BITSAT 2025 Logical Reasoning Questions with Solutions

BITSAT 2025 Logical Reasoning Questions with Solutions

Edited By Team Careers360 | Updated on May 20, 2025 07:59 AM IST | #BITSAT
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BITSAT 2025 Logical Reasoning Questions - Aspirants aiming for the BITSAT 2025 exam must check the sample BITSAT logical reasoning questions given here. The logical reasoning section is a crucial part of BITSAT, designed to assess student's logical and analytical abilities. Aspirants often find this section challenging due to its diverse range of question types and the need for quick thinking. Careers360 provides some important types of BITSAT logical reasoning questions 2025. BITS Pilani conducts BITSAT 2025 as a computer-based test for admission to undergraduate engineering programs at its campuses in Pilani, Hyderabad, and Goa.

This Story also Contains
  1. BITSAT 2025 Logical Reasoning Questions
  2. BITSAT 2025 logical Reasoning Topics (Weightage-wise)
  3. BITSAT 2025 Preparation Tips
BITSAT 2025 Logical Reasoning Questions with Solutions
BITSAT 2025 Logical Reasoning Questions with Solutions

BITSAT 2025 will be held in two sessions as a computer-based test - Session 1 (May 26 to 30, 2025) and Session 2 (June 22 to 26, 2025).

Preparing for the logical reasoning questions of BITSAT 2025 requires understanding fundamental concepts and practising BITSAT sample papers. Regular practice of previous years' question papers, BITSAT LR questions and BITSAT 2025 mock tests will help candidates familiarize themselves with various types of BITSAT LR questions and improve their speed and accuracy.

BITSAT 2025 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 LR questions. Each question will be worth 3 marks, thus making the logical reasoning section a significant component. The BITSAT 2025 logical reasoning questions will test candidates on various aspects of logical reasoning, analytical thinking, and problem-solving skills. Check the BITSAT logical reasoning verbal reasoning questions below.

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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.

1715836013187

  1. 18

  2. 24

  3. 36

  4. 58

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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.

1715836012737

  1. 6

  2. 8

  3. 12

  4. 10

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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.

1715836012479

  1. 1715836013794

  2. 1715836011077

  3. 1715836014022

  4. 1715836013123

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.

1715836013429

  1. 1715836011213

  2. 1715836012393

  3. 1715836011619

  4. 1715836011503

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.

1715836013344

  1. 1715836012038

  2. 1715836011736

  3. 1715836010966

  4. 1715836012923

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.

1715836014262

  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?

1715836011332

  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.

1715836012809

  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.

1715836012153

  1. 1715836014081

  2. 1715836011681

  3. 1715836011802

  4. 1715836011451

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.

1715836011271

  1. 1715836013049

  2. 1715836014326

  3. 1715836014202

  4. 1715836014137

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.

1715836013898

  1. 1715836012215

  2. 1715836011857

  3. 1715836011918

  4. 1715836013962

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.

1715836012541

  1. 1715836011149

  2. 1715836011563

  3. 1715836012096

  4. 1715836012333

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.

1715836013716

  1. 1715836013242

  2. 1715836012603

  3. 1715836012667

  4. 1715836012869

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.

1715836013514

  1. 1715836011390

  2. 1715836012277

  3. 1715836012992

  4. 1715836011977

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.

BITSAT 2025 logical Reasoning Topics (Weightage-wise)

Candidates can check some important topics for the BITSAT 2025 logical reasoning sections along with their expected weightage below.

Logical Reasoning Topics

Important Topics

Weightage of topics in the exam

Figure Matrix, Figure Formation and Analysis

40%

  • Logical Deduction

  • Analogy Test

  • Series Test Numerical and Alphabetical

10-20%

  • Paper folding and cutting

  • Detection of the rule

  • Figure Completion Test

5-7%

Related links:

BITSAT 2025 Preparation Tips

Here are some preparation tips for BITSAT 2025. Candidates can refer to the following points to ace the exam.

  • Know the Syllabus- Candidates must understand the BITSAT 2025 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 2025, 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 before time.

  • Important Books- Candidates must generally start the BITSAT preparations with 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 analyze 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.

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Questions related to BITSAT

Have a question related to BITSAT ?

Writing a BITSAT mock test is a great way to prepare! Here’s how you can do it:

Visit the Official BITSAT Website:
Go to bitsadmission.com (https://bitsadmission.com/) , where BITS Pilani provides official mock tests.

Find the Mock Test Link:
Look for the “Mock Test” or “Sample Test” section on the homepage.

