IBPS Clerk Mains Mock Test - 10 - Banking Exams MCQ

1) F is sitting third to the right of J, who has scored 59 more runs than F.

2) The person sitting third to the right of F has scored 70 runs.

(As it is a unidirectional circular arrangement, we can randomly place J and then place F accordingly.)

3) L has scored 25% less than J and he is sitting third to the left of K.

4) G has scored 5 fewer runs than L and he is one of the immediate neighbors of F, also he is sitting second to the left of L.

(Let us assume that G is to the immediate right of F, then it means that L will be second to the left of J and it also means that L scored 70 runs.

If L scored 70 runs which is 75% of J, implies J scored 93.33 runs, which is not possible as runs cannot be in decimals.

Hence, our assumption was wrong. G is to the immediate left of F and L is sitting opposite J.)

5) H is sitting third to the left of E and he has scored 10 more runs than the person who is 2nd to his left.

(There is only one way for this to be possible, i.e. if H is sitting to the immediate left of K and E is sitting to the immediate right of J.

It also means that H scored 70 runs and L scored 60 runs.

L scored 25% fewer runs than J, implies J scored 80 runs (acc. to statement 3).

F scored 59 runs less than J, implies F scored 21 runs (acc. to statement 1).

G scored 55 runs (acc. to statement 4).)

6) K scored twice as many runs as the person who is sitting opposite him but seven more than I.

(As I only person is left to be mapped, we can say that I is sitting opposite E.

K is sitting opposite F who scored 21 runs, implies K scored 42 runs.

It further implies that I scored 35 runs.)

7) The average score of E and the person sitting on the opposite end is 25.

(I is sitting opposite E and his score id 35, implies the score of E is 15.)

E is the lowest scorer and J and G are his immediate neighbors.

1) F is sitting third to the right of J, who has scored 59 more runs than F.

2) The person sitting third to the right of F has scored 70 runs.

(As it is a unidirectional circular arrangement, we can randomly place J and then place F accordingly.)

3) L has scored 25% less than J and he is sitting third to the left of K.

4) G has scored 5 fewer runs than L and he is one of the immediate neighbors of F, also he is sitting second to the left of L.

(Let us assume that G is to the immediate right of F, then it means that L will be second to the left of J and it also means that L scored 70 runs.

If L scored 70 runs which is 75% of J, implies J scored 93.33 runs, which is not possible as runs cannot be in decimals.

Hence, our assumption was wrong. G is to the immediate left of F and L is sitting opposite J.)

5) H is sitting third to the left of E and he has scored 10 more runs than the person who is 2nd to his left.

(There is only one way for this to be possible, i.e. if H is sitting to the immediate left of K and E is sitting to the immediate right of J.

It also means that H scored 70 runs and L scored 60 runs.

L scored 25% fewer runs than J, implies J scored 80 runs (acc. to statement 3).

F scored 59 runs less than J, implies F scored 21 runs (acc. to statement 1).

G scored 55 runs (acc. to statement 4).)

6) K scored twice as many runs as the person who is sitting opposite him but seven more than I.

(As I only person is left to be mapped, we can say that I is sitting opposite E.

K is sitting opposite F who scored 21 runs, implies K scored 42 runs.

It further implies that I scored 35 runs.)

7) The average score of E and the person sitting on the opposite end is 25.

(I is sitting opposite E and his score id 35, implies the score of E is 15.)

The score of I is 35.

1) F is sitting third to the right of J, who has scored 59 more runs than F.

2) The person sitting third to the right of F has scored 70 runs.

(As it is a unidirectional circular arrangement, we can randomly place J and then place F accordingly.)

3) L has scored 25% less than J and he is sitting third to the left of K.

4) G has scored 5 fewer runs than L and he is one of the immediate neighbors of F, also he is sitting second to the left of L.

(Let us assume that G is to the immediate right of F, then it means that L will be second to the left of J and it also means that L scored 70 runs.

If L scored 70 runs which is 75% of J, implies J scored 93.33 runs, which is not possible as runs cannot be in decimals.

Hence, our assumption was wrong. G is to the immediate left of F and L is sitting opposite J.)

5) H is sitting third to the left of E and he has scored 10 more runs than the person who is 2nd to his left.

