 and (4 x 6) = 24.

Hence option B is correct.

Q2. In the following question, a matrix of certain characters is given. These characters follow a certain trend, row-wise or column-wise. Find out this trend and choose the missing character accordingly.

1. 6

2. 8

3. 12

4. 10

Solution:

In each column, the fourth value is obtained by taking the cube root of the result of multiplying the first three values of the same column.

3*6*12=216=2161/3=6

2*4*64=512=5121/3=8

Hence option B is correct.

Q3. Select a suitable figure from the four alternatives that would complete the figure matrix.

Solution: In each row, the third figure is the uncommon part of the first two figures.

Hence option D is correct.

Q4. Select a suitable figure from the four alternatives that would complete the figure matrix.

Solution: There are 3 types of faces, 3 types of bodies, 3 types of hands and 3 types of legs, each of which is used only once in a single row. So, the features which have not been used in the first two figures of the third row would combine to produce the missing figure.

Hence option B is correct.

Q5. Find out which of the answer figures (a), (b), (c) and (d) complete the figure matrix.

Solution: In each row, the second figure is obtained from the first figure by reversing the direction of the RHS arrow and the third figure is obtained from the second figure by reversing the direction of both arrows.

Hence option A is correct.

Q6. Count the number of triangles and squares in the given figure.

1. 44 triangles, 10 squares

2. 14 triangles, 16 squares

3. 27 triangles, 6 squares

4. 36 triangles, 9 squares

Solution:

Triangles:

• The simplest triangles are AEI, EOI, OHI, HAI, EBJ, BFJ, FOJ, OEJ, HOL, OGL, GDL, DHL, OFK, FCK, CGK and GOK i.e. 16 in number.
• The triangles composed of two components each are HAE, AEO, EOH, OHA, OEB, EBF, BFO, FOE, DHO, HOG, OGD, GDH, GOF, OFC, FCG and CGO i.e. 16 in number.
• The triangles are composed of four components each HEF, EFG, FGH, GHE, ABO, BGO, CDO and DAO i.e. 8 in number.
• The triangles are composed of eight components each are DAB, ABC, BCD and CDA i.e. 4 in number.
• Total number of triangles in the figure = 16 + 16 + 8 + 4 = 44.

Squares:

• The squares composed of two components are HIOL, IEJO, JFKO and KGLO i.e. 4 in number.
• The squares composed of four components are AEOH, EBFO, OFGC and HOGD i.e. 4 in number.
• There is only one square EFGH which is composed of eight components.
• There is only one square ABCD which is composed of sixteen components.
• Total number of squares in the figure = 4 + 4 + 1 + 1 = 10

Hence option A is correct.

Q7. How many rectangles are there in the given figure?

1. 10

2. 9

3. 8

4. 7

Solution: The simplest rectangles are ABJI, BCKJ, IJFG and JKEF i.e. 4 in number.

The rectangle composed of two components each is ACKI, BCEF, IKEG and ABFG i.e. 4 in number.

The only rectangle composed of four components is ACEG.

Thus, there are 4+4+1=9 rectangles in the given figure.

Hence option B is correct.

Q8. Find the number of triangles in the given figure.

1. 18

2. 20

3. 24

4. 27

Solution: The simplest triangles are IJO, BCJ, CDK, KQL, MLQ, GFM, GHN and NIO i.e. 8 in number.

The triangles composed of two components each are ABO, AHO, NIJ, IGP, ICP, DEQ, FEQ, KLM, LCP and LGP i.e. 10 in number.

The triangles are composed of four components each HAB, DEF, LGI, GIC, ICL and GLC i.e. 6 in number.

Total triangles= 8+10+6= 24

Hence option C is correct.

Q9. Choose the alternative which closely resembles the water image of the given combination.

Solution: In mirror images, the characters and numbers are horizontally inverted. In mirror images right side appears as a left side in the same way the left appears in the right side. The water image is only reversed for all letters. The water image is the anticlockwise reverse of the mirror image. Firstly make a mirror image and then rotate anti-clockwise your notebook. It will be a water image. Hence option D is correct.

Q10. Choose the alternative which closely resembles the water image of the given combination.

Solution: In mirror images, the characters and numbers are horizontally inverted. In mirror images right side appears as a left side in the same way the left appears in the right side. The water image is only reversed for all letters. The water image is the anticlockwise reverse of the mirror image. Firstly make a mirror image and then rotate anti-clockwise your notebook. It will be a water image. Hence option B is correct.

Q 11. Choose the alternative which closely resembles the mirror image of the given combination.

Solution: In mirror images, the characters and numbers are horizontally inverted. In mirror images right side appears as the left side in the same way the left appears in the right side. The water image is only reversed for all letters. The water image is the anticlockwise reverse of the mirror image. Firstly make a mirror image and then rotate anti-clockwise your notebook. It will be a water image. Hence option D is correct.

Q12. Choose the alternative which closely resembles the mirror image of the given combination.

Solution: In mirror images, the characters and numbers are horizontally inverted. In mirror images, the right side appears as the left side in the same way the left appears on the right side. The water image is only reversed for all letters. The water image is the anticlockwise reverse of the mirror image. Firstly make a mirror image and then rotate anti-clockwise your notebook. It will be a water image. Hence option D is correct.

Q13. Choose the alternative which closely resembles the mirror image of the given combination.

Solution: In mirror images, the characters and numbers are horizontally inverted. In mirror images right side appears as a left side in the same way the left appears in the right side. The water image is only reversed for all letters. The water image is the anticlockwise reverse of the mirror image. Firstly make a mirror image and then rotate anti-clockwise your notebook. It will be a water image. Hence option B is correct.

