Mastering Syllogism in Reasoning

Mastering Syllogism in Reasoning: 20 Unique Examples with Explanations

Syllogism is an essential part of logical reasoning in various competitive exams like SSC CGL, Bank PO, and UPSC. It involves deductive reasoning where conclusions are drawn from given statements. Understanding syllogism helps in developing analytical thinking and problem-solving skills. In this blog, we will explore 20 unique syllogism questions along with explanations.

Basic Rules of Syllogism

1. All A are B – This means every element of A belongs to B.

2. Some A are B – This means at least one A is B.

3. No A is B – This means A and B are completely separate.

4. Some A are not B – This means at least one A does not belong to B.

Let’s dive into the questions and their detailed explanations:

20 Unique Syllogism Questions

1. Basic Concept

Statements:

1. All cats are animals.

2. Some animals are dogs.
   Conclusion:
   A) Some cats are dogs.
   B) All dogs are animals.
   Answer: Only B follows. (Cats and dogs are not directly related.)

2. Negative Conclusion

Statements:

1. No apple is banana.

2. Some bananas are mangoes.
   Conclusion:
   A) No apple is mango.
   B) Some mangoes are not apples.
   Answer: Only B follows.

3. Possibility Case

Statements:

1. Some chairs are tables.

2. All tables are wood.
   Conclusion:
   A) Some chairs may be wood.
   B) All wood are tables.
   Answer: Only A follows.

4. Universal Affirmative

Statements:

1. All students are smart.

2. All smart people are intelligent.
   Conclusion:
   A) Some students are intelligent.
   B) All students are intelligent.
   Answer: Both A and B follow.

5. Reverse Statement

Statements:

1. Some roses are flowers.

2. All flowers are beautiful.
   Conclusion:
   A) Some beautiful things are roses.
   B) No rose is beautiful.
   Answer: Only A follows.

6. Contradictory Conclusion

Statements:

1. Some birds are sparrows.

2. No sparrow is crow.
   Conclusion:
   A) Some birds are not crows.
   B) Some crows are birds.
   Answer: Only A follows.

7. Inference-Based Question

Statements:

1. All men are human.

2. Some humans are scientists.
   Conclusion:
   A) Some scientists are men.
   B) All scientists are men.
   Answer: No conclusion follows.

8. Either-Or Case

Statements:

1. Some dogs are cats.

2. No cat is an elephant.
   Conclusion:
   A) Some dogs are not elephants.
   B) Some elephants are dogs.
   Answer: Either A or B follows.

9. Conflicting Statements

Statements:

1. All teachers are educated.

2. Some educated people are not teachers.
   Conclusion:
   A) Some teachers are not educated.
   B) Some educated people are teachers.
   Answer: Only B follows.

10. Real-Life Application

Statements:

1. All doctors are professionals.

2. Some professionals are engineers.
   Conclusion:
   A) Some engineers are doctors.
   B) Some professionals may be doctors.
   Answer: Only B follows.

11. Inclusion Relationship

Statements:

1. Some books are novels.

2. All novels are fiction.
   Conclusion:
   A) Some books are fiction.
   B) All fiction are books.
   Answer: Only A follows.

12. Direct Negation

Statements:

1. No fish is an amphibian.

2. Some amphibians are reptiles.
   Conclusion:
   A) Some fish are reptiles.
   B) No reptile is a fish.
   Answer: Only B follows.

13. Mutual Exclusivity

Statements:

1. Some actors are singers.

2. No singer is a dancer.
   Conclusion:
   A) Some actors are not dancers.
   B) All dancers are actors.
   Answer: Only A follows.

14. Combination Case

Statements:

1. All squares are rectangles.

2. Some rectangles are circles.
   Conclusion:
   A) Some squares are circles.
   B) All circles are rectangles.
   Answer: No conclusion follows.

15. Contradictory Possibility

Statements:

1. Some pencils are pens.

2. No pen is an eraser.
   Conclusion:
   A) Some pencils are not erasers.
   B) Some erasers are pencils.
   Answer: Only A follows.

16. Specific Grouping

Statements:

1. All scientists are researchers.

2. Some researchers are teachers.
   Conclusion:
   A) Some scientists may be teachers.
   B) No teacher is a scientist.
   Answer: Only A follows.

17. Hierarchical Classification

Statements:

1. Some mammals are carnivores.

2. All carnivores are animals.
   Conclusion:
   A) Some mammals are animals.
   B) Some animals are mammals.
   Answer: Both A and B follow.

18. Abstract Relationship

Statements:

1. Some beliefs are traditions.

2. All traditions are cultural.
   Conclusion:
   A) Some beliefs are cultural.
   B) All cultural things are traditions.
   Answer: Only A follows.

