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MAH-CET MBA Mock Test- 10


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MAH-CET MBA Mock Test- 10 - Question 1

Directions: Read the following information carefully and answer the question given below-

Eleven persons A, B, C, D, E, F, G, H, I, J and K were living in a building with 11 floors. The floors were numbered 1-11 from bottom to top. None of the floors was vacant and only one person lived on each floor.

The floor number of H was thrice as that of the floor number of J. G lived just below J. There are 2 floors between A and B. A lived on one of the floors above B. Sum of the floor numbers of F and G was equal to the floor number of A. K lived just above E. I lived on one of the prime numbered floors. Only 2 persons lived between D and I. I lived on one of the floors above D. At least 4 persons lived below E. A lives one of the floor above floor 6.

How many persons lived above A?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 1
  • The floor number of H was thrice as that of the floor number of J. Thus, possible combinations can be (9, 3) (6, 2) (3, 1)
  • G lived just below J. Thus, J cannot be on floor 1

  • I lived on one of the prime numbered floors
  • Only 2 persons lived between D and I. I lived on one of the floors above D

  • A and B lived at a gap of 2 floors. A lived on one of the floors above B
  • Sum of the floor numbers of F and G was equal to the floor number of A
  • A lives one of the floor above floor 6.

  • K lived just above E, thus, case 1 and 2a become invalid.
  • At least 4 persons lived below E, thus case 2 becomes invalid.

Thus, the final arrangement is as follows:

3 persons lived above A

MAH-CET MBA Mock Test- 10 - Question 2

Directions: Read the following information carefully and answer the question given below-

Eleven persons A, B, C, D, E, F, G, H, I, J and K were living in a building with 11 floors. The floors were numbered 1-11 from bottom to top. None of the floors was vacant and only one person lived on each floor.

The floor number of H was thrice as that of the floor number of J. G lived just below J. There are 2 floors between A and B. A lived on one of the floors above B. Sum of the floor numbers of F and G was equal to the floor number of A. K lived just above E. I lived on one of the prime numbered floors. Only 2 persons lived between D and I. I lived on one of the floors above D. At least 4 persons lived below E. A lives one of the floor above floor 6.

What was the floor number of B?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 2
  • The floor number of H was thrice as that of the floor number of J. Thus, possible combinations can be (9, 3) (6, 2) (3, 1)
  • G lived just below J. Thus, J cannot be on floor 1

  • I lived on one of the prime numbered floors
  • Only 2 persons lived between D and I. I lived on one of the floors above D

  • A and B lived at a gap of 2 floors. A lived on one of the floors above B
  • Sum of the floor numbers of F and G was equal to the floor number of A
  • A lives one of the floor above floor 6.

  • K lived just above E, thus, case 1 and 2a become invalid.
  • At least 4 persons lived below E, thus case 2 becomes invalid.

Thus, the final arrangement is as follows:

B lived on floor number 5

MAH-CET MBA Mock Test- 10 - Question 3

Directions: Read the following information carefully and answer the question given below-

Eleven persons A, B, C, D, E, F, G, H, I, J and K were living in a building with 11 floors. The floors were numbered 1-11 from bottom to top. None of the floors was vacant and only one person lived on each floor.

The floor number of H was thrice as that of the floor number of J. G lived just below J. There are 2 floors between A and B. A lived on one of the floors above B. Sum of the floor numbers of F and G was equal to the floor number of A. K lived just above E. I lived on one of the prime numbered floors. Only 2 persons lived between D and I. I lived on one of the floors above D. At least 4 persons lived below E. A lives one of the floor above floor 6.

Who among the following lived on floor 7?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 3
  • The floor number of H was thrice as that of the floor number of J. Thus, possible combinations can be (9, 3) (6, 2) (3, 1)
  • G lived just below J. Thus, J cannot be on floor 1

  • I lived on one of the prime numbered floors
  • Only 2 persons lived between D and I. I lived on one of the floors above D

  • A and B lived at a gap of 2 floors. A lived on one of the floors above B
  • Sum of the floor numbers of F and G was equal to the floor number of A
  • A lives one of the floor above floor 6.

  • K lived just above E, thus, case 1 and 2a become invalid.
  • At least 4 persons lived below E, thus case 2 becomes invalid.

Thus, the final arrangement is as follows:

I lived on floor 7

MAH-CET MBA Mock Test- 10 - Question 4

Directions: Read the following information carefully and answer the question given below-

Eleven persons A, B, C, D, E, F, G, H, I, J and K were living in a building with 11 floors. The floors were numbered 1-11 from bottom to top. None of the floors was vacant and only one person lived on each floor.

The floor number of H was thrice as that of the floor number of J. G lived just below J. There are 2 floors between A and B. A lived on one of the floors above B. Sum of the floor numbers of F and G was equal to the floor number of A. K lived just above E. I lived on one of the prime numbered floors. Only 2 persons lived between D and I. I lived on one of the floors above D. At least 4 persons lived below E. A lives one of the floor above floor 6.

Who lived on the lowest floor?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 4
  • The floor number of H was thrice as that of the floor number of J. Thus, possible combinations can be (9, 3) (6, 2) (3, 1)
  • G lived just below J. Thus, J cannot be on floor 1

  • I lived on one of the prime numbered floors
  • Only 2 persons lived between D and I. I lived on one of the floors above D

  • A and B lived at a gap of 2 floors. A lived on one of the floors above B
  • Sum of the floor numbers of F and G was equal to the floor number of A
  • A lives one of the floor above floor 6.

  • K lived just above E, thus, case 1 and 2a become invalid.
  • At least 4 persons lived below E, thus case 2 becomes invalid.

Thus, the final arrangement is as follows:

C lived on the lowest floor

MAH-CET MBA Mock Test- 10 - Question 5

Directions: Read the following information carefully and answer the question given below-

Eleven persons A, B, C, D, E, F, G, H, I, J and K were living in a building with 11 floors. The floors were numbered 1-11 from bottom to top. None of the floors was vacant and only one person lived on each floor.

The floor number of H was thrice as that of the floor number of J. G lived just below J. There are 2 floors between A and B. A lived on one of the floors above B. Sum of the floor numbers of F and G was equal to the floor number of A. K lived just above E. I lived on one of the prime numbered floors. Only 2 persons lived between D and I. I lived on one of the floors above D. At least 4 persons lived below E. A lives one of the floor above floor 6.