(There is only one way for this to be possible, i.e. if H is sitting to the immediate left of K and E is sitting to the immediate right of J.

It also means that H scored 70 runs and L scored 60 runs.

L scored 25% fewer runs than J, implies J scored 80 runs (acc. to statement 3).

F scored 59 runs less than J, implies F scored 21 runs (acc. to statement 1).

G scored 55 runs (acc. to statement 4).)

6) K scored twice as many runs as the person who is sitting opposite him but seven more than I.

(As I only person is left to be mapped, we can say that I is sitting opposite E.

K is sitting opposite F who scored 21 runs, implies K scored 42 runs.

It further implies that I scored 35 runs.)

7) The average score of E and the person sitting on the opposite end is 25.

(I is sitting opposite E and his score id 35, implies the score of E is 15.)

Score of J = 80

Score of L = 60

Average = 70

1) F is sitting third to the right of J, who has scored 59 more runs than F.

2) The person sitting third to the right of F has scored 70 runs.

(As it is a unidirectional circular arrangement, we can randomly place J and then place F accordingly.)

3) L has scored 25% less than J and he is sitting third to the left of K.

4) G has scored 5 fewer runs than L and he is one of the immediate neighbors of F, also he is sitting second to the left of L.

(Let us assume that G is to the immediate right of F, then it means that L will be second to the left of J and it also means that L scored 70 runs.

If L scored 70 runs which is 75% of J, implies J scored 93.33 runs, which is not possible as runs cannot be in decimals.

Hence, our assumption was wrong. G is to the immediate left of F and L is sitting opposite J.)

5) H is sitting third to the left of E and he has scored 10 more runs than the person who is 2nd to his left.

(There is only one way for this to be possible, i.e. if H is sitting to the immediate left of K and E is sitting to the immediate right of J.

It also means that H scored 70 runs and L scored 60 runs.

L scored 25% fewer runs than J, implies J scored 80 runs (acc. to statement 3).

F scored 59 runs less than J, implies F scored 21 runs (acc. to statement 1).

G scored 55 runs (acc. to statement 4).)

6) K scored twice as many runs as the person who is sitting opposite him but seven more than I.

(As I only person is left to be mapped, we can say that I is sitting opposite E.

K is sitting opposite F who scored 21 runs, implies K scored 42 runs.

It further implies that I scored 35 runs.)

7) The average score of E and the person sitting on the opposite end is 25.

(I is sitting opposite E and his score id 35, implies the score of E is 15.)

Clearly, G is sitting to the immediate right of E.

1) F is sitting third to the right of J, who has scored 59 more runs than F.

2) The person sitting third to the right of F has scored 70 runs.

(As it is a unidirectional circular arrangement, we can randomly place J and then place F accordingly.)

3) L has scored 25% less than J and he is sitting third to the left of K.

4) G has scored 5 fewer runs than L and he is one of the immediate neighbors of F, also he is sitting second to the left of L.

(Let us assume that G is to the immediate right of F, then it means that L will be second to the left of J and it also means that L scored 70 runs.

If L scored 70 runs which is 75% of J, implies J scored 93.33 runs, which is not possible as runs cannot be in decimals.

Hence, our assumption was wrong. G is to the immediate left of F and L is sitting opposite J.)

5) H is sitting third to the left of E and he has scored 10 more runs than the person who is 2nd to his left.

(There is only one way for this to be possible, i.e. if H is sitting to the immediate left of K and E is sitting to the immediate right of J.

It also means that H scored 70 runs and L scored 60 runs.

L scored 25% fewer runs than J, implies J scored 80 runs (acc. to statement 3).

F scored 59 runs less than J, implies F scored 21 runs (acc. to statement 1).

G scored 55 runs (acc. to statement 4).)

6) K scored twice as many runs as the person who is sitting opposite him but seven more than I.

(As I only person is left to be mapped, we can say that I is sitting opposite E.

K is sitting opposite F who scored 21 runs, implies K scored 42 runs.

It further implies that I scored 35 runs.)

7) The average score of E and the person sitting on the opposite end is 25.

(I is sitting opposite E and his score id 35, implies the score of E is 15.)

Clearly, J is the highest scorer.

Based on the given information we can draw family tree:

Hence B is the Mother-in-law of M

Based on the given information we can draw family tree:

Here we doesn't know the gender of T.

