Reasoning Quiz for Indian Bank PO Mains : 16th October 2018

Dear Aspirants,

Reasoning Questions for Indian Bank PO Mains

Reasoning Ability is an onerous section. With the increasing complexity of questions, it becomes hard for one to give it the cold shoulder. The only way to make the grade in this particular section in the forthcoming banking exams like Indian Bank PO Mains is to practice continuously with all your heart and soul. And, to let you practice with the best of the latest pattern questions, here is the Adda247 Reasoning Quiz based on the exact same pattern of questions that are being asked in the exams.

Directions (1-5): Study the information carefully and answer the questions given below 

In a zoo there are six spots A, B, C, D, E, F which have inhabited different animals i.e. lion, tiger, chimpanzee, monkey, giraffe and bear and in different number 2, 3, 4, 5, 6, 10. The zoo are divided equally in two categories α and β.
E has giraffe. β category zoo does not has chimpanzee. C has the 2nd least number of animals. F is not α category zoo. C and E are in same category. The number of chimpanzee which is not an odd number, is equal to the sum of the number of lion and bear. Monkeys are in highest number. B and D are in same category. C and F do not belong to same category. B has 4 animals. A and the zoo which has giraffe belong to same category. The number of lion is more than the number of tiger. A and D does not have lion and bear.

Q1. What is the number of giraffe in zoo E?

Solution:

E has giraffe. C has the 2nd least number of animals. F is not α category zoo So, F is β. C and F do not belong to same category, so C is α. B has 4 animals

The number of chimpanzee, which is not an odd number, is equal to the sum of the number of lion and bear. Monkeys are in highest number. So only one possibility is there, Lion/bear- 2/4 and chimpanzee-6. So, either A can have monkey or D. A and D does not has lion and bear. So in case 1 D has chimpanzee and in case2 A has chimpanzee. In both the cases C has tiger and E has 5 giraffe. Case1-

Case2-

C and E are in same category. A and the zoo which has giraffe belong to same category. The number of lion is more than the number of tiger. B and D are in same category. So, lion are 4 in number. Case1-

Case2-

β category zoo does not has chimpanzee. So, case1 gets eliminated. The final arrangement is:

Q2. In which of the following zoo is tiger inhabited?

Solution:

E has giraffe. C has the 2nd least number of animals. F is not α category zoo So, F is β. C and F do not belong to same category, so C is α. B has 4 animals

The number of chimpanzee, which is not an odd number, is equal to the sum of the number of lion and bear. Monkeys are in highest number. So only one possibility is there, Lion/bear- 2/4 and chimpanzee-6. So, either A can have monkey or D. A and D does not has lion and bear. So in case 1 D has chimpanzee and in case2 A has chimpanzee. In both the cases C has tiger and E has 5 giraffe. Case1-

Case2-

C and E are in same category. A and the zoo which has giraffe belong to same category. The number of lion is more than the number of tiger. B and D are in same category. So, lion are 4 in number. Case1-

Case2-

β category zoo does not has chimpanzee. So, case1 gets eliminated. The final arrangement is:

Q3. Which of the following combination is true?

Solution:

E has giraffe. C has the 2nd least number of animals. F is not α category zoo So, F is β. C and F do not belong to same category, so C is α. B has 4 animals

The number of chimpanzee, which is not an odd number, is equal to the sum of the number of lion and bear. Monkeys are in highest number. So only one possibility is there, Lion/bear- 2/4 and chimpanzee-6. So, either A can have monkey or D. A and D does not has lion and bear. So in case 1 D has chimpanzee and in case2 A has chimpanzee. In both the cases C has tiger and E has 5 giraffe. Case1-

Case2-

C and E are in same category. A and the zoo which has giraffe belong to same category. The number of lion is more than the number of tiger. B and D are in same category. So, lion are 4 in number. Case1-

Case2-

β category zoo does not has chimpanzee. So, case1 gets eliminated. The final arrangement is:

Q4. Four of the following belongs to a group find the one that does not belong to that group?