What was the difference between the floor numbers of A and F?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 5
  • The floor number of H was thrice as that of the floor number of J. Thus, possible combinations can be (9, 3) (6, 2) (3, 1)
  • G lived just below J. Thus, J cannot be on floor 1

  • I lived on one of the prime numbered floors
  • Only 2 persons lived between D and I. I lived on one of the floors above D

  • A and B lived at a gap of 2 floors. A lived on one of the floors above B
  • Sum of the floor numbers of F and G was equal to the floor number of A
  • A lives one of the floor above floor 6.

  • K lived just above E, thus, case 1 and 2a become invalid.
  • At least 4 persons lived below E, thus case 2 becomes invalid.

Thus, the final arrangement is as follows:

The floor number of A was 8 and the floor number of F was 6 required difference = 2

MAH-CET MBA Mock Test- 10 - Question 6

In the question below are given two statements (I) and (II). These statements may be either independent causes or may be effects of independent causes or a common cause. One of these statements may be the effect of the other statement. Read both the statements and decide which of the following answer choice correctly depicts the relationship between these two statements.

I. There have been three cases of road accidents after dark in neighbourhood Y.

II. Most of the streetlamps in neighbourhood Y do not work.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 6 It is clear that the lack of streetlamps in neighbourhood Y are responsible for the occurrence of road accidents in the area after dark. So, II is the cause, and I, its effect.
MAH-CET MBA Mock Test- 10 - Question 7

Directions: Read the following information carefully and answer the question given below-

6 persons A, B, C, D, E and F have different heights as shown below.

D > F > A > C > B > E

Each one plays different games among Cricket, Hockey, Tennis, Badminton, Volleyball and Basketball not necessarily in the same order. Each one has different weights also.

Only 2 persons are lighter than tallest person.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

Shortest person is lightest whereas B is 2nd heaviest.

A is heavier than all the any person who is taller than him, but A is not heaviest.

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Who plays Tennis?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 7 Only 2 persons are lighter than tallest person.

So, only 2 persons are lighter than D.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

So, D = basketball

B = Badminton

Shortest person is lightest whereas B is 2nd heaviest.

So, E = lightest and B = 2nd heaviest

So, we get: -

A is heavier than all the any person who is taller than him, but A is not heaviest.

Now, D and F are taller than A. So, A is heavier than D and F.

So, we get: -

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Final Solution:

E plays Tennis.

MAH-CET MBA Mock Test- 10 - Question 8

Directions: Read the following information carefully and answer the question given below-

6 persons A, B, C, D, E and F have different heights as shown below.

D > F > A > C > B > E

Each one plays different games among Cricket, Hockey, Tennis, Badminton, Volleyball and Basketball not necessarily in the same order. Each one has different weights also.

Only 2 persons are lighter than tallest person.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

Shortest person is lightest whereas B is 2nd heaviest.

A is heavier than all the any person who is taller than him, but A is not heaviest.

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Who among the following is heavier than A?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 8 Only 2 persons are lighter than tallest person.

So, only 2 persons are lighter than D.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

So, D = basketball

B = Badminton

Shortest person is lightest whereas B is 2nd heaviest.

So, E = lightest and B = 2nd heaviest

So, we get:-

A is heavier than all the any person who is taller than him, but A is not heaviest.

Now, D and F are taller than A. So, A is heavier than D and F.

So, we get:-

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Final Solution:

Only B is heavier than A among the above-mentioned persons.

MAH-CET MBA Mock Test- 10 - Question 9

Directions: Read the following information carefully and answer the question given below-

6 persons A, B, C, D, E and F have different heights as shown below.

D > F > A > C > B > E

Each one plays different games among Cricket, Hockey, Tennis, Badminton, Volleyball and Basketball not necessarily in the same order. Each one has different weights also.

Only 2 persons are lighter than tallest person.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

Shortest person is lightest whereas B is 2nd heaviest.

A is heavier than all the any person who is taller than him, but A is not heaviest.

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Who plays Volleyball?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 9 Only 2 persons are lighter than tallest person.

So, only 2 persons are lighter than D.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

So, D = basketball

B = Badminton

Shortest person is lightest whereas B is 2nd heaviest.

So, E = lightest and B = 2nd heaviest

So, we get:-

A is heavier than all the any person who is taller than him, but A is not heaviest.

Now, D and F are taller than A. So, A is heavier than D and F.

So, we get:-

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Final Solution:

F plays Volleyball.

MAH-CET MBA Mock Test- 10 - Question 10

Directions: Read the following information carefully and answer the question given below-

6 persons A, B, C, D, E and F have different heights as shown below.

D > F > A > C > B > E

Each one plays different games among Cricket, Hockey, Tennis, Badminton, Volleyball and Basketball not necessarily in the same order. Each one has different weights also.

Only 2 persons are lighter than tallest person.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

Shortest person is lightest whereas B is 2nd heaviest.

A is heavier than all the any person who is taller than him, but A is not heaviest.

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Who plays Hockey?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 10 Only 2 persons are lighter than tallest person.

So, only 2 persons are lighter than D.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

So, D = basketball

B = Badminton

Shortest person is lightest whereas B is 2nd heaviest.

So, E = lightest and B = 2nd heaviest

So, we get:-

A is heavier than all the any person who is taller than him, but A is not heaviest.

Now, D and F are taller than A. So, A is heavier than D and F.

So, we get:-

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Final Solution:

A plays Hockey.

MAH-CET MBA Mock Test- 10 - Question 11

Directions: Read the following information carefully and answer the question given below-

6 persons A, B, C, D, E and F have different heights as shown below.

D > F > A > C > B > E

Each one plays different games among Cricket, Hockey, Tennis, Badminton, Volleyball and Basketball not necessarily in the same order. Each one has different weights also.

Only 2 persons are lighter than tallest person.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

Shortest person is lightest whereas B is 2nd heaviest.

A is heavier than all the any person who is taller than him, but A is not heaviest.

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Who among the following is lighter than F?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 11 Only 2 persons are lighter than tallest person.

So, only 2 persons are lighter than D.

Tallest person plays Basketball whereas 2nd shortest person plays badminton.

So, D = basketball

B = Badminton

Shortest person is lightest whereas B is 2nd heaviest.

So, E = lightest and B = 2nd heaviest

So, we get:-

A is heavier than all the any person who is taller than him, but A is not heaviest.

Now, D and F are taller than A. So, A is heavier than D and F.

So, we get:-

Heaviest person plays Cricket.

Neither E nor A plays Volleyball.

Hockey player is heavier than Tennis player.

Final Solution:

E, who plays Tennis, is lighter than F.