Hence we cannot determine the relation between T and R.

Based on the given information we can draw family tree:

Hence J and R are cousins.

8 players: P, Q, R, S, T, U, V & W standing in a row facing south or north.

Condition:

Junior Football team – 3 Players (16 ≤ Age ≤ 18)

Rugby team – 3 players (18 ≤ Age)

Hockey team – 2 players (16 ≤ Age ≤ 21)

Only three players are facing towards the south. (5 facing North)

(1) Only V and X is standing between S & W.

(2) R is not an immediate neighbor of T and X is not an immediate neighbor of S.

(3) V is facing towards the south and his both the immediate neighbors are Rugby players.

(4) W is sitting 2nd to the right of T and S is standing 2nd to the right of P.

(5) Both the immediate neighbors of Q are facing towards south.

(Here case – 2 will gets eliminated as there R is not an immediate neighbor of T)

(6) W is the oldest among them and he is 24 years old.

(As maximum age to play Hockey is 21 year and football is 18 year, therefore, W must be playing Rugby)

(7) P & T don’t belong to Junior Football team.

(Therefore, P and T belong to Hockey team and R, V, Q belong to Football Team)

(8) V & Q are the youngest players among the eight players.

(9) V has just attained the minimum age of eligibility for the game he plays.

(as V plays Football there V’s and Q’s age is 16 year)

(10) T has also just attained the minimum age to play Rugby.

(Minimum age to Play Rugby is 18 years therefore T is 18-years-old.)

(11) S is three years older to P.

(12) Also, P is 4 years older to V.

(as V = 16 year, P will be 20 year old and S will be 23 year old)

(13) R is one year younger to T.

(R = 17 year)

(14) One of the Rugby players has the same age as that of one of the Junior Hockey players.

(only X’s age is undetermined i.e. X can be either 20 or 18 years old)

(P plays hockey and is 20 years old and T plays hockey and is 18 years old as they attained the minimum age for Rugby)

Above combination will be the final combination.

Except in P, R, S all another group two person face north one face south. But P, R, S all are faing toward North.

8 players: P, Q, R, S, T, U, V & W standing in a row facing south or north.

Condition:

Junior Football team – 3 Players (16 ≤ Age ≤ 18)

Rugby team – 3 players (18 ≤ Age)

Hockey team – 2 players (16 ≤ Age ≤ 21)

Only three players are facing towards the south. (5 facing North)

(1) Only V and X is standing between S & W.

(2) R is not an immediate neighbor of T and X is not an immediate neighbor of S.

(3) V is facing towards the south and his both the immediate neighbors are Rugby players.

(4) W is sitting 2nd to the right of T and S is standing 2nd to the right of P.

(5) Both the immediate neighbors of Q are facing towards south.

(Here case – 2 will gets eliminated as there R is not an immediate neighbor of T)

(6) W is the oldest among them and he is 24 years old.

(As maximum age to play Hockey is 21 year and football is 18 year, therefore, W must be playing Rugby)

(7) P & T don’t belong to Junior Football team.

(Therefore, P and T belong to Hockey team and R, V, Q belong to Football Team)

(8) V & Q are the youngest players among the eight players.

(9) V has just attained the minimum age of eligibility for the game he plays.

(as V plays Football there V’s and Q’s age is 16 year)

(10) T has also just attained the minimum age to play Rugby.

(Minimum age to Play Rugby is 18 years therefore T is 18-years-old.)

(11) S is three years older to P.

(12) Also, P is 4 years older to V.

(as V = 16 year, P will be 20 year old and S will be 23 year old)

(13) R is one year younger to T.

(R = 17 year)

(14) One of the Rugby players has the same age as that of one of the Junior Hockey players.

(only X’s age is undetermined i.e. X can be either 20 or 18 years old)

(P plays hockey and is 20 years old and T plays hockey and is 18 years old as they attained the minimum age for Rugby)

Above combination will be the final combination.

S is 23 years old and he plays Rugby.

8 players: P, Q, R, S, T, U, V & W standing in a row facing south or north.

Condition:

Junior Football team – 3 Players (16 ≤ Age ≤ 18)

Rugby team – 3 players (18 ≤ Age)

Hockey team – 2 players (16 ≤ Age ≤ 21)

Only three players are facing towards the south. (5 facing North)

(1) Only V and X is standing between S & W.