Q14. Find out which of the answer figures (a), (b), (c) and (d) complete the figure matrix.

Solution: In each row(as well as column), the figure in the third box is a combination of all the elements of the first and second figures.

Hence option D is correct.

Q14. 3,10,29,66,127,?

1. 164

2. 187

3. 216

4. 218

Solution: The given series is a triangular pattern series.

So we have,

3 10 29 66 127 ?

7 19 37 61

12 18 24

6 6

so missing term is = 127+ (61+24+6)=127+91=218

Q15. Earth is to Venus as Mercury is to

1. Sun

2. Pluto

3. Mars

4. Moon

Solution: Venus is the planet nearest to the Earth. Likewise, Mercury is the planet nearest to the Sun.

Q16. Find the missing number in the following series:

5, 10, 8, 16, 14, ?

Solution:

At first glance, the series does not follow a simple increasing or decreasing pattern. Therefore, split it into two separate sequences.

Sequence A: 5 → 8 → 14

Sequence B: 10 → 16 → ?

For Sequence A:

5 + 3 = 8

8 + 6 = 14

The increase is +3 and then +6.

For Sequence B:

10 + 6 = 16

Following the increasing pattern, the next increment is +8.

Therefore,

16 + 8 = 24

Answer: 24

Q17. Pointing towards a woman, a man says,

"She is the daughter of my mother's only daughter."

How is the woman related to the man?

Solution:

Break the statement into smaller parts.

"My mother's only daughter" refers to the man's sister.

The woman is the daughter of his sister.

Therefore, the woman is the man's niece.

Answer: Niece

Q18. If CAT = DBU, then DOG = ?

Solution:

Observe the pattern:

C → D (+1)

A → B (+1)

T → U (+1)

Every letter is shifted one position forward in the English alphabet.

Now apply the same rule:

D → E

O → P

G → H

Therefore,

DOG = EPH

Answer: EPH

Q19. Ravi walks 20 metres north, turns right, walks 10 metres, then turns right again and walks 20 metres.

Which direction is Ravi facing now?

Solution:

Step-by-step movement:

Initially facing North

Right turn → East

Another right turn → South

Therefore, Ravi is now facing South.

Answer: South

Q20. A is the father of B, but B is not the son of A.

What is B's relation to A?

Solution:

Since B is not the son, the only possible relation is that B is the daughter of A.

Answer: Daughter

Q21. Statements:

All roses are flowers.

Some flowers are red.

Conclusion:

Some roses are red.

Solution:

The second statement only tells us that some flowers are red.

It does not specify that those red flowers are roses.

Hence, the conclusion cannot be confirmed.

Answer: The conclusion does not follow.

Q22. Find the missing letter.

A, C, F, J, O, ?

Solution:

Convert letters into positions.

A = 1

C = 3 (+2)

F = 6 (+3)

J = 10 (+4)

O = 15 (+5)

The increments increase by one each time.

Next increment = +6

15 + 6 = 21

The 21st letter is U.

Answer: U

Q23. If RED = 27, then find the value of GREEN.

(Given: A = 1, B = 2, ..., Z = 26)

Solution:

Calculate the alphabetical values.

G = 7

R = 18

E = 5

E = 5

N = 14

Total = 7 + 18 + 5 + 5 + 14 = 49

Answer: 49

Q24. A person walks 10 metres north, then 5 metres east, and finally 10 metres south.

How far is the person from the starting point?

Solution:

After moving north and then south by equal distances, the person returns to the original horizontal level.

Only the movement towards the east remains.

Therefore, the final distance from the starting point is 5 metres.

Answer: 5 metres

Q25. Choose the odd one out.

1) Lion

2) Tiger

3) Elephant

4) Dog

Solution:

• Lion, tiger, and Dog are mammals that primarily consume meat (dogs are omnivorous but commonly grouped with carnivorous mammals in such reasoning questions).
• Elephant is a herbivore.

Therefore, Elephant differs from the others.

Answer: Elephant

BITSAT Logical Reasoning Previous Year Trends

Previous years' analysis of BITSAT papers shows that the Logical Reasoning part often turns out to be quite scoring, especially when there is a lot of figure-based reasoning, then series-style questions, analogy comparisons, and the deductive kind of logic.

BITSAT 2027 Logical Reasoning Topics (Weightage-Wise)

Candidates can check some important topics for the BITSAT 2027 logical reasoning sections along with their expected weightage below.

Logical Reasoning Topics

Best Books for BITSAT 2027 Logical Reasoning Preparation

Picking the right book is pretty essential if you’re aiming to score well in the BITSAT 2027 Logical Reasoning part.

BITSAT 2027 Preparation Tips

Here are some preparation tips for BITSAT 2027. Candidates can refer to the following points to ace the exam.

• Go through the syllabus: Candidates must understand the BITSAT 2027 syllabus well before starting their preparation. Knowledge of the correct syllabus will help the candidates to study the right topics for the exam.

• Make a timetable: Candidates must manage all the available time for preparation in a better way by preparing a timetable and following the BITSAT books. As we have already mentioned, the important topics for BITSAT 2027, candidates can also take up the important topics first. The timetable should be made in such a way that the entire syllabus of each section of the exam is covered within the time.

• Important Books: Candidates must generally start the BITSAT preparations with the NCERT books. After this, they should refer to the important reference books mentioned in the article.

• Mock tests and sample papers- Candidates must practice more and more mock tests of BITSAT, as these are based on the actual pattern of the exam. These also help the candidates to analyse their performance.

• Revision: As revision is the key, candidates should give proper time to revision in their BITSAT preparation plan. The candidates must revise whatever they have studied.