MAH-CET MBA Mock Test- 10 - Question 12

Directions: Read the following information carefully and answer the question given below-

10 friends A, B, C, D, E, F, G, H, I & J are sitting in 2 parallel rows such that each person in 1 row faces one person of other row. There are 5 people sitting in each row.

C is 2nd to right of A. E is not adjacent to H. G doesn't face I. Person sitting immediate left of H faces J. J & D faces opposite direction. E is 3rd to right of D. G sits 3rd to left of F. Neither E nor D is adjacent to C.E and B are not diagonally opposite to each other.

Four of the following are similar in some way. Find the odd one.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 12 C is 2nd to right of A.

  • G sits 3rd to left of F.
  • E is 3rd to right of D.
  • Neither E nor D is adjacent to C.

This means G & F will be in one row & E & D will be in another row.

So case 3 is eliminated.

  • Person sitting immediate left of H faces J.
  • J & D faces opposite direction.

  • E is 3rd to right of D.
  • E is not adjacent to H.

So case 2 is eliminated.

  • G doesn't face I.

In option (e), F & I are sitting in different rows, while in other options both the persons are sitting in same row.

MAH-CET MBA Mock Test- 10 - Question 13

Directions: Read the following information carefully and answer the question given below-

10 friends A, B, C, D, E, F, G, H, I & J are sitting in 2 parallel rows such that each person in 1 row faces one person of other row. There are 5 people sitting in each row.

C is 2nd to right of A. E is not adjacent to H. G doesn't face I. Person sitting immediate left of H faces J. J & D faces opposite direction. E is 3rd to right of D. G sits 3rd to left of F. Neither E nor D is adjacent to C.E and B are not diagonally opposite to each other.

Who is 2nd to right of H?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 13 C is 2nd to right of A.

  • G sits 3rd to left of F.
  • E is 3rd to right of D.
  • Neither E nor D is adjacent to C.

This means G & F will be in one row & E & D will be in another row.

So case 3 is eliminated.

  • Person sitting immediate left of H faces J.
  • J & D faces opposite direction.

  • E is 3rd to right of D.
  • E is not adjacent to H.

So case 2 is eliminated.

  • G doesn't face I.

E is 2nd to right of H

MAH-CET MBA Mock Test- 10 - Question 14

Directions: Read the following information carefully and answer the question given below-

10 friends A, B, C, D, E, F, G, H, I & J are sitting in 2 parallel rows such that each person in 1 row faces one person of other row. There are 5 people sitting in each row.

C is 2nd to right of A. E is not adjacent to H. G doesn't face I. Person sitting immediate left of H faces J. J & D faces opposite direction. E is 3rd to right of D. G sits 3rd to left of F. Neither E nor D is adjacent to C.E and B are not diagonally opposite to each other.

Who sits 3rd to right of the person who is sitting opposite to F?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 14 C is 2nd to right of A.

  • G sits 3rd to left of F.
  • E is 3rd to right of D.
  • Neither E nor D is adjacent to C.

This means G & F will be in one row & E & D will be in another row.

So case 3 is eliminated.

  • Person sitting immediate left of H faces J.
  • J & D faces opposite direction.

  • E is 3rd to right of D.
  • E is not adjacent to H.

So case 2 is eliminated.

  • G doesn't face I.

B sits 3rd to right of the person who is sitting opposite to F.

MAH-CET MBA Mock Test- 10 - Question 15

Directions: Read the following information carefully and answer the question given below-

10 friends A, B, C, D, E, F, G, H, I & J are sitting in 2 parallel rows such that each person in 1 row faces one person of other row. There are 5 people sitting in each row.

C is 2nd to right of A. E is not adjacent to H. G doesn't face I. Person sitting immediate left of H faces J. J & D faces opposite direction. E is 3rd to right of D. G sits 3rd to left of F. Neither E nor D is adjacent to C.E and B are not diagonally opposite to each other.

Which of the following statements is/are correct?

I. A & I are sitting in same row

II. B & D are adjacent to each other

III. A faces H.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 15
  • C is 2nd to right of A.

  • G sits 3rd to left of F.
  • E is 3rd to right of D.
  • Neither E nor D is adjacent to C.

This means G & F will be in one row & E & D will be in another row.

So case 3 is eliminated.

  • Person sitting immediate left of H faces J.
  • J & D faces opposite direction.

  • E is 3rd to right of D.
  • E is not adjacent to H.

So case 2 is eliminated.

  • G doesn't face I.

None of these statements are correct.

MAH-CET MBA Mock Test- 10 - Question 16

Directions: Read the following information carefully and answer the question given below-

10 friends A, B, C, D, E, F, G, H, I & J are sitting in 2 parallel rows such that each person in 1 row faces one person of other row. There are 5 people sitting in each row.

C is 2nd to right of A. E is not adjacent to H. G doesn't face I. Person sitting immediate left of H faces J. J & D faces opposite direction. E is 3rd to right of D. G sits 3rd to left of F. Neither E nor D is adjacent to C.E and B are not diagonally opposite to each other.

D faces ____

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 16
  • C is 2nd to right of A.

  • G sits 3rd to left of F.
  • E is 3rd to right of D.
  • Neither E nor D is adjacent to C.

This means G & F will be in one row & E & D will be in another row.

So case 3 is eliminated.

  • Person sitting immediate left of H faces J.
  • J & D faces opposite direction.

  • E is 3rd to right of D.
  • E is not adjacent to H.

So case 2 is eliminated.

  • G doesn't face I.

D faces J.

MAH-CET MBA Mock Test- 10 - Question 17

A 2014 peer-reviewed study in the Journal of the American Medical Association found that habitual violent video game playing had a causal link with increased, long-term, aggressive behavior. Several peer-reviewed studies have shown that children who play M-rated games are more likely to bully and cyberbully their peers, get into physical fights, be hostile, argue with teachers, and show aggression towards their peers throughout the school year.

Which of the following statements is the author least likely to agree with?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 17
  • The author is likely to agree with Options B and E, as they support the author's view that violence in video games increases aggressive behaviour. Options B and E are eliminated.

  • The author is not likely to agree with Option 3. However, this statement is about a court ruling. As the author is primarily concerned with providing findings of a scientific study, a better answer would entail a scientific report or study that claims that violence in video games does not increase aggression, as the author would be more likely to disagree with this. Thus, we keep Option 3 on hold, for want of a better option.

  • The author is unlikely to agree with Option 4, however, this statement is not supported by any scientific information or evidence. Therefore, the author would not agree with it, but we keep this on hold for want of a better option.