(2) R is not an immediate neighbor of T and X is not an immediate neighbor of S.

(3) V is facing towards the south and his both the immediate neighbors are Rugby players.

(4) W is sitting 2nd to the right of T and S is standing 2nd to the right of P.

(5) Both the immediate neighbors of Q are facing towards south.

(Here case – 2 will gets eliminated as there R is not an immediate neighbor of T)

(6) W is the oldest among them and he is 24 years old.

(As maximum age to play Hockey is 21 year and football is 18 year, therefore, W must be playing Rugby)

(7) P & T don’t belong to Junior Football team.

(Therefore, P and T belong to Hockey team and R, V, Q belong to Football Team)

(8) V & Q are the youngest players among the eight players.

(9) V has just attained the minimum age of eligibility for the game he plays.

(as V plays Football there V’s and Q’s age is 16 year)

(10) T has also just attained the minimum age to play Rugby.

(Minimum age to Play Rugby is 18 years therefore T is 18-years-old.)

(11) S is three years older to P.

(12) Also, P is 4 years older to V.

(as V = 16 year, P will be 20 year old and S will be 23 year old)

(13) R is one year younger to T.

(R = 17 year)

(14) One of the Rugby players has the same age as that of one of the Junior Hockey players.

(only X’s age is undetermined i.e. X can be either 20 or 18 years old)

(P plays hockey and is 20 years old and T plays hockey and is 18 years old as they attained the minimum age for Rugby)

Above combination will be the final combination.

Only one person sits to the left of R.

Therefore No one sits 3rd to the left of R.

Hence, None of these is the correct option

Here mathematical operation in followed in every step:

In first step:

5 × 1 = 5, 1 × 3 = 3

2 × 4 = 8, 1 × 3 = 3

2 × 2 = 4, 1 × 4 = 4

In second step:

Adding all the first digits: 5 + 8 + 4 =17

Adding all the second digits: 3 + 3 + 4 = 10

In third step:

1 + 7 = 8, 8 / 2 = 4

1 + 0 = 1, 1 / 2 = 0.5

In fourth step:

4 - 0.5 = 3.5

Similarly in the given input:

In first step:

6 × 1 = 6, 2 × 1 = 2

2 × 1 = 2, 4 × 2 = 8

7 × 1 = 7, 1 × 4 = 4

In second step:

Adding all the first digits: 6 + 2 + 7 = 15

Adding all the second digits: 2 + 8 + 4 = 14

In third step:

1 + 5 = 6, 6 / 2 = 3

1 + 4 = 5, 5 / 2 = 2.5

In fourth step:

3 - 2.5 = 0.5

On arranging all the letters in English alphabetical order

ABCCCCCCCDEEEFIJJKKLLMMMMNNOSSUU

the letters left to right at the positions

10 = C, 5 = A, 13 = E ,27 = O, 6 = L, 5 = A, 4 = D, and 19 = C

ACCOLADE

Means, An award or privilege granted as a special honor or as an acknowledgment of merit.

If every letter beginning from position 7, counted from left, is replaced by its successor in the English alphabet

Modified Series: BCCDALMNNDDDFVNLOODJDLNTTFPGVKKF

Hence there are three L.

Five of them are facing north and the rest of them is facing south. Also, their age lies between 19- 47 and none of them has the same age. Not more than two adjacent persons are facing the same direction.

1) B faces north direction. I sit either immediate left or third to the right of B.

2) I and B are facing the opposite direction. Only one person is sitting between B and D, both are facing the opposite direction.

3) The number of persons sitting to the left of D is the same as to the left of C. D sits at the extreme end.

From point 3 we can say that there are three possibilities so we take 3 cases.

4) Only two persons are sitting between the one whose age is 37 and C.

5) The immediate neighbors of I are facing the same direction. A is facing the same direction as I.

6) A sits immediate right of H, neither of them is adjacent to C nor B. H and D are facing the opposite direction.

From point 5, case 2 will be eliminated. And from point 6, case 3 will be eliminated because we cannot place them according to the statement. So we will apply these statements to case 1 only. As A is facing the same direction as I, there is only one place left for A and H.

7) F neither sits adjacent to A nor H. F and the one whose age is 19 are not sitting adjacent to each other.