19. Definite vs Indefinite

Statements:

1. No flower is a tree.

2. Some trees are plants.
   Conclusion:
   A) No flower is a plant.
   B) Some plants are trees.
   Answer: Only B follows.

20. Overlapping Sets

Statements:

1. All squares are polygons.

2. Some polygons are triangles.
   Conclusion:
   A) Some triangles are squares.
   B) Some polygons are squares.
   Answer: Only B follows.

Conclusion

Understanding syllogism is key to scoring well in reasoning sections of competitive exams. By practicing different types of syllogism questions, you can improve your logical thinking and enhance problem-solving abilities. Keep practicing and master the art of logical deduction!

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Calendar Problems in Reasoning: 20 Unique Questions with Solutions

The calendar is an essential topic in logical reasoning, especially for competitive exams like SSC CGL, Banking, and UPSC. It includes questions related to finding the day of the week, leap years, odd days, and other calendar-based problems. In this blog, we will explore 20 unique calendar problems with detailed explanations to help you master this topic.

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1. Finding the Day of a Given Date

Question: What was the day of the week on 15th August 1947?

Solution:

• Base year: 1900 (1st Jan 1900 was a Monday)

• Counting odd days:

  • 1900-1947 = 47 years (11 leap years + 36 normal years) → (11×2 + 36×1) = 58 odd days → 58 ÷ 7 = 2 odd days

  • January to August 15th, 1947 → 3 + 0 + 3 + 2 + 3 + 2 + 3 + 5 = 21 odd days → 21 ÷ 7 = 0 odd days

  • Total odd days = 2

  • Monday + 2 days = Friday

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2. Leap Year Identification

Question: Which of the following is a leap year? A) 1700
B) 1800
C) 1900
D) 2000

Solution: A year is a leap year if it is divisible by 4, but if it is a century year, it must be divisible by 400. Only 2000 is a leap year.

Answer: D) 2000

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3. Finding Odd Days

Question: How many odd days are there in 400 years?

Solution:

• 1 normal year = 365 days = 1 odd day

• 1 leap year = 366 days = 2 odd days

• 400 years = 100 leap years + 300 normal years

• Odd days = (100×2 + 300×1) = 200 + 300 = 500 days

• 500 ÷ 7 = 0 odd days

Answer: 0 odd days

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4. Next Same Calendar Year

Question: When will the calendar of 2024 repeat?

Solution:

• Leap years repeat after 28 years.

• The next year with the same calendar as 2024 will be 2052.

Answer: 2052

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5. Previous Same Calendar Year

Question: Which year had the same calendar as 2023?

Solution:

• Normal years repeat after 6 or 11 years.

• The last same calendar was in 2017.

Answer: 2017

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6. First Day of Any Century

Question: What is the first day of the 21st century (1st January 2001)?

Solution:

• 1st Jan 1901 was Tuesday.

• 100 years = 0 odd days.

• So, 1st Jan 2001 is also Monday.

Answer: Monday

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7. Day Calculation with Odd Days

Question: If today is Wednesday, what day will it be after 60 days?

Solution:

• 60 ÷ 7 = 4 odd days.

• Wednesday + 4 days = Sunday

Answer: Sunday

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8. Last Day of a Century

Question: What is the last day of the 20th century (31st December 2000)?

Solution:

• 31st Dec 1899 was Sunday.

• 100 years = 0 odd days.

• So, 31st Dec 2000 is also Sunday.

Answer: Sunday

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9. Repeating Calendar within a Decade

Question: When will the calendar of 2022 repeat?

Solution:

• 2022 is a normal year, and such years repeat after 6 or 11 years.

• The next similar calendar will be in 2033.

Answer: 2033

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10. Counting Fridays in a Year

Question: How many Fridays are there in a leap year?

Solution:

• 366 days = 52 weeks + 2 extra days.

• These extra days can make 53 Fridays if the year starts on a Thursday or Friday.

Answer: 52 or 53

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11. Odd Days in a Month

Question: How many odd days are there in February of a leap year?

Solution:

• February (leap year) = 29 days

• 29 ÷ 7 = 1 odd day

Answer: 1 odd day

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12. Identifying a Leap Century Year

Question: Which of the following is a leap century year? A) 1600
B) 1700
C) 1800
D) 1900

Solution: Only centuries divisible by 400 are leap years.

Answer: A) 1600

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13. Number of Sundays in a Year

Question: How many Sundays are there in a normal year?

Solution:

• 365 days = 52 weeks + 1 extra day

• Extra day can be any weekday, so Sundays remain 52 or 53.

Answer: 52 or 53

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14. Checking Leap Year

Question: Is the year 2028 a leap year?

Solution:

• 2028 ÷ 4 = yes, so it’s a leap year.

Answer: Yes

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15. 1000th Day from Today

Question: If today is Monday, what day will it be after 1000 days?