  • Option 1 is the right answer, as it gives an example of a study that claims that violence in video games does not increase aggression. The author is least likely to agree with this option than Options C and D.

MAH-CET MBA Mock Test- 10 - Question 18

Directions: Read the following information carefully and answer the question given below-

Eight members - A, B, C, D, E, F, G and H of a family consists of three generations in which either both or none of the parents are alive. Every married couple has a child.

B is married to the only brother of H. F is daughter-in-law of B. C is nephew of G, who is unmarried. D is brother-in-law of G. C and E are siblings. E is the daughter of D, who is not married to B.

Who is daughter-in-law of A?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 18 B is married to the only brother of H. F is daughter-in-law of B.

C is nephew of G, who is unmarried. D is brother-in-law of G. C and E are siblings. E is the daughter of D, who is not married to B.

F is daughter-in-law of A.

MAH-CET MBA Mock Test- 10 - Question 19

Directions: Read the following information carefully and answer the question given below-

Eight members - A, B, C, D, E, F, G and H of a family consists of three generations in which either both or none of the parents are alive. Every married couple has a child.

B is married to the only brother of H. F is daughter-in-law of B. C is nephew of G, who is unmarried. D is brother-in-law of G. C and E are siblings. E is the daughter of D, who is not married to B.

If H is married to P and have daughter Q, then how Q is related to D?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 19 B is married to the only brother of H. F is daughter-in-law of B.

C is nephew of G, who is unmarried. D is brother-in-law of G. C and E are siblings. E is the daughter of D, who is not married to B.

D and Q are cousin

MAH-CET MBA Mock Test- 10 - Question 20

Directions: Read the following information carefully and answer the question given below-

Eight members - A, B, C, D, E, F, G and H of a family consists of three generations in which either both or none of the parents are alive. Every married couple has a child.

B is married to the only brother of H. F is daughter-in-law of B. C is nephew of G, who is unmarried. D is brother-in-law of G. C and E are siblings. E is the daughter of D, who is not married to B.

How C is related to F?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 20 B is married to the only brother of H. F is daughter-in-law of B.

C is nephew of G, who is unmarried. D is brother-in-law of G. C and E are siblings. E is the daughter of D, who is not married to B.

C is son of F.

MAH-CET MBA Mock Test- 10 - Question 21

In each of the following questions, two statements I and II are given. There may be cause and effect relationships between the two statements. These two statements may be the effect of same cause or independent causes. These statements may be independent causes without having any relationship. Read both the statements in each question and mark your answer accordingly

Statement I: The food prices touched an all time high during this weekend.

Statement II: Many shops were raided and adulterated food items were seized.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 21 Both the statements are effects of different causes. Food prices might have touched an all time high because of inflation and statement II could be because of the complaints against the shops.
MAH-CET MBA Mock Test- 10 - Question 22

Directions: Study the following information carefully and answer the questions given below:

In a certain code language:

"shirt coat call jacket" is written as "tpr spr bpr kpr"

"jacket coat call date" is written as "tpr kpr ppr spr"

"coat shirt page paper" is written as "bpr apr tpr mpr"

"kpr" is the code of which of the following word?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 22 "shirt coat call jacket" = "tpr spr bpr kpr" ----(1)

"jacket coat call date" = "tpr kpr ppr spr" ---(2)

"coat shirt page paper" = "bpr apr tpr mpr" ----(3)

From (1) and (2), "date" = ppr

From (1), (2) and (3), "coat" = tpr

From (1) and (3), "shirt" = bpr

From (1) and (2), "call/jacket"="kpr/spr"

From (3), "page/paper" = apr/mpr

"kpr" is the code of "call/jacket".

MAH-CET MBA Mock Test- 10 - Question 23

Directions: Study the following information carefully and answer the questions given below:

In a certain code language:

"shirt coat call jacket" is written as "tpr spr bpr kpr"

"jacket coat call date" is written as "tpr kpr ppr spr"

"coat shirt page paper" is written as "bpr apr tpr mpr"

What is the code of "date"?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 23 "shirt coat call jacket" = "tpr spr bpr kpr" ----(1)

"jacket coat call date" = "tpr kpr ppr spr" ---(2)

"coat shirt page paper" = "bpr apr tpr mpr" ----(3)

From (1) and (2), "date" = ppr

From (1), (2) and (3), "coat" = tpr

From (1) and (3), "shirt" = bpr

From (1) and (2), "call/jacket"="kpr/spr"

From (3), "page/paper" = apr/mpr

The code of "date" is "ppr".

MAH-CET MBA Mock Test- 10 - Question 24

Directions: Study the following information carefully and answer the questions given below:

In a certain code language:

"shirt coat call jacket" is written as "tpr spr bpr kpr"

"jacket coat call date" is written as "tpr kpr ppr spr"

"coat shirt page paper" is written as "bpr apr tpr mpr"

If "page call shirt" is written as "apr spr bpr" then what is the code of "paper jacket coat"?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 24 "shirt coat call jacket" = "tpr spr bpr kpr" ----(1)

"jacket coat call date" = "tpr kpr ppr spr" ---(2)

"coat shirt page paper" = "bpr apr tpr mpr" ----(3)

From (1) and (2), "date" = ppr

From (1), (2) and (3), "coat" = tpr

From (1) and (3), "shirt" = bpr

From (1) and (2), "call/jacket" = "kpr/spr"

From (3), "page/paper" = apr/mpr

"page call shirt" = "apr spr bpr"

It means "page" = apr, "call" = spr, "shirt" = bpr

Then "paper" = mpr, "jacket" = kpr and "coat" = tpr from the given codes in the directions.

Hence, option 3.

MAH-CET MBA Mock Test- 10 - Question 25

Directions: Read the following information carefully and answer the question given below-

Below are given passages followed by several possible inferences which can be drawn from the facts stated in the passage. You have to examine each inference separately in context of the passage and decide upon its degree of truth or falsity and mark the correct answer.

Fourteen new models in the next 15 months. And all of them are bringing in spanking new releases—not ageing models that are being discontinued in western markets. While no manufacturer wants to give away details about exactly which variant within each model it plans to launch, the price it will charge, and when exactly the models will be launched.

Model of a car plays an important role in its selection by customers.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 25 It can be inferred that model of a car plays an important role in its selection by customers as the companies are planning to launch many new models.
MAH-CET MBA Mock Test- 10 - Question 26

Directions: Read the following information carefully and answer the question given below-

Below are given passages followed by several possible inferences which can be drawn from the facts stated in the passage. You have to examine each inference separately in context of the passage and decide upon its degree of truth or falsity and mark the correct answer.