8) The one whose age is 47 sits third to the left of E.

9) The one whose age is 47 sits to the right of A.

Now it is given that five of them face north and not more than two adjacent persons are facing the same direction. So the seat between C and H cannot face north.

10) The one whose age is 45 sits immediate right of the one whose age is 38.

11) The age of I is 6 more than the age of E.

12) The age of B is one year more than A. The age of B is twice the age of C.

The one whose age is 45 and the one whose age is 38 cannot be the age of E, I and D because the difference between 45 and 38 is not 6 and twice of D is more than 47. It is also given that B’s age is 1 more than A, therefore 38 and 45 will be the age of B and F respectively. C’s age will be 19.

13) E’s age is an even number which is two years more than D's age.

14) The age difference between B and A is the same as between C and D.

15) H’s age is twice of D's age.

From point 14 it is clear that the age of D will be 20 because it cannot be 18 (Since the age range is 19-47). Therefore E’s age will be 22 and H’s age will be 40. and after these only G is left whose age will be 47. And I’s age will be 28.

Therefore the final arrangement will be,

sum of the age of H and B is 78.

Conclusions:

I. No 1 is 3 → It’s not definitely sure, hence false.

II. Some 4 is 2→ It’s not definitely sure, hence false.

III. Some 3 is not 6→ It’s not definitely sure, hence false

IV. All 5 is 4 → It’s not definitely sure, hence false.

Conclusions:

I. Some cars are locks → False (It is possible but not definite)

II. Some cars are cakes → False (It is possible but not definite)

III. No car is locks → False (It is possible but not definite)

Conclusion I and III form a complementary pair.

Thus, either I or III follows.

(ii) For the letter in the code- middle letter of the word.

Conclusions:

I. All ocean are sky is a possibility - False (only sky are ocean which means all oceans are sky. So, all oceans are sky is definitely true. When any conclusion is definitely true then its possibility is always false.)

II. Some water are sea is a possibility - True (As only a few water are sky and Some sky are sea)

III. No Ocean are sea – True (As Only sky are ocean which means all oceans are sky but ocean can never have any positive relation with any other element. So, no ocean are sea is definitely true.)

The speed of bus P is 50 km/hr and that of bus Q is 60 km/hr.

After first hour bus P which is moving from west to east reached point A and from there takes a left turn, runs for 30 minutes and reaches point F whereas the other bus which was moving from east to west, keeps running on the road till 2 hours to reach point B and then takes a left turn, runs for half an hour and reaches point L.

The bottommost floor is numbered as 1 and the floor above it is 2 and so on. Flat A, B and C have 5, 4 and 3 floors respectively.

1) O lives three floors above the one who has 4 laptops but O does not live on the 5th floor and the one who has 4 laptops lives in N's Flat. This means O cannot be placed on the 5th floor and both O and the one who has 4 laptops are from different flats.

2) N lives North West of P, who lives on the topmost floor in his flat and O neither lives in N's flat nor lives in P's flat.

3) N lives below M but a different type of Flat.

From the above points, it is clear that O has two possibilities: O can be placed in Flat A or Flat B but from point 3 we can conclude that M lives in Flat A on the 5th floor. Therefore N cannot live in flat A and this clears that N lives in flat B and O lives in flat A.

4) In same type of flat only two person lives between S and M, who lives above of S and has 16 laptops

6) In the same flat, S lives immediately above the one who has 8 laptops. S has 24 laptops.

7) In a different flat, M lives two floors above the one who has 2 laptops but P does not have 2 laptops.

There is only one place left for the one who has 2 laptops.

8) W lives immediately above T in the same type of flat, T has 12 laptops. W and X do not live in flat A.

9) R does not have 2 laptops but he lives north east of the U. R lives on an even number floor but not in flat C.

10) The one who has 10 laptops lives two floors above T, who does not live on an even floor.

T lives on odd number floor and have 12 laptops so there is only one place left for T, that is flat C and floor 1.

11) Q lives west of the one who has 10 laptops but Q does not have 2 laptops.

12) V does not live below X. The one who has 18 laptops lives North West of the one who has 12 laptops here there are two possibilities of the one who have 18 laptops that is R and O.

13) The one who has 14 laptops, lives north-east of the one who has 6 laptops. O has more laptops than R but less than W.