Register or Log In:
You might need to create an account or just enter basic details to access the mock test.

Read Instructions Carefully:
Before starting, read the guidelines about time limits, question types, and navigation.

Start the Test:
The mock test simulates the real exam with questions in Physics, Chemistry, Mathematics/Biology, English, and Logical Reasoning.

If your Aadhaar card only displays your year of birth and lacks the full date (day, month, and year), it may pose a challenge at the BITSAT exam center. The BITSAT admit card requires a complete date of birth to verify your identity. Discrepancies between the date of birth on your Aadhaar and the one on your admit card could lead to issues during the verification process.

Hello Swetha,

Yes, you're eligible for JEE Main 2026 & BITSAT 2026 with your 2025 Board improvement exam marks.

JEE Main: Your 2025 marks count for eligibility. These improved marks will be used for the 75% criteria for NIT/IIT admissions.

BITSAT: Your 2025 improved marks are accepted. Ensure they meet the 75% aggregate in PCM (or PCB) and 60% individually in each. Your "Pass" status must remain.

Ensure results are available by deadlines.


I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.

Thank you, and I wish you all the best in your bright future.

hi,

Yes, you can still apply for engineering courses even if your PCM aggregate is less than 75%, as long as you have scored above the minimum required marks in each individual subject. Many colleges and entrance exams consider individual subject marks and overall eligibility criteria differently. However, some top engineering colleges or specific programs might require a minimum PCM aggregate of 75%. So, check the specific eligibility criteria of the colleges or exams you are targeting to be sure.

Hello ,

I hope you are absolutely fine. As per your mentioned query , student should have passed their class 12th from any recognised board having PCM ( physics, chemistry and mathematics) or PCB ( physics, chemistry and biology) as main subjects with minimum aggregate of 75% .

To know more , refer this :

https://engineering.careers360.com/articles/bitsat-eligibility-criteria

Revert for further query!

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By mistake i wrote my birthdate wrong in the form and i alreay confromed it. It use birthdate as a password 

A photon of wavelength 6000 Å has an energy E. The wavelength of the photon of light which corresponds to an energy equal to 2E is

  • Option 1)

    300 Å

  • Option 2)

    3000 Å

  • Option 3)

    30 Å

  • Option 4)

    3 Å

 

A reduction in atomic size with increase in atomic number is a characteristic of elements of:

  • Option 1)

    f-block

  • Option 2)

    radioactive series

  • Option 3)

    high atomic mass

  • Option 4)

    d-block

 

An element has the configuration 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{2} : To which block in the long form of the periodic table, does this belong?

  • Option 1)

    s-block

  • Option 2)

    p-block

  • Option 3)

    d-block

  • Option 4)

    f-block

 

An element R forms the highest oxide R_{2}O_{5} . R belong to

  • Option 1)

    4 group

  • Option 2)

    6 group

  • Option 3)

    5 group

  • Option 4)

    8th group

 

An element X occurs in short period having configuration ns2 np1 . The formula and nature of its oxide are:

  • Option 1)

    X03 , basic

  • Option 2)

    XO3 ,acidic

  • Option 3)

    X2O3 , amphoteric

  • Option 4)

    X2O3, Basic

 

Elements of group IB and IIB are called:

  • Option 1)

    Normal Elements

  • Option 2)

    Transition elements

  • Option 3)

    Alkaline earth metals 

  • Option 4)

    Alkali metals

 

Gases begin to conduct electricity at low pressure because

  • Option 1)

    At low pressure gases turn to plasma

  • Option 2)

    Atoms break up into electrons and protons

  • Option 3)

    Electrons in atoms can move freely at low pressure

  • Option 4)

    Colliding electrons can acquire higher kinetic energy due to the increased mean free path leading to ionisation of atoms.

 

Halogens are placed in same group because they:

  • Option 1)

    Are electronegative

  • Option 2)

    Are most reactive

  • Option 3)

    Are not metals

  • Option 4)

    Have 7 electrons in outermost orbit.

 

Identify the pairs which are not of isotopes

(i)  _{6}^{12}\textrm{X}, \:_{6}^{13}\textrm{Y}

(ii) _{17}^{35}\textrm{X}, \:_{17}^{37}\textrm{Y}

(iii) _{6}^{14}\textrm{X}, \:_{7}^{14}\textrm{Y}

(iv) _{4}^{8}\textrm{X}, \:_{5}^{8}\textrm{Y}

  • Option 1)

    i and ii

  • Option 2)

    ii and iii

  • Option 3)

    iii and iv

  • Option 4)

    i and iv 

 
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