Fourteen new models in the next 15 months. And all of them are bringing in spanking new releases—not ageing models that are being discontinued in western markets. While no manufacturer wants to give away details about exactly which variant within each model it plans to launch, the price it will charge, and when exactly the models will be launched.

Customers prefer cheap cars.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 26 There is no sufficient information in the passage that customers prefer cheap cars.
MAH-CET MBA Mock Test- 10 - Question 27

In the question below, three statements are given followed by five conclusions in the options. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read the conclusions and decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.

Statements:

Only a few stairs are fogs.

Some tastes are joints.

All fogs are tastes.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 27 Only a few stairs are fogs, i.e., some stairs will definitely not be fogs. Some tastes are joints. All fogs are tastes.

Some stairs are joints is not a definite truth. No taste being stair is a possibility is not true. So, option (a) does not follow.

All joints can never be fogs is not true. At least some stairs are tastes is true. So, option (b) does not follow.

Some joints are stairs is not a definite truth. All joints being tastes is a possibility is true. So, option (c) does not follow.

Some stairs are tastes is true. No joint being fog is a possibility is true. So, option (d) follows.

All tastes can never be joints is not true. At least some joints are fogs is not true. So, option (e) does not follow.

MAH-CET MBA Mock Test- 10 - Question 28

In the question below, three statements are given followed by five conclusions in the options. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read the conclusions and decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.

Statements:

Only a few boxes are closures.

All boxes are answers.

No queries are answers.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 28 Only a few boxes are closures, i.e., some boxes will definitely not be closures. All boxes are answers. No queries are answers.

All queries can never be closures is not true.

Some closures are queries is not a definite truth.

All closures can never be answers is not true.

Some queries being boxes is a possibility is not true as all boxes are answers.

All closures can never be queries is true as no query is an answer and some closures are answers as some closures are boxes.

So, (e) is the conclusion that follows.

MAH-CET MBA Mock Test- 10 - Question 29

Should the use of performance enhancing drugs be permitted in professional sports?

I. No, because it would be unfair on athletes who do not use these drugs, and excel on their merit.

II. Yes, because adult athletes should have the freedom to take drugs that might have adverse impacts on health.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 29 I is strong as it gives a valid reason for not permitting these drugs. II is not strong as it is absurd; even in a democracy, one cannot be permitted to injure or harm oneself in the name of freedom. Option A is the right answer.
MAH-CET MBA Mock Test- 10 - Question 30

The question below consists of a question and two statements numbered I and II . You have to decide whether the data provided in the statements is sufficient to answer the question.

Charu, Farah, Shalini, Gunjan, Taru, Jayant and Pankaj stand in a row. What is the position of Shalini with respect to Gunjan?

Statement I: Shalini stands neighbouring Charu and Farah. Gunjan stands equidistant from Shalini and Jayant, where Shalini stands to the left of Jayant. Charu stands at the first position from the left side.

Statement II: Charu stands third to the left of Taru. Pankaj stands second to the right of Gunjan. Taru stands to the left of Gunjan.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 30 From statement I

Shalini stands neighbouring Charu and Farah.

Gunjan stands equidistant from Shalini and Jayant, where Shalini stands to the left of Jayant. Charu stands at first position from the left side.

Only possible case is:

Shalini's position with respect to Gunjan is second to her left.

From statement II: In statement II, position of Shalini is not given, we cannot determine position of Shalini with respect to Gunjan.

MAH-CET MBA Mock Test- 10 - Question 31

Directions: Read the following information carefully and answer the question given below-

Six persons - A, B, C, D, E and F are born in different years and each has some amount of money ranging between Rs.100 to Rs.300 (both included). No-one has same amount.

  • The youngest one is born in 1970 and the eldest in 1939. Both these persons are not E.

  • E is born twelve years before the one who has Rs.150.

  • C is four years elder to B who is not born in year 1957.

  • The one born in 1957 has Rs.225 which is Rs.125 more than E had.

  • B has Rs.200 and the one born in 1952 has Rs.50 less than B.

  • C is born twenty-six years after E.

  • E is born in just immediate next year to when A is born.

  • One of them had Rs.300 who is not A.

  • D is younger than F.

Find the sum of amount that F and D had.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 31 Since, the years in which persons are born is given in the hints only, so we draw them out from hints.

Youngest is born in 1970, and eldest in 1939

From the hints we get,

One person is born in 1957 and one in 1952.

One born in 1957 has Rs.225

Amount with E = 225 - 125 = Rs.100

B has Rs.200

One born in 1952 has 200 - 50 = Rs.150

E is born 12 years before the one who has Rs.150

So, E is born in year = 1952 - 12 = 1940

C is born in year = 1940 + 26 = 1966

So, we can arrange the years in order as follows,

Now,

E is born just the next year in which A is born,

So, A is born in 1939

C is four years elder to B, So, B is born in year = 1966 + 4 = 1970

B has Rs.200

Now,

D is younger than F, so D is born in 1957 and F in 1952

One of them had Rs.300 who is not A, so is C

And, we cannot determine the amount which A had.

Final arrangement is,

F had Rs.150 and D had Rs.225

Sum = 150 + 225 = 375

MAH-CET MBA Mock Test- 10 - Question 32

Directions: Read the following information carefully and answer the question given below-

Six persons - A, B, C, D, E and F are born in different years and each has some amount of money ranging between Rs.100 to Rs.300 (both included). No-one has same amount.

  • The youngest one is born in 1970 and the eldest in 1939. Both these persons are not E.

  • E is born twelve years before the one who has Rs.150

  • C is four years elder to B who is not born in year 1957.

  • The one born in 1957 has Rs.225 which is Rs.125 more than E had.

  • B has Rs.200 and the one born in 1952 has Rs.50 less than B.

  • C is born twenty-six years after E.

  • E is born in just immediate next year to when A is born.

  • One of them had Rs.300 who is not A.

  • D is younger than F.

If the given persons are arranged in increasing order of their present ages (assuming the present year to be 2018), then which if the following represents the correct order.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 32 Since, the years in which persons are born is given in the hints only, so we draw them out from hints.

Youngest is born in 1970, and eldest in 1939

From the hints we get,

One person is born in 1957 and one in 1952.