Now there are 3 persons left who are 18, 20 and 22 and these persons are O, R and W. From point 13, W > O >R (22 > 20 > 18) therefore it is clear that W, O and R have 22, 20 and 18 laptops respectively.

Therefore, the final arrangement will be,

The bottommost floor is numbered as 1 and the floor above it is 2 and so on. Flat A, B and C have 5, 4 and 3 floors respectively.

1) O lives three floors above the one who has 4 laptops but O does not live on the 5th floor and the one who has 4 laptops lives in N's Flat. This means O cannot be placed on the 5th floor and both O and the one who have 4 laptops are from different flats.

2) N lives North West of P, who lives on the topmost floor in his flat and O neither lives in N's flat nor lives in P's flat.

3) N lives below M but a different type of Flat.

From the above points, it is clear that O has two possibilities: O can be placed in Flat A or Flat B but from point 3 we can conclude that M lives in Flat A on the 5th floor. Therefore N cannot live in flat A and this clears that N lives in flat B and O lives in flat A.

4) In the same type of flat only two person live between S and M, who lives above S and has 16 laptops.

6) In the same flat, S lives immediately above the one who has 8 laptops. S has 24 laptops.

7) In different flat, M lives two floors above of the one who has 2 laptops but P does not have 2 laptops.

There is only one place left for the one who has 2 laptops.

8) W lives immediately above T in the same type of flat, T has 12 laptops. W and X do not live in flat A.

9) R does not have 2 laptops but he lives north east of U. R lives on an even number floor but not in flat C.

10) The one who has 10 laptops lives two floors above T, who does not live on an even floor.

T lives on odd number floor and has 12 laptops so there is only one place left for T, that is flat C and floor 1.

11) Q lives west of the one who has 10 laptops but Q does not have 2 laptops.

12) V does not live below X. The one who has 18 laptops, lives North West of the one who has 12 laptops. Here there are two possibilities of the one who has 18 laptops that is R and O.

13) The one who has 14 laptops, lives north-east of the one who has 6 laptops. O has more laptops than R but less than W.

Now there are 3 persons left who are 18, 20 and 22 and these persons are O, R and W. From point 13, W > O >R (22 > 20 > 18) therefore it is clear that W, O and R have 22, 20 and 18 laptops respectively.

Therefore, the final arrangement will be,

The bottommost floor is numbered as 1 and the floor above it is 2 and so on. Flat A, B and C have 5, 4 and 3 floors respectively.

1) O lives three floors above the one who has 4 laptops but O does not live on the 5th floor and the one who has 4 laptops lives in N's Flat. This means O cannot be placed on the 5th floor and both O and the one who has 4 laptops are from different flats.

2) N lives North West of P, who lives on the topmost floor in his flat and O neither lives in N's flat nor lives in P's flat.

3) N lives below M but a different type of Flat.

From the above points, it is clear that O has two possibilities: O can be placed in Flat A or Flat B but from point 3 we can conclude that M lives in Flat A on the 5th floor. Therefore N cannot live in flat A and this clears that N lives in flat B and O lives in flat A.

4) In the same type of flat only two person live between S and M, who lives above S and has 16 laptops.

6) In the same flat, S lives immediately above the one who has 8 laptops. S has 24 laptops.

7) In a different flat, M lives two floors above the one who has 2 laptops but P does not have 2 laptops.

There is only one place left for the one who has 2 laptops.

8) W lives immediately above T in the same type of flat, T has 12 laptops. W and X do not live in flat A.

9) R does not have 2 laptops but he lives north east of the U. R lives on an even number floor but not in flat C.

10) The one who has 10 laptops lives two floors above T, who does not live on an even floor.

T lives on odd number floor and has 12 laptops so there is only one place left for T, that is flat C and floor 1.

11) Q lives west of the one who has 10 laptops but Q does not have 2 laptops.

12) V does not live below X. The one who has 18 laptops, lives North West of the one who has 12 laptops. Here there are two possibilities of the one who has 18 laptops that is R and O.

13) The one who has 14 laptops, lives north-east of the one who has 6 laptops. O has more laptops than R but less than W.

Now there are 3 persons left who have 18, 20 and 22 and these persons are O, R and W. From point 13, W > O >R (22 > 20 > 18) therefore it is clear that W, O and R have 22, 20 and 18 laptops respectively.

