Reasoning PYQs

**Given

Statement:

** U ≤ F ≤ G ≤ Y; Y ≤ R > I < O

Conclusions:

I. R > U → False (Given U ≤ F ≤ G ≤ Y and Y ≤ R > I < O so by combining both statement U ≤ F ≤ G ≤ Y ≤ R > I < O. Thus R > U is false)

II. U = R → False (Given U ≤ F ≤ G ≤ Y and Y ≤ R > I < O so by combining both statement U ≤ F ≤ G ≤ Y ≤ R > I < O. Thus U = R is false)

As U ≤ R  → R > U or R = U

Hence, Either I or II follow is the correct answer.

Given

Statements:

P > Q ≤ S ≥ T; V = W ≤T

Conclusions:

I. P > T - False (Given P > Q ≤ S ≥ T and V = W ≤ T  so by combining  P > Q ≤ S ≥ T ≥ W = V  so the relation between P and T is not clear thus P > T is false)

II. S ≥ V - True (Given P > Q ≤ S ≥ T and  V = W ≤T  so by combining  P > Q ≤ S ≥ T ≥ W = V so the relation between S and V is clear thus S ≥ V is true )

 Hence, Only conclusion​ II is true.

Six people: A, B, C, D, E, and F eat different chocolates on six days of week starting from Monday to Saturday.

(1) Only three people eat chocolate between E and the one who eat 5-star. (2) E eats before A who eats Dairy Milk on Thursday. 

 

(3) Three people eat chocolate between the one who eats Kitkat and the one who eats Snicker. (4) B eats chocolate immediately before the one who eats Amul.

 

(5) D eats Perk two days before the one who eats Kitkat. Therefore, Case (I) eliminates here. (6) C does not like 5-Star. Thus, the final arrangement is as follows:

 

Hence, E eats Amul.

Six people: A, B, C, D, E, and F eat different chocolates on six days of week starting from Monday to Saturday.

(1) Only three people eat chocolate between E and the one who eat 5-star. (2) E eats before A who eats Dairy Milk on Thursday. 

 

(3) Three people eat chocolate between the one who eats Kitkat and the one who eats Snicker. (4) B eats chocolate immediately before the one who eats Amul.

 

(5) D eats Perk two days before the one who eats Kitkat. Therefore, Case (I) eliminates here. (6) C does not like 5-Star. Thus, the final arrangement is as follows:

 

Hence, B eats the chocolate on Monday.

Six people: A, B, C, D, E, and F eat different chocolates on six days of week starting from Monday to Saturday.

(1) Only three people eat chocolate between E and the one who eat 5-star. (2) E eats before A who eats Dairy Milk on Thursday. 

 

(3) Three people eat chocolate between the one who eats Kitkat and the one who eats Snicker. (4) B eats chocolate immediately before the one who eats Amul.

 

(5) D eats Perk two days before the one who eats Kitkat. Therefore, Case (I) eliminates here. (6) C does not like 5-Star. Thus, the final arrangement is as follows:

 

Hence, B-Monday-Snicker is correct.

Six people: A, B, C, D, E, and F eat different chocolates on six days of week starting from Monday to Saturday.

(1) Only three people eat chocolate between E and the one who eat 5-star. (2) E eats before A who eats Dairy Milk on Thursday. 

 

(3) Three people eat chocolate between the one who eats Kitkat and the one who eats Snicker. (4) B eats chocolate immediately before the one who eats Amul.

 

(5) D eats Perk two days before the one who eats Kitkat. Therefore, Case (I) eliminates here. (6) C does not like 5-Star. Thus, the final arrangement is as follows:

 

Hence, four persons eats chocolate before C.

Six people: A, B, C, D, E, and F eat different chocolates on six days of week starting from Monday to Saturday.

(1) Only three people eat chocolate between E and the one who eat 5-star. (2) E eats before A who eats Dairy Milk on Thursday. 

 

(3) Three people eat chocolate between the one who eats Kitkat and the one who eats Snicker. (4) B eats chocolate immediately before the one who eats Amul.

 

(5) D eats Perk two days before the one who eats Kitkat. Therefore, Case (I) eliminates here. (6) C does not like 5-Star. Thus, the final arrangement is as follows:

 

We can see that F eats 5-Star on Saturday.

Hence, option (4) is false.

Given : TELEVISION

After changing the vowels to the next letter and the consonants to the previous letter;

SFKFUJRJPM

Hence, there are two letters (F, J) in the given words which are repeated more than once.

According to the given information,

 

Hence, Southeast is the correct answer.

According to the given information,

Here the distance between House and final point is = 17 - (8 + 7) = 2 km.

Hence, 2 km is the correct answer.

According to the given information,

Here the shortest distance between point D and B is = BD2 = BC2 + CD2

BD2 = 82 + 102

BD2 = 64 + 100 = 164

BD = √164

Hence, √164 km is the correct answer.

According to the given information,

Hence, Southwest is the correct answer.

From the given information,

1. There are three married couples in the family.

2. Shyam is the brother-in-law of Shubham, who is the daughter-in-law of Himanshu.

3. Aditi and Shyam are unmarried members.

4. Himanshu is the brother-in-law of Aditi.

(from the above statements it is clear that brother of Shyam is married to Shubham because Shyam is an unmarried member and Shubham is daughter in law of Himanshu, which means Himanshu is the father of Shyam)

5. Himanshu is the brother-in-law of Aditi. ( It means Himanshu is married to the sibling of Aditi because Aditi is an unmarried member)

6. Piyush is the maternal grandfather of Neha and father of Aditi. ( This means Neha belongs to the third generation)

7. Ram is the paternal uncle of Neha and Ragini is the grandmother of Shyam. ( This means Ragini belongs to first-generation and she is the wife of Piyush and Ram belongs to the second generation that means Himanshu is the father of Neha and brother of Ram) 

8. There are five males and four females in the family. ( this implies that Aditi is a female member of the family).

The final arrangement

Hence, Anjali is the Mother-in-law of Shubham.

From the given information,

1. There are three married couples in the family.

2. Shyam is the brother-in-law of Shubham, who is the daughter-in-law of Himanshu.

3. Aditi and Shyam are unmarried members.

4. Himanshu is the brother-in-law of Aditi.

(from the above statements it is clear that brother of Shyam is married to Shubham because Shyam is an unmarried member and Shubham is daughter in law of Himanshu, which means Himanshu is the father of Shyam)

5. Himanshu is the brother-in-law of Aditi. ( It means Himanshu is married to the sibling of Aditi because Aditi is an unmarried member)

6. Piyush is the maternal grandfather of Neha and father of Aditi. ( This means Neha belongs to the third generation)

7. Ram is the paternal uncle of Neha and Ragini is the grandmother of Shyam. ( This means Ragini belongs to first-generation and she is the wife of Piyush and Ram belongs to the second generation that means Himanshu is the father of Neha and brother of Ram) 

8. There are five males and four females in the family. ( this implies that Aditi is a female member of the family).

The final arrangement

Hence, Anjali is the mother of Shyam.

From the given information,

1. There are three married couples in the family.

2. Shyam is the brother-in-law of Shubham, who is the daughter-in-law of Himanshu.

3. Aditi and Shyam are unmarried members.

4. Himanshu is the brother-in-law of Aditi.

(from the above statements it is clear that brother of Shyam is married to Shubham because Shyam is an unmarried member and Shubham is daughter in law of Himanshu, which means Himanshu is the father of Shyam)

5. Himanshu is the brother-in-law of Aditi. ( It means Himanshu is married to the sibling of Aditi because Aditi is an unmarried member)

6. Piyush is the maternal grandfather of Neha and father of Aditi. ( This means Neha belongs to the third generation)

7. Ram is the paternal uncle of Neha and Ragini is the grandmother of Shyam. ( This means Ragini belongs to first-generation and she is the wife of Piyush and Ram belongs to the second generation that means Himanshu is the father of Neha and brother of Ram) 

8. There are five males and four females in the family. ( this implies that Aditi is a female member of the family).

The final arrangement

Hence, Himanshu is the Son-in-law of Ragini.

Nine boxes: A, B, C, D, E, F, G, H and I.

(1) Only five boxes are placed between D and C. (2) E is placed immediately above C. (3) D is neither at the topmost nor at the bottommost position.

 

(4) Only three boxes are placed between E and A. (5) As many boxes are placed between A and E as between C and F.

 

(6) One box is placed between F and G. (7) More than five boxes are placed between E and B. Here, case (1) eliminates.

 

(8) I is placed above H but not immediately above G. Therefore, case (3) eliminates here and the final arrangement is as follows:

 

We can see that there are five boxes above H.

Hence, number of boxes above H is equal to the number of boxes below A.

Nine boxes: A, B, C, D, E, F, G, H and I.

(1) Only five boxes are placed between D and C. (2) E is placed immediately above C. (3) D is neither at the topmost nor at the bottommost position.

 

(4) Only three boxes are placed between E and A. (5) As many boxes are placed between A and E as between C and F.

 

(6) One box is placed between F and G. (7) More than five boxes are placed between E and B. Here, case (1) eliminates.

 

(8) I is placed above H but not immediately above G. Therefore, case (3) eliminates here and the final arrangement is as follows:

 

Hence, there are three boxes between D and G.

Nine boxes: A, B, C, D, E, F, G, H and I.

(1) Only five boxes are placed between D and C. (2) E is placed immediately above C. (3) D is neither at the topmost nor at the bottommost position.

 

(4) Only three boxes are placed between E and A. (5) As many boxes are placed between A and E as between C and F.

 

(6) One box is placed between F and G. (7) More than five boxes are placed between E and B. Here, case (1) eliminates.

 

(8) I is placed above H but not immediately above G. Therefore, case (3) eliminates here and the final arrangement is as follows:

 

Hence, option (5) is correct.

Nine boxes: A, B, C, D, E, F, G, H and I.

(1) Only five boxes are placed between D and C. (2) E is placed immediately above C. (3) D is neither at the topmost nor at the bottommost position.

 

(4) Only three boxes are placed between E and A. (5) As many boxes are placed between A and E as between C and F.

 

(6) One box is placed between F and G. (7) More than five boxes are placed between E and B. Here, case (1) eliminates.

 

(8) I is placed above H but not immediately above G. Therefore, case (3) eliminates here and the final arrangement is as follows:

 

Hence, there are seven boxes below I.

Nine boxes: A, B, C, D, E, F, G, H and I.

(1) Only five boxes are placed between D and C. (2) E is placed immediately above C. (3) D is neither at the topmost nor at the bottommost position.

 

(4) Only three boxes are placed between E and A. (5) As many boxes are placed between A and E as between C and F.

 

(6) One box is placed between F and G. (7) More than five boxes are placed between E and B. Here, case (1) eliminates.

 

(8) I is placed above H but not immediately above G. Therefore, case (3) eliminates here and the final arrangement is as follows:

 

Hence, H is placed at the sixth position from the top.