One born in 1957 has Rs.225

Amount with E = 225 - 125 = Rs.100

B has Rs.200

One born in 1952 has 200 - 50 = Rs.150

E is born 12 years before the one who has Rs.150

So, E is born in year = 1952 - 12 = 1940

C is born in year = 1940 + 26 = 1966

So, we can arrange the years in order as follows,

Now,

E is born just the next year in which A is born,

So, A is born in 1939

C is four years elder to B, So, B is born in year = 1966 + 4 = 1970

B has Rs.200

Now,

D is younger than F, so D is born in 1957 and F in 1952

One of them had Rs.300 who is not A, so is C

And, we cannot determine the amount which A had.

Final arrangement is,

Age of B < c="" />< f="" />< e.="" rest="" are="" not="" in="" correct="" />

MAH-CET MBA Mock Test- 10 - Question 33

Directions: Read the following information carefully and answer the question given below-

Six persons - A, B, C, D, E and F are born in different years and each has some amount of money ranging between Rs.100 to Rs.300 (both included). No-one has same amount.

  • The youngest one is born in 1970 and the eldest in 1939. Both these persons are not E.

  • E is born twelve years before the one who has Rs.150

  • C is four years elder to B who is not born in year 1957.

  • The one born in 1957 has Rs.225 which is Rs.125 more than E had.

  • B has Rs.200 and the one born in 1952 has Rs.50 less than B.

  • C is born twenty-six years after E.

  • E is born in just immediate next year to when A is born.

  • One of them had Rs.300 who is not A.

  • D is younger than F.

Who is the youngest person?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 33 Since, the years in which persons are born is given in the hints only, so we draw them out from hints.

Youngest is born in 1970, and eldest in 1939

From the hints we get,

One person is born in 1957 and one in 1952.

One born in 1957 has Rs.225

Amount with E = 225 - 125 = Rs.100

B has Rs.200

One born in 1952 has 200 - 50 = Rs.150

E is born 12 years before the one who has Rs.150

So, E is born in year = 1952 - 12 = 1940

C is born in year = 1940 + 26 = 1966

So, we can arrange the years in order as follows,

Now,

E is born just the next year in which A is born,

So, A is born in 1939

C is four years elder to B, So, B is born in year = 1966 + 4 = 1970

B has Rs.200

Now,

D is younger than F, so D is born in 1957 and F in 1952

One of them had Rs.300 who is not A, so is C

And, we cannot determine the amount which A had.

Final arrangement is,

B is born in 1970 and is youngest

MAH-CET MBA Mock Test- 10 - Question 34

Directions: Read the following information carefully and answer the question given below-

Six persons - A, B, C, D, E and F are born in different years and each has some amount of money ranging between Rs.100 to Rs.300 (both included). No-one has same amount.

  • The youngest one is born in 1970 and the eldest in 1939. Both these persons are not E.

  • E is born twelve years before the one who has Rs.150

  • C is four years elder to B who is not born in year 1957.

  • The one born in 1957 has Rs.225 which is Rs.125 more than E had.

  • B has Rs.200 and the one born in 1952 has Rs.50 less than B.

  • C is born twenty-six years after E.

  • E is born in just immediate next year to when A is born.

  • One of them had Rs.300 who is not A.

  • D is younger than F.

E is born in ______.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 34 Since, the years in which persons are born is given in the hints only, so we draw them out from hints.

Youngest is born in 1970, and eldest in 1939

From the hints we get,

One person is born in 1957 and one in 1952.

One born in 1957 has Rs.225

Amount with E = 225 - 125 = Rs.100

B has Rs.200

One born in 1952 has 200 - 50 = Rs.150

E is born 12 years before the one who has Rs.150

So, E is born in year = 1952 - 12 = 1940

C is born in year = 1940 + 26 = 1966

So, we can arrange the years in order as follows,

Now,

E is born just the next year in which A is born,

So, A is born in 1939

C is four years elder to B, So, B is born in year = 1966 + 4 = 1970

B has Rs.200

Now,

D is younger than F, so D is born in 1957 and F in 1952

One of them had Rs.300 who is not A, so is C

And, we cannot determine the amount which A had.

Final arrangement is,

E is born in 1940

MAH-CET MBA Mock Test- 10 - Question 35

Directions: Read the following information carefully and answer the question given below-

Six persons - A, B, C, D, E and F are born in different years and each has some amount of money ranging between Rs.100 to Rs.300 (both included). No-one has same amount.

  • The youngest one is born in 1970 and the eldest in 1939. Both these persons are not E.

  • E is born twelve years before the one who has Rs.150

  • C is four years elder to B who is not born in year 1957.

  • The one born in 1957 has Rs.225 which is Rs.125 more than E had.

  • B has Rs.200 and the one born in 1952 has Rs.50 less than B.

  • C is born twenty-six years after E.

  • E is born in just immediate next year to when A is born.

  • One of them had Rs.300 who is not A.

  • D is younger than F.

Who among the following are born in 1950s (i.e. between 1950 to 1959)?

I: A

II: F

III: E

IV: D

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 35 Since, the years in which persons are born is given in the hints only, so we draw them out from hints.

Youngest is born in 1970, and eldest in 1939

From the hints we get,

One person is born in 1957 and one in 1952.

One born in 1957 has Rs.225

Amount with E = 225 - 125 = Rs.100

B has Rs.200

One born in 1952 has 200 - 50 = Rs.150

E is born 12 years before the one who has Rs.150

So, E is born in year = 1952 - 12 = 1940

C is born in year = 1940 + 26 = 1966

So, we can arrange the years in order as follows,

Now,

E is born just the next year in which A is born,

So, A is born in 1939

C is four years elder to B, So, B is born in year = 1966 + 4 = 1970

B has Rs.200

Now,

D is younger than F, so D is born in 1957 and F in 1952

One of them had Rs.300 who is not A, so is C

And, we cannot determine the amount which A had.

Final arrangement is,

F is born in 1952 and D in 1957.

MAH-CET MBA Mock Test- 10 - Question 36

The question below consists of 2 statements. You have to decide whether the data provided in the statements are sufficient to answer the question.

Seven members - A, B, C, D, E, F and G of a family consists of two generations in which either both or none of the parents are alive. D have two children. E is not married to F. How E is related to A?

Statement 1: A is the father of C, who is brother in law of F. F is the daughter of B. G is mother in law of E and not married to D.

Statement 2: B is the mother of E. C is son in law of D. F is sister in law of C. G is the mother of C.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 36 If the data either in statement 1 or in statement 2 alone is sufficient to answer the question

From statement 1:

A is the father of C, who is brother in law of F. F is the daughter of B. G is mother in law of E and not married to D. E is not married to F.

E is daughter in law of A

So, statement 1 alone is sufficient to answer the question

From statement 2:

B is the mother of E. C is son in law of D. F is sister in law of C. G is the mother of C.

E is daughter in law of A

So, statement 2 alone is sufficient to answer the question

Hence, the data either in statement 1 or in statement 2 alone are sufficient to answer the question.

MAH-CET MBA Mock Test- 10 - Question 37

The question below consists of a question and two statements numbered I and II. You have to decide whether the data provided in the statements is sufficient to answer the question.

Aman, Biren, Charu, Darshan, Esha and Fiza sit around a circular table. Each of them faces towards the table. How many persons are sitting between Aman and Darshan?

Statement I: Aman sits opposite to Fiza. Biren and Charu does not sit adjacent to each other.

Statement II: Number of persons sitting between Aman and Esha is equal to the number of persons sitting between Charu and Darshan.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 37 From statement I

Biren and Charu does not sit adjacent to each other. That means Biren and Charu sit to the either side of Fiza or Aman. But we cannot determine the exact number of persons sitting between Aman and Darshan.

From statement II: From the statement II we cannot determine the exact number of persons sitting between Aman and Darshan.

From statement I and statement II:

Even using statement I and II together, we cannot determine the answer.

MAH-CET MBA Mock Test- 10 - Question 38

In the given question, a set of conclusions is given. There are five options comprising of three or more statements. You need to choose the option that contains the set of statements from which the given conclusions logically follow.

All pockets being jasmines is not a possibility. All jasmines being stoves is not a possibility.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 38 (1)

This condition is one of the possibilities. In this condition neither of the conclusions follows.

(2)

This condition is one of the possibilities. In this condition neither of the conclusions follows.

(3)

This condition is one of the possibilities. In this condition neither of the conclusions follows.

(4)

In this scenario both the conclusions are true. Some pockets are not stores and all jasmines are stores, hence all pockets can never be jasmines. As no stoves are stores and all jasmines are stores, hence all jasmines being stoves is not a possibility.

The statements given in option 4 leads to the required conclusions.

MAH-CET MBA Mock Test- 10 - Question 39

Each of the questions below consist of a question and two statements 1 and 2 given below it.

Six boxes were kept one above the other in the form of stack. Box U is kept above box T. Which box is kept immediately above box P?

Statement 1: Three boxes were kept between box Q and box T. One box is kept between box R and box T. One box is kept between box U and P. Box P is kept below box Q.

Statement 2: More than two boxes were kept between box S and box U. One box is kept between box T and box R. Box Q is kept above box U.

Read both the statements and mark your answer as:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 39 If the data given in both statements 1 and 2 together are not sufficient to answer the question

From Statement 1:

Three boxes were kept between box Q and box T. One box is kept between box R and box T. One box is kept between box U and P. Box P is kept below box Q.

From three cases, it is not sufficient to answer the question

So, statement 1 alone is not sufficient to answer the question.

From statement 2:

More than two boxes were kept between box S and box U. One box is kept between box T and box R. Box Q is kept above box U.

From the above cases it is not sufficient to answer the question.

So, statement 2 alone is not sufficient to answer the question.

From statement 1 and statement 2 together:

Using 'Three boxes were kept between box Q and box T' in statement 2, we get

From the above cases it is not sufficient to answer the question.

So, statement 1 and statement 2 together are not sufficient to answer the question.

MAH-CET MBA Mock Test- 10 - Question 40

The question below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question:

In a certain code word "You will go there" means 'YO JO KO NO', then how "go" is coded?

I. "Come home will joy" means 'PO KA JO ZA' and "You come to Joy" means 'YO PO KU ZA'.

II. "You come there Happiness" means 'YO PO NO FO' and "You will Joy Forever" means 'YO JO ZA KU'.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 40 From I

You will go there → YO JO KO NO

Come home will joy → PO KA JO ZA

You come to Joy → YO PO KU ZA

We get following results:

You → YO; Come/Joy → PO/ZA; to → KU; will → JO; go/there → KO/NO

Hence, Statement I is not sufficient.

From II:

You will go there → YO JO KO NO

You come there Happiness → YO PO NO FO

You will Joy Forever → YO JO ZA KU

We get following result.

You → YO; will → JO; there → NO; go → KO

Hence, statement II is alone sufficient.

MAH-CET MBA Mock Test- 10 - Question 41

Directions: Read the following information carefully and answer the question given below-

A word arrangement machine when given an input line of words

Rearranges them following a particular rule in each step. The following is an illustration of the input and various steps of rearrangement

INPUT: India Peru Africa Bangladesh China England Uganda Norway

Step I: Africa India Peru Bangladesh China England Uganda Norway

Step II: Africa Peru India Bangladesh China England Uganda Norway

Step III: Africa Peru England India Bangladesh China Uganda Norway

Step IV: Africa Peru England China India Bangladesh Uganda Norway

Step V: Africa Peru England China India Norway Bangladesh Uganda

Step VI: Africa Peru England China India Norway Uganda Bangladesh

As per the rules followed in above steps, answer the questions for the following input:

INPUT: Ice captain umbrella rain glass owl man age

What is the final step?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 41 Looking at the solved input, we can observe from the last step that the words at odd places start with a vowel and arranged alphabetically(A,E,I,O,U).

on the other hand, if we look at the words at even position, we can find that they are consonants and arranged according to the no. of alphabets in a words in increasing order.

So, for the given input:

INPUT: Ice captain umbrella rain glass owl man age

First, we bring the word starting with a vowel alphabetically i.e. age

Step I: age ice captain umbrella rain glass owl man now the consonant with the least no of words brought at the next position.

Step II: age man ice captain umbrella rain glass owl

Step III: age man ice rain captain umbrella glass owl

Step IV: age man ice rain owl captain umbrella glass

Step V: age man ice rain owl glass captain umbrella

Step VI: age man ice rain owl glass umbrella captain age man ice rain owl glass umbrella captain is the final step

MAH-CET MBA Mock Test- 10 - Question 42

Directions: Read the following information carefully and answer the question given below-

A word arrangement machine when given an input line of words

Rearranges them following a particular rule in each step. The following is an illustration of the input and various steps of rearrangement

INPUT: India Peru Africa Bangladesh China England Uganda Norway

Step I: Africa India Peru Bangladesh China England Uganda Norway

Step II: Africa Peru India Bangladesh China England Uganda Norway

Step III: Africa Peru England India Bangladesh China Uganda Norway

Step IV: Africa Peru England China India Bangladesh Uganda Norway

Step V: Africa Peru England China India Norway Bangladesh Uganda

Step VI: Africa Peru England China India Norway Uganda Bangladesh

As per the rules followed in above steps, answer the questions for the following input:

INPUT: Ice captain umbrella rain glass owl man age

In which step is "captain" fourth from the right end?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 42 Looking at the solved input, we can observe from the last step that the words at odd places start with a vowel and arranged alphabetically(A,E,I,O,U).

on the other hand, if we look at the words at even position, we can find that they are consonants and arranged according to the no. of alphabets in a words in increasing order.

So, for the given input:

INPUT: Ice captain umbrella rain glass owl man age

First, we bring the word starting with a vowel alphabetically i.e. age

Step I: age ice captain umbrella rain glass owl man now the consonant with the least no of words brought at the next position.

Step II: age man ice captain umbrella rain glass owl

Step III: age man ice rain captain umbrella glass owl

Step IV: age man ice rain owl captain umbrella glass

Step V: age man ice rain owl glass captain umbrella

Step VI: age man ice rain owl glass umbrella captain

In step III is "captain" fourth from the right end

MAH-CET MBA Mock Test- 10 - Question 43

Directions: Read the following information carefully and answer the question given below-

A word arrangement machine when given an input line of words

Rearranges them following a particular rule in each step. The following is an illustration of the input and various steps of rearrangement

INPUT: India Peru Africa Bangladesh China England Uganda Norway

Step I: Africa India Peru Bangladesh China England Uganda Norway

Step II: Africa Peru India Bangladesh China England Uganda Norway

Step III: Africa Peru England India Bangladesh China Uganda Norway

Step IV: Africa Peru England China India Bangladesh Uganda Norway

Step V: Africa Peru England China India Norway Bangladesh Uganda

Step VI: Africa Peru England China India Norway Uganda Bangladesh

As per the rules followed in above steps, answer the questions for the following input:

INPUT: Ice captain umbrella rain glass owl man age

"age man ice rain owl glass captain umbrella" is which step for the given input?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 43 Looking at the solved input, we can observe from the last step that the words at odd places start with a vowel and arranged alphabetically(A,E,I,O,U).

on the other hand, if we look at the words at even position, we can find that they are consonants and arranged according to the no. of alphabets in a words in increasing order.

So, for the given input:

INPUT: Ice captain umbrella rain glass owl man age

First, we bring the word starting with a vowel alphabetically i.e. age

Step I: age ice captain umbrella rain glass owl man now the consonant with the least no of words brought at the next position.

Step II: age man ice captain umbrella rain glass owl

Step III: age man ice rain captain umbrella glass owl

Step IV: age man ice rain owl captain umbrella glass

Step V: age man ice rain owl glass captain umbrella

Step VI: age man ice rain owl glass umbrella captain

"age man ice rain owl glass captain umbrella" is step V.

MAH-CET MBA Mock Test- 10 - Question 44

Directions: Read the following information carefully and answer the question given below-

A word arrangement machine when given an input line of words

Rearranges them following a particular rule in each step. The following is an illustration of the input and various steps of rearrangement

INPUT: India Peru Africa Bangladesh China England Uganda Norway

Step I: Africa India Peru Bangladesh China England Uganda Norway

Step II: Africa Peru India Bangladesh China England Uganda Norway

Step III: Africa Peru England India Bangladesh China Uganda Norway

Step IV: Africa Peru England China India Bangladesh Uganda Norway

Step V: Africa Peru England China India Norway Bangladesh Uganda

Step VI: Africa Peru England China India Norway Uganda Bangladesh

As per the rules followed in above steps, answer the questions for the following input:

INPUT: Ice captain umbrella rain glass owl man age

What is the second element from right end in step IV?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 44 Looking at the solved input, we can observe from the last step that the words at odd places start with a vowel and arranged alphabetically(A,E,I,O,U).

on the other hand, if we look at the words at even position, we can find that they are consonants and arranged according to the no. of alphabets in a words in increasing order.

So, for the given input:

INPUT: Ice captain umbrella rain glass owl man age

First, we bring the word starting with a vowel alphabetically i.e. age

Step I: age ice captain umbrella rain glass owl man now the consonant with the least no of words brought at the next position.

Step II: age man ice captain umbrella rain glass owl

Step III: age man ice rain captain umbrella glass owl

Step IV: age man ice rain owl captain umbrella glass

Step V: age man ice rain owl glass captain umbrella

Step VI: age man ice rain owl glass umbrella captain

Umbrella is the second element from right end in step IV.

MAH-CET MBA Mock Test- 10 - Question 45

Directions: Read the following information carefully and answer the question given below-

A word arrangement machine when given an input line of words

Rearranges them following a particular rule in each step. The following is an illustration of the input and various steps of rearrangement

INPUT: India Peru Africa Bangladesh China England Uganda Norway

Step I: Africa India Peru Bangladesh China England Uganda Norway

Step II: Africa Peru India Bangladesh China England Uganda Norway

Step III: Africa Peru England India Bangladesh China Uganda Norway

Step IV: Africa Peru England China India Bangladesh Uganda Norway

Step V: Africa Peru England China India Norway Bangladesh Uganda

Step VI: Africa Peru England China India Norway Uganda Bangladesh

As per the rules followed in above steps, answer the questions for the following input:

INPUT: Ice captain umbrella rain glass owl man age

How many steps are required to find the output for the given input?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 45 Looking at the solved input, we can observe from the last step that the words at odd places start with a vowel and arranged alphabetically(A,E,I,O,U).

on the other hand, if we look at the words at even position, we can find that they are consonants and arranged according to the no. of alphabets in a words in increasing order.

So, for the given input:

INPUT: Ice captain umbrella rain glass owl man age

First, we bring the word starting with a vowel alphabetically i.e. age

Step I: age ice captain umbrella rain glass owl man now the consonant with the least no of words brought at the next position.

Step II: age man ice captain umbrella rain glass owl

Step III: age man ice rain captain umbrella glass owl

Step IV: age man ice rain owl captain umbrella glass

Step V: age man ice rain owl glass captain umbrella

Step VI: age man ice rain owl glass umbrella captain

6 steps are there in total for the given input.

MAH-CET MBA Mock Test- 10 - Question 46

The company P has a Mon-Fri work schedule, while company Q has a Mon-Sat work schedule. Due to this