Therefore, the final arrangement will be,

The bottommost floor is numbered as 1 and the floor above it is 2 and so on. Flat A, B and C have 5, 4 and 3 floors respectively.

1) O lives three floors above the one who has 4 laptops but O does not live on the 5th floor and the one who has 4 laptops lives in N's Flat. This means O cannot be placed on the 5th floor and both O and the one who has 4 laptops are from different flats.

2) N lives North West of P, who lives on the topmost floor in his flat and O neither lives in N's flat nor lives in P's flat.

3) N lives below M but a different type of Flat.

From the above points, it is clear that O has two possibilities: O can be placed in Flat A or Flat B but from point 3 we can conclude that M lives in Flat A on the 5th floor. Therefore N cannot live in flat A and this clears that N lives in flat B and O lives in flat A.

4) In the same type of flat only two person live between S and M, who live above S and has 16 laptops.

6) In the same flat, S lives immediately above the one who has 8 laptops. S has 24 laptops.

7) In a different flat, M lives two floors above the one who has 2 laptops but P does not have 2 laptops.

There is only one place left for the one who has 2 laptops.

8) W lives immediately above T in the same type of flat, T has 12 laptops. W and X do not live in flat A.

9) R does not have 2 laptops but he lives north east of the U. R lives on an even number floor but not in flat C.

10) The one who has 10 laptops lives two floors above T, who does not live on an even floor.

T lives on odd number floor and have 12 laptops so there is only one place left for T, that is flat C and floor 1.

11) Q lives west of the one who has 10 laptops but Q does not have 2 laptops.

12) V does not live below X. The one who has 18 laptops lives North West of the one who has 12 laptops here there are two possibilities of the one who have 18 laptops that is R and O.

13) The one who has 14 laptops, lives north-east of the one who has 6 laptops. O has more laptops than R but less than W.

Now there are 3 persons left who are 18, 20 and 22 and these persons are O, R and W. From point 13, W > O > R (22 > 20 > 18) therefore it is clear that W, O and R have 22, 20 and 18 laptops respectively.

Therefore, the final arrangement will be,

The bottommost floor is numbered as 1 and the floor above it is 2 and so on. Flat A, B and C have 5, 4 and 3 floors respectively.

1) O lives three floors above the one who has 4 laptops but O does not live on the 5th floor and the one who has 4 laptops lives in N's Flat. This means O cannot be placed on the 5th floor and both O and the one who has 4 laptops are from different flats.

2) N lives North West of P, who lives on the topmost floor in his flat and O neither lives in N's flat nor lives in P's flat.

3) N lives below M but a different type of Flat.

From the above points, it is clear that O has two possibilities: O can be placed in Flat A or Flat B but from point 3 we can conclude that M lives in Flat A on the 5th floor. Therefore N cannot live in flat A and this clears that N lives in flat B and O lives in flat A

4) In the same type of flat only two people live between S and M, who live above S and have 16 laptops.

6) In the same flat, S lives immediately above the one who has 8 laptops. S has 24 laptops.

7) In a different flat, M lives two floors above the one who has 2 laptops but P does not have 2 laptops.

There is only one place left for the one who has 2 laptops.

8) W lives immediately above T in the same type of flat, T has 12 laptops. W and X do not live in flat A.

9) R does not have 2 laptops but he lives north east of OUR lives on an even number floor but not in flat C.

10) The one who has 10 laptops lives two floors above T, who does not live on an even floor.

T lives on an odd number floor and have 12 laptops so there is only one place left for T, that is flat C and floor 1.

11) Q lives west of the one who has 10 laptops but Q does not have 2 laptops.

12) V does not live below X. The one who has 18 laptops, lives North West of the one who has 12 laptops. Here there are two possibilities of the one who has 18 laptops that is R and O.

13) The one who has 14 laptops, lives north-east of the one who has 6 laptops. O has more laptops than R but less than W.

Now there are 3 persons left who are 18, 20 and 22 and these persons are O, R and W. From point 13, W > O > R (22 > 20 > 18) therefore it is clear that W, O and R have 22, 20 and 18 laptops respectively.

Therefore, the final arrangement will be,

N does not enroll in April. Only two candidates have enrolled between N and Q

The one who is preparing for the Railway department and likes G.A subject enrolling on an odd number date of the month which has 31 days.