Solution:

• 1000 ÷ 7 = 6 odd days

• Monday + 6 = Sunday

Answer: Sunday

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16. Month Having Same Calendar as August

Question: Which month has the same calendar as August in a non-leap year?

Answer: February

---

17. Counting Leap Years Between Two Dates

Question: How many leap years are there between 1901 and 2000?

Solution:

• Leap years: 1904, 1908, ..., 2000

• Total = 25

Answer: 25

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18. 365th Day from Today

Question: If today is Monday, what will be the day after 365 days?

Solution:

• 365 ÷ 7 = 1 odd day

• Monday + 1 = Tuesday

Answer: Tuesday

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19. Century Not a Leap Year

Question: Which of the following is not a leap year? A) 1800
B) 2000
C) 1600
D) 2400

Answer: A) 1800

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20. Calendar Repetition Rule

Question: After how many years does a leap year repeat?

Solution: 28 years.

Answer: 28 years

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These 20 questions cover all important concepts of calendar problems in reasoning. Keep practicing to master this topic!

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Clock-Based Reasoning Questions for Competitive Exams

Clock-based reasoning is a crucial topic in logical reasoning for exams like SSC CGL, Banking, Railway, and other competitive exams. These questions test a candidate’s ability to understand the movement of the clock’s hands and calculate angles, time loss or gain, mirror images, etc. Here, we present 20 unique multiple-choice questions (MCQs) along with their explanations.

Clock-Based Reasoning Questions for Competitive Exams

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1. Angle Between Hands of the Clock

Q1. What is the angle between the hands of the clock at 4:20? 

A) 100°
B) 110°
C) 120°
D) 140°

Answer: B) 110°
Explanation: Formula:
= 110°

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2. Overlapping Hands

Q2. At what time between 2:00 and 3:00 do the hour and minute hands coincide? 

A) 2:10
B) 2:20
C) 2:21.8
D) 2:30

Answer: C) 2:21.8
Explanation: The hands coincide at: minutes

= 21.8 minutes

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3. Reflex Angle Calculation

Q3. Find the reflex angle between the hands of the clock at 5:40. 

A) 100°
B) 200°
C) 220°
D) 240°

Answer: C) 220°
Explanation: Reflex angle = 360° - smaller angle
Smaller angle = 120° (using formula)
So, 220° is the reflex angle.

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4. Mirror Image of Clock Time

Q4. What will be the mirror image of 5:15? 

A) 6:45
B) 7:45
C) 6:15
D) 6:50 

Answer: A) 6:45
Explanation: Mirror image formula = 6:45

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5. Time Between Two Same Positions

Q5. How many times do the hands of the clock coincide in 12 hours? 

A) 10
B) 11
C) 12
D) 24

Answer: B) 11
Explanation: Hands coincide 11 times in 12 hours.

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6. Hands in a Straight Line

Q6. At what time between 1:00 and 2:00 do the hands of the clock form a straight line? A) 1:30
B) 1:32.73
C) 1:40
D) 1:50

Answer: B) 1:32.73
Explanation: Formula:

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7. Hands at Right Angle

Q7. At what time between 3:00 and 4:00 do the hands form a right angle? 

A) 3:15
B) 3:20
C) 3:27.3
D) 3:40

Answer: C) 3:27.3
Explanation: Formula:

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8. Fast and Slow Clocks

Q8. A clock gains 5 minutes every hour. How much will it gain in a day? 

A) 60 minutes
B) 100 minutes
C) 120 minutes
D) 150 minutes

Answer: C) 120 minutes
Explanation: 5 minutes/hour × 24 hours = 120 minutes

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9. Time Loss Calculation

Q9. A clock loses 2 minutes every 6 hours. How much time will it lose in a week? 

A) 24 minutes
B) 48 minutes
C) 56 minutes
D) 56 hours

Answer: C) 56 minutes
Explanation: 2 minutes × (24/6) × 7 days = 56 minutes

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10. Identifying Wrong Clock Time

Q10. A clock shows 9:15 but the actual time is 9:30. How much does the clock lose in a day if it continues at the same rate? 

A) 30 minutes
B) 60 minutes
C) 120 minutes
D) 150 minutes

Answer:* B) 60 minutes*
Explanation:* 15 minutes in 6 hours means 60 minutes in 24 hours.

11. Angle Between Hands of the Clock

Question: What is the angle between the hour and minute hand at 4:20?

Options:
A) 120°
B) 110°
C) 130°
D) 105°

Answer: B) 110°
Explanation: The angle formula is:

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12. Right Angle Time Calculation

Question: At what time between 3:00 and 4:00 will the hands form a right angle (90°)?

Options:
A) 3:15
B) 3:30
C) 3:40
D) 3:22

Answer: D) 3:22
Explanation: The formula for when hands are at 90°: So, the answer is 3:22.

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13. Coincidence of Hands

Question: At what time between 2:00 and 3:00 do the hands coincide?

Options:
A) 2:05
B) 2:10
C) 2:11
D) 2:12

Answer: C) 2:11
Explanation: The formula for coincidence is: So, the hands coincide at 2:11.

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14. Reflex Angle Calculation

Question: Find the reflex angle between the hands at 7:45.

Options:
A) 150°
B) 170°
C) 200°
D) 220°

Answer: D) 220°
Explanation: The smaller angle: The reflex angle = 360° - 37.5° = 220°.

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15. When Are the Hands Opposite to Each Other?

Question: At what time between 6:00 and 7:00 are the hands opposite to each other?

Options:
A) 6:40
B) 6:50
C) 6:55
D) 6:45

Answer: A) 6:40
Explanation: Hands are opposite when the angle is 180°. So, the time is 6:40.

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16. Overlapping of Hands

Question: Between 8:00 and 9:00, when will the hands overlap?

Options:
A) 8:36
B) 8:40
C) 8:43
D) 8:45

Answer: C) 8:43
Explanation: Hands overlap when: So, hands overlap at 8:43.

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17. Hands Forming 120°

Question: At what time between 5:00 and 6:00 do the hands form an angle of 120°?

Options:
A) 5:10
B) 5:20
C) 5:30
D) 5:40

Answer: B) 5:20
Explanation: So, the answer is 5:20.

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18. Hands Being 45° Apart

Question: At what time between 10:00 and 11:00 do the hands form a 45° angle?

Options:
A) 10:05
B) 10:10
C) 10:15
D) 10:20

Answer: B) 10:10
Explanation: Using the formula: So, the answer is 10:10.

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19. Angle at 12:15

Question: Find the angle between the hands at 12:15.

Options:
A) 75°
B) 90°
C) 82.5°
D) 97.5°

Answer: C) 82.5°
Explanation:

20. Clock Reflection

Question: If a clock shows 9:10, what will be its mirror image?

Options:
A) 2:50
B) 3:50
C) 3:40
D) 2:40

Answer: A) 2:50
Explanation: Mirror image formula:

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Conclusion

Clock-based reasoning is a crucial topic in SSC CGL and other competitive exams. Mastering these questions through formulas and practice will improve accuracy and speed in the exam. Keep practicing and applying these techniques to enhance your problem-solving skills!

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Seating Arrangement Questions for SSC CGL with solutions

Seating arrangement is an important topic in logical reasoning for competitive  exams like SSC CGL. These questions test a candidate’s ability to analyse and arrange people or objects according to given conditions. Below are 10 unique seating arrangement questions along with detailed explanations and solutions.

1. Circular Arrangement

Question: Six friends A, B, C, D, E, and F are sitting in a circle facing the center. B is to the immediate right of C. A is third to the left of B. E is between A and F. Who is sitting opposite to C?

Detailed Solution:

1. Since all are facing the center, the right and left directions remain the same.

2. B is to the immediate right of C, so placing them accordingly.

3. A is third to the left of B, which helps in fixing A’s position.

4. E is between A and F.

5. The final arrangement reveals that E is opposite to C.

Circular Arrangement Question

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2. Linear Arrangement (Single Row)

Question: Seven students P, Q, R, S, T, U, and V are sitting in a row facing north. Q is sitting third from the left. P is sitting at one of the extreme ends. U is sitting exactly between T and S. Who is sitting at the rightmost end?

Detailed Solution:

1. Placing Q in the third position from the left.

2. P is at one of the extreme ends.

3. U is between T and S, which helps in their positioning.

4. After arranging all, V is found at the rightmost end.

Linear Arrangement

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3. Linear Arrangement (Double Row)

Question: Eight persons are sitting in two parallel rows, facing each other. A, B, C, and D are sitting in row 1, facing south. P, Q, R, and S are sitting in row 2, facing north. C is sitting at one of the extreme ends. B is sitting second to the left of C. The person facing B is Q. Who is sitting opposite to A?

Detailed Solution:

1. Arranging row 1 with A, B, C, and D facing south.

2. Placing row 2’s persons opposite to them.

3. C is at the extreme end and B is second to the left.

4. The arrangement shows that R is opposite to A.

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4. Circular Arrangement (Facing Outwards)

Question: Five persons L, M, N, O, and P are sitting in a circle, facing outward. M is to the immediate left of O. P is second to the right of L. Who is sitting exactly between L and O?

Detailed Solution:

1. Since they are facing outward, left and right directions are reversed.

2. Placing M to the immediate left of O.

3. Placing P second to the right of L.

4. The final arrangement shows that N is between L and O.

Circular Seating Arrangement 

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5. Square Seating Arrangement

Question: Eight friends A, B, C, D, E, F, G, and H are sitting at the corners and middle of the sides of a square table. The ones sitting at the corners are facing the center, while the ones sitting at the sides are facing outward. E is sitting to the immediate right of H, who is sitting at a corner. Who is sitting opposite to H?

Detailed Solution:

1. Placing H at a corner.

2. E is to H’s immediate right.

3. Arranging others accordingly.

4. C is found sitting opposite to H.

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6. Mixed Arrangement

Question: Six persons J, K, L, M, N, and O are sitting in a row. Some are facing north while some are facing south. J is sitting at an extreme end. L is sitting second to the left of M. N is sitting to the immediate right of O, who is facing north. Who is sitting exactly in the middle?

Detailed Solution:

1. J is at an extreme end.

2. L is positioned second to the left of M.

3. N is immediately right of O.

4. M is found to be exactly in the middle.

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7. Linear Arrangement with Conditions

Question: Five people A, B, C, D, and E are sitting in a row facing north. A is sitting to the left of B but not next to C. D is sitting at an extreme end. Who is sitting second to the right of A?

Seating Arrangement Questions

Detailed Solution:

1. Placing D at an extreme end.

2. Arranging A to the left of B but not next to C.

3. The final arrangement shows that C is second to the right of A.

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8. Triangular Arrangement

Question: Seven people are sitting around a triangular table. Three sit at the corners, and four sit along the sides. P is sitting at a corner. R is sitting to the immediate right of Q. S is sitting opposite to P. Who is sitting to the left of S?

Detailed Solution:

1. Placing P at a corner.

2. R is immediately right of Q.

3. Arranging others accordingly.

4. T is found sitting to the left of S.

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9. Rectangular Seating Arrangement

Question: Ten people are sitting around a rectangular table, with four sitting at the corners and six sitting along the sides. X is sitting to the immediate left of Y. The person opposite to X is Z. Who is sitting diagonally opposite to Z?

Detailed Solution:

1. Arranging four people at corners and six along the sides.

2. X is to the left of Y.

3. Finding who sits opposite and diagonal to Z.

4. W is diagonally opposite to Z.

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10. Row Arrangement with Height Order

Question: Five boys A, B, C, D, and E are sitting in a row according to their heights. A is taller than C but shorter than B. D is shorter than C but taller than E. Who is the second tallest?

Detailed Solution:

1. Arranging them based on height order.

2. A is taller than C but shorter than B.

3. D is shorter than C but taller than E.

4. A is found to be the second tallest.

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Conclusion

Seating arrangement questions require careful reading and logical thinking. The best way to solve these is by drawing diagrams and arranging positions step by step. Regular practice will help in quick solving during exams like SSC CGL. Developing a habit of noting key points and testing different positions logically will boost confidence and accuracy in solving such questions efficiently.

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Logical Reasoning for SSC CGL: Important Questions and Explanations

 Logical reasoning is an essential part of the SSC CGL exam, testing a candidate's ability to analyze patterns, understand relationships, and solve problems efficiently. Here, we will discuss some important logical reasoning questions along with simple explanations to help you prepare effectively.

1. Number Series

Question: Find the missing number: 2, 6, 12, 20, ?, 42
Options:
A) 30
B) 28
C) 32
D) 36

Answer: B) 30
Explanation: The pattern follows +4, +6, +8, +10... So, 20 + 10 = 30

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2. Blood Relation

Question: Pointing to a boy, Rita said, "He is the son of my grandfather’s only son." How is the boy related to Rita?
Options:
A) Brother
B) Cousin
C) Nephew
D) Uncle

Answer: A) Brother
Explanation: Grandfather’s only son is Rita’s father. His son is her brother.

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3. Odd One Out

Question: Find the odd one out:
Options:
A) Apple
B) Mango
C) Potato
D) Banana

Answer: C) Potato
Explanation: All others are fruits, but potato is a vegetable.

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4. Direction Sense

Question: A man walks 10m North, turns right and walks 5m, then turns right again and walks 10m. In which direction is he now facing?
Options:
A) East
B) West
C) North
D) South

Answer: B) West
Explanation: He initially faced North, then East, then South. Now he faces West.

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5. Coding-Decoding

Question: If CAT is coded as 3120, how will DOG be coded?
Options:
A) 4150
B) 4815
C) 4740
D) 4950

Answer: C) 4740
Explanation: Each letter’s position in the alphabet is multiplied:
C (3 × 1), A (1 × 2), T (20 × 1) → 3120
D (4 × 1), O (15 × 2), G (7 × 1) → 4740

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6. Syllogism

Question: Statements:

1. All cats are animals.

2. Some animals are wild.

Conclusion:
A) Some cats are wild.
B) All animals are cats.

Options:
A) Only A follows
B) Only B follows
C) Both follow
D) None follow

Answer: D) None follow
Explanation: The given statements do not guarantee that some cats are wild or that all animals are cats.

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7. Mirror Image

Question: What will be the mirror image of 1234 in a vertical mirror?
Options:
A) 4321
B) 2143
C) 1234
D) 3412

Answer: A) 4321
Explanation: The order of digits reverses in a mirror image.

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8. Calendar Problem

Question: If 15th August 2023 was a Tuesday, what day will 15th August 2024 be?
Options:
A) Wednesday
B) Thursday
C) Friday
D) Saturday

Answer: B) Thursday
Explanation: 2024 is a leap year, so the extra day shifts it by 2 days.

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9. Puzzle-Based Question

Question: A is the father of B, but B is not the son of A. Who is B?
Options:
A) Brother
B) Daughter
C) Uncle
D) Cousin

Answer: B) Daughter
Explanation: If B is not the son, then B must be the daughter.

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10. Clock-Based Question

Question: What is the angle between the hands of the clock at 3:30?
Options:
A) 75°
B) 90°
C) 105°
D) 120°

Answer: C) 105°
Explanation: The formula is |(30×H - 5.5×M)|
= |(30×3 - 5.5×30)| = |90 - 82.5| = 105°

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Conclusion

Logical reasoning is an important section of the SSC CGL exam, and practicing different types of questions will help improve problem-solving skills. By understanding these fundamental concepts and practicing regularly, candidates can enhance their reasoning abilities and score well in the exam. Keep practicing and stay confident!

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Most Important circle Problems of Reasoning for different competitive Exams

Problem   1

----- Explanation   ----  

(11 × 7) - (5 × 3) = 77 - 15 = 62 

(18 × 3) - (5 × 4) = 54 - 20 = 34

(9 × 6) - (4 × 8) = 54 - 32 = 22    

 Problem  2

 ----- Explanation   ----  

2² + 5² + 3² + 7² = 4 + 25 + 9 + 49 = 87

8² + 5² + 4² + 3² = 64 + 25 + 16 + 9 = 114

9² + 4² + 1² + 6² = 81 + 16 + 1 + 36 = 134

  Problem   3

 ----- Explanation   ----  

 2 x (5 + 7 + 3) = 2 x 15 = 30 

8 x (5 + 4 + 3) = 8 x 12 = 96 

3 x (4 + 9 + 6) =  3 x 19 = 57  

 Problem   4

 ----- Explanation   ----  

 1² + (5 × 3 × 7) =1 + 105 = 106

4² + (5 × 3 × 6) =16 + 90 = 106

3² + (3 × 2 × 7) = 9 + 42 = 51 

 Problem 5

----- Explanation   ----  

   1² + (5 + 3 + 7) = 1 + 15 = 16

   4² + (5 + 3 + 6) = 16 + 14 = 30

   3² + (3 + 2 + 7)  = 9 + 12 = 21

 Problem 6

----- Explanation   ----  

(2 + 5)² + (7 - 3)² = 7² +4² =  49 + 16 = 65

(1 + 5)² + (6 - 3)² = 6² + 3² = 36 + 9 = 45 = ?

(2 + 3)² + (7 - 2)² = 5² + 5² = 25 + 25 = 50    

  Problem 7

----- Explanation   ----  

(1+2+3+4)² = 10² = 100

(2+3+4+5)² = 14² = 196

(4+5+6+7)² = 22² = 484 

 Problem 8

----- Explanation   ----  

LCM Of  1, 4, 3 and 2 = 12

LCM Of 3, 2, 5 and 4 = 60

LCM Of  4 , 5 ,6 and 2 = 60 = ? 

  Problem 9

----- Explanation   ----  

 19 + 16 + 15 + 12 = 62 ⇔26

23 + 22 + 27 + 19 = 91 ⇔19

24 + 23 + 20 + 22 = 89 ⇔98 

Problem 10

----- Explanation   ----  

√81+√16+√256+√25 = 9+4+16+5 = 34

√9+√49+√1+√64 = 3+7+1+8 = 19

√4+√36+√121+√289 = 2+6+11+7 = 26 = ?   

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Ten important Questions of Reasoning series completion for competitive exams

Formula:- 

Any Term = n² × (n-1), where 'n' is natural number less than equal to 6

1st Term = 1² × (0) = 1 × 0 = 0

2nd Term = 2² × (1) = 4 × 1 = 4

3rd Term = 3² × (2) = 9 × 2 = 18

4th Term = 4² × (3) = 16 × 3 = 48

5th Term = 5² × (4) = 25 × 4 = 100

6th Term = 6² × (5) = 36 × 5 = 180

Hence Option (4)100 is correct.

Formula:-

Any Term =  Previous Term + ( sum of digits of Previous Term)

1st Term =17

2nd Term =17 + (1+7) = 17+ 8 = 25
3rd Term = 25 + (2+5) = 25 + 7 = 32

4th Term =32 + (3+2) = 32 + 5 = 37

5th Term =37 + (3+7) = 37+ 10 = 47

6th Term =47 + (4+7) = 47+ 11 = 58 = Question mark

Hence Option (2)58 is correct.

Formula:-

Difference of any two  consecutive terms = 4 +  Difference of next two consecutive terms

Difference of 2nd term and 1st term = 7 - 5 = 2
Difference of 3rd term and 2nd term = 13 - 7 = 6

Difference of 4th term and 3rd term = 23 - 13 = 10

Difference of 5th term and 4th term = 37 - 23 = 14

Difference of 6th term and 5th term = 55 - 37 = 18

Difference of 7th term and 6th term = ? - 55 = must be 4 greater than previous difference i.e. 18 + 4 = 22

? - 55 = 22

? = 22 + 55

? = 77
Hence Option (3)77 is correct.

1st term = 0.15 × 2 = 0.30 = 0.3

2nd  Term = 0.3 × 2 =  0.60 = 0.6

Also 5th Term =  1.2  × 2 = 2.4

Similarly 3rd =  0.3  × 2 = 0.6= 0.60

4th Term = 0.6 × 2 = 1.2

5th Term = 1.2 × 2 = 2.4 

Hence Option (1)0.6 is correct.

3rd term = 1 + 1 = 2

4th Term = 1 + 2 = 3

5thTerm = 2 + 3 = 5

6thTerm = 3 + 5 = 8

7th Term = 5 + 8 = 13

8th Term = 8 + 13 = 21

missing Term = 13 + 21 = ?

                           ? = 34

Hence Option 12)34 is correct.

Formula:-

Any Term = Square of Previous Term + 1

1st Term = 3

2nd  Term = (1st Term)² + 1

2nd Term =3² + 1 = 9 + 1 = 10

3rd  Term = (2nd Term)² + 1

3rd Term = 10² + 1 = 100 + 1 = 101

4th  Term = (3rd Term)² + 1

4th Term = 101² + 1 = 10201 + 1 = 10202 = The value of question mark

3² +1 , 10² + 1 , 101² + 1

Hence Option (1)19202 is correct.

Formula:-

Any Term = Twice of Previous Term + Rank of previous term

1st Term = 13

2nd Term = (13 × 2) + 1

3rd Term = (27 × 2) + 2

4th Term = (56 × 2) + 3

5th Term = (115 × 2) + 4 = 230 + 4 = 234

Hence Option (4)234 s correct.

Formula:-

Any Term = Four times of Previous Term - 1

2nd  Term = (2 × 4) - 1 = 7

3rd Term = (7 × 4) - 1 = 27

4th Term = (27 × 4) - 1 = 107

                = (107 × 4) - 1 

                = 428 - 1 = 427 

Hence Option (2)427 s correct.

Formula:-

Any Term = Twice of Previous Term  1 

2nd  Term = (2 × 2) + 1 = 4 + 1 = 5

3rd Term = (5 × 2) - 1 = 10 - 1 = 9

4th Term = (9 × 2) + 1 = 18 + 1 = 19

(19 × 2) - 1 = 38 - 1 = 37

(37 × 2) + 1 = 74 + 1 = 75

Hence Option (2)75 s correct.

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Ten Logical Reasoning questions and answers for competitive exams

Ten Logical Reasoning questions and answers for competitive  exams  like Bank PO, Bank clerk, SSC CGL, ssc chsl, RRB NTPC , group D etc have been discussed in this post in very easy style and shortcut way .

Logical Reasoning questions and answers for competitive Exams 

Problem # 1

In this reasoning problem 1st number (2) is associated to 10 with the help of any rule , in the same rule we have to associate 8  to a number out of four given options.

Look carefully the given numbers consists of three digits. These three digits can be utilised with the help a formula given below.

Formula :- 

2nd Number =  a² + (2×a) + 2, where 'a' is 1st number. 

1st Number

2² + (2×2) + 2 = 4 + 4 + 2 = 10

2nd Number

8² + (2×8) + 2 = 64 + 16 + 2 = 82

Therefore option (1)2 is correct option.

Problem # 2

In this reasoning problem 1st number (3) is associated to 39 with the help of any rule , in the same rule we have to associate 6 to a number out of four given options.

Look carefully the given numbers consists of three digits. These three digits can be utilised with the help a formula given below.

Formula :- 

2nd Number = a³ + (3×a) + a, where 'a' is 1st number.  

1st Number

3³ + (3×3) + 3 = 27 + 9 + 3 = 39

2nd Number

6³ + (3×6) + 3 = 216 + 18 + 6 = 240.  

Therefore option (1)240 is correct option.

Problem # 3

In this reasoning problem 1st number (4) is associated to 60 with the help of any rule , in the same rule we have to associate 9 to any one  number out of four given options.

Look carefully the given numbers consists of three digits. These three digits can be utilised with the help a formula given below.

Formula :- 

2nd Number =  (a-1) × (a) × (a + 1), where 'a' is 1st number.  

1st Number

3 × 4 × 5 = 60

2nd Number

8 × 9 × 10 = 720

Therefore option (4)720 is correct option.

Problem # 4

In this problem of reasoning we have to combine two given  numbers in such a way that after applying any mathematical operation/operations we could get the number in the right hand side of all the three problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.

Formula : - 

(a×b) + (a+b),

where 'a' and 'b' are two given numbers.

1st Number

(27×4) + (27+4) = 108 + 31 = 139

2nd Number

(31×9) + (31+9) = 279 + 40 = 319

3rd Number

(21×6) + (21+6) = 126 + 27 = 133

Therefore option (1)153 is correct option.

Problem # 5

In this problem of reasoning we have to combine all the three given  numbers in such a way that after applying any mathematical operation/operations we could get the number on the right hand side of all the four problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.

Formula : - 

(a×b×c)  + (a+b+c),

where 'a' , 'b' and 'c' are three given numbers.

1st Number

(5×2×3) + (5+2+3) = 30 + 10 = 40

2nd Number

(6×3×4) + (6+3+4) = 72 + 13 = 85

3rd Number

(4×1×8) + (4+1+8) = 32 + 13 = 45

Required Number

(3×2×8) + (3+2+8) = 48 + 13 = 61

Therefore option (4)61 is correct option.

Problem # 6

1st Method

Formula : - 

Any term = ( 2 × Previous term ) + 1

1st Term = 6

2nd Term = ( 2 × 6 ) + 1 = 12 + 1 = 13

3rd Term = ( 2 × 3 ) + 1 = 26 + 1 = 27

4th Term = ( 2 × 27 ) + 1 = 54 + 1 = 55

5th Term = ( 2 × 55 ) + 1 = 110 + 1 = 111

6th Term = ( 2 × 111 ) + 1 = 222 + 1 = 223

2nd Method

Formula : -

 Any term - Preceding Term =  Difference is multiple of 7 

13 - 6 = 7

27 - 13 = 14 = 2 × 7 = Double of 7

55 - 27 = 28 = 2 × 14 = Double of 14

111 - 55 = 56 = 2 × 28 = Double of 14

? - 111 =  It must be  2 × 56 = 112

 Hence    

? - 111 =  112

? = 112 + 111

? =  223

Therefore option (1)223 is correct option.

Problem # 7

Formula : - 

Any term =  one less or one more than the square of natural numbers greater than equal to 6 and smaller than equal to 11.

35 = 6² - 1

50 = 7² + 1

63 = 8² - 1

82 = 9² + 1

?  = 10² - 1 = 100 - 1 = 99

122 = 11² + 1

Therefore option (4)99 is correct option.

Problem # 8

 Split this number series into two series by picking alternate numbers.

13, 19 , 25  and 11, 17, 23 , ? 

1st series

13, 19 , 25 

Now considering the 2nd series. In this series every term can be found by adding 6 to its previous term.

1st Term = 13

2nd Term = 1st Term + 6 = 13  + 6 = 19

3rd Term = 2nd Term + 6 = 19  + 6 = 25

2nd series

11, 17, 23 , ?

Now considering the 1st series. In this series every term can be found by adding 6 to its previous term.

1st Term = 11

2nd Term = 1st Term + 6 = 11  + 6 = 17

3rd Term = 2nd Term + 6 = 17  + 6 = 23

4th Term = 3rd Term + 6 = 23  + 6 = 29 = ? (The value of question mark).

Therefore option (3) 29 is correct option.

In this reasoning problem three out of four options have been calculated by Multiplying  the 1st number with 12 to get 2nd number. This rule is applicable to only three out of four option. And one which do not follow this rule will be the correct option.

10 × 12 = 120

20 × 12 = 240

14 × 12 = 168(196)

12 × 12 = 120

Therefore option (3)14,196 is odd option.

In this problem of reasoning we have to combine both the given  numbers in such a way that after applying any mathematical operation/operations we could get the number on the right hand side of all the three problems given above. Because in these types of reasoning problems we can change mathematical sign according to our requirements.

Formula : - 

(a×b) - (a+b),

where 'a' and 'b' are three given numbers.

1st Number

(27×4) - (27+ 4) = 108 - 31 = 77

2nd Number

(31×9) - (31+9) = 279 - 40 = 239

3rd Number

(21×6) - (21+6) = 126 - 27 = 99

Therefore option (3)99 is correct option.

Conclusion

So these were the ten  problems regarding the post Ten Logical Reasoning questions and answers pdf for competitive exams like SSC CGL ,SSC CHSL ,CPO ,Bank exams and RRB NTPC etc which were explained in this post. Feel free to comment your valuable suggestions.

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