Solution:

E has giraffe. C has the 2nd least number of animals. F is not α category zoo So, F is β. C and F do not belong to same category, so C is α. B has 4 animals

The number of chimpanzee, which is not an odd number, is equal to the sum of the number of lion and bear. Monkeys are in highest number. So only one possibility is there, Lion/bear- 2/4 and chimpanzee-6. So, either A can have monkey or D. A and D does not has lion and bear. So in case 1 D has chimpanzee and in case2 A has chimpanzee. In both the cases C has tiger and E has 5 giraffe. Case1-

Case2-

C and E are in same category. A and the zoo which has giraffe belong to same category. The number of lion is more than the number of tiger. B and D are in same category. So, lion are 4 in number. Case1-

Case2-

β category zoo does not has chimpanzee. So, case1 gets eliminated. The final arrangement is:

Q5. Which of the following animal is inhabited in zoo A?

Solution:

E has giraffe. C has the 2nd least number of animals. F is not α category zoo So, F is β. C and F do not belong to same category, so C is α. B has 4 animals

The number of chimpanzee, which is not an odd number, is equal to the sum of the number of lion and bear. Monkeys are in highest number. So only one possibility is there, Lion/bear- 2/4 and chimpanzee-6. So, either A can have monkey or D. A and D does not has lion and bear. So in case 1 D has chimpanzee and in case2 A has chimpanzee. In both the cases C has tiger and E has 5 giraffe. Case1-

Case2-

C and E are in same category. A and the zoo which has giraffe belong to same category. The number of lion is more than the number of tiger. B and D are in same category. So, lion are 4 in number. Case1-

Case2-

β category zoo does not has chimpanzee. So, case1 gets eliminated. The final arrangement is:

Directions (6-7): Study the following information carefully and answer the questions given below: 

A is the daughter of E. F has only two children – A and G. H is the brother of I. G is married to K. F has only two daughters. M is the mother of E. H is married to A. M is married to N. F is the son of O.

Q6. Who among the following is the father of E?

Solution:

Q7. Who among the following is the sister-in-law of I?

Solution:

Directions (8-10) : Study the following information to answer the given questions: 

In a certain code, 
‘za la ka ga’ is code for ‘cat key pen mobile’, 
‘ za fa sa na' is code for ‘we key the boys’ , 
‘ na la da sa’ is a code for ‘ we cat be boys’ , 
and ‘wa sa za da’ is code for ‘be boys key girls’. 

Q8. Which of the following is the code for ‘cat’?

Solution:

Q9. What does the code ‘za’ stand for?

Solution:

Q10. Which of the following is the code for ‘boys’?

Solution:

Directions (11-15) : Some statements are given followed by some conclusions. You have to consider the --statements to be true even if they seem to be at variance from commonly known facts. You have to decide which of the following conclusions if any, follow from the given statements: 

Q11. Statements: All boys are girls. 
                              Some girls are men. 
                              No men is a women 
         Conclusion: (i) Some girls are not women 
                              (ii) Some girls are women 
                              (iii) Some boys are not women

Solution:

Q12. Statements: Some birds are animals. 
                              Some animals are idiots. 
                              All idiots are humans. 
                              No idiots are bat. 
                              Some bats are Animals 
       Conclusion: (i) Some animals are bats 
                            (ii) Some humans being bat is a possibility 
                            (iii) some humans are definitely not bat

Solution:

Q13. Statements: All boys are girls. 
                              Some girls are not men. 
                              No men is a women. 
                              Some women are girls. 
                              No women are human. 
         Conclusion : (i) Some girls are not human. 
                               (ii) Some girls being men is a possibility. 
                               (iii) All boys being men is a possibility.

Solution:

Q14. Statements: All beetels are nuts. 
                              All nuts are leaves. 
                              All leaves are trash. 
                              Some trash is ball. 
                              No leave is important. 
                              Some important are trash. 
        Conclusion : (i) All ball being important is a possibility 
                              (ii) Some trash being beetel is a possibility 
                              (iii) Some important being beetel is a possibility

Solution:

Q15. Statements: All fun is revenge. 
                              No fun is entertainment. 
                              All entertainment is awesome. 
                              Some awesome is revenge. 
                              No revenge is heart. 
         Conclusion : (i) Some awesome is heart 
                               (ii) Some awesome is not entertainment 
                               (iii) Some awesome is not heart

Solution:

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