The one who likes G. A subject enrolled on a date just after the one who likes Math subject who does not prepare for the Delhi police department.

S and Q enrolling dates are the same but in different months. O enrolled on one of the days before S’ enrolled. Only one person enrolled between O and P.

Less than four candidates have enrolled between P and U.

Case1 eliminated. In case 1, O enrolled before S, but S enrolled on 16th of March and no date is prior to that.

There are two candidates who are different between P and the one who is preparing for Delhi police department. The one who likes Physics subject enrolled on a date just before the one who is preparing for Delhi police department.

There are two candidates who are different between O and the one who likes the Reasoning subject who enrolled on date just after the one who is preparing for the post office department.

The one who is preparing for the LIC department also likes an English subject who does not enrolled on an odd number date. The one who likes Hindi subjects enrolled on a date just after the one who is preparing for the LIC department.

There are two candidates different between the one who preparation for SSC department and Electricity department who just enrolled on date just after the one who likes Biology subject who does not preparation for post office department. Case2 eliminated

There is one candidate who is different between the one who is preparing for UPSC and the one who likes Chemistry . The one candidate in this group who is preparing for the banking department.

N does not enroll in April. Only two candidates have enrolled between N and Q

The one who is preparing for the Railway department and likes G.A subject enrolling on an odd number date of the month which has 31 days.

The one who likes G.A subject enrolled on a date just after the one who likes Math subject who does not prepare for the Delhi police department.

S and Q enrolling dates are the same but in different months. O enrolled on one of the days before S’ enrolled. Only one person enrolled between O and P.

Less than four candidates have enrolled between P and U.

Case1 eliminated. In case 1, O enrolled before S, but S enrolled on 16th of March and no date is prior to that.

There are two candidates who are different between P and the one who is preparing for Delhi police department. The one who likes Physics subject enrolled on a date just before the one who is preparing for Delhi police department.

There are two candidates who are different between O and the one who likes the Reasoning subject who enrolled on date just after the one who is preparing for the post office department.

The one who is preparing for the LIC department also likes English subjects who does not enroll on odd number dates. The one who likes Hindi subjects enrolled on a date just after the one who is preparing for the LIC department.

There are two candidates different between the one who preparation for SSC department and Electricity department who just enrolled on date just after the one who likes Biology subject who does not preparation for post office department. Case2 eliminated

There is one candidate who is different between the one who is preparing for UPSC and the one who likes Chemistry. The one candidate in this group who is preparing for the banking department.

N does not enroll in April. Only two candidates have enrolled between N and Q

The one who is preparing for the Railway department and likes G.A subject enrolling on an odd number date of the month which has 31 days.

The one who likes G.A subject enrolled on a date just after the one who likes Math subject who does not prepare for the Delhi police department.

S and Q enrolling dates are same but in different months. O enrolled on one of the days before S’ enrolled. Only one person enrolled between O and P.

Less than four candidates have enrolled between P and U.

Case1 eliminated. In case 1, O enrolled before S, but S enrolled on 16th of March and no date is prior to that.

There are two candidates who are different between P and the one who is preparing for Delhi police department. The one who likes Physics subject enrolled on a date just before the one who is preparing for Delhi police department.

There are two candidates who are different between O and the one who likes the Reasoning subject who enrolled on date just after the one who is preparing for the post office department.

The one who preparation for LIC department also likes an English subject who does not enroll on an odd number date. The one who likes Hindi subjects enrolled on a date just after the one who is preparing for the LIC department.

There are two candidates different between the one who preparation for SSC department and Electricity department who just enrolled on date just after the one who likes Biology subject who does not preparation for post office department. Case2 eliminated

There is one candidate who is different between the one who preparation for UPSC and the one who likes Chemistry subject. The one candidate in this group who is preparing for the banking department.

On combining: P < Q < R > S < T = U > V > W = X < Y < Z

Conclusions:

I. P < T → False (as P < Q < R > S < T à therefore we cannot determine the relation between P and T)

II. R > Z → False (as R > S < T = U > V > W = X < Y < Z à therefore we cannot determine the relation between R and Z)

III. U < Z → False (as U > V > W = X < Y < Z à therefore we cannot determine the relation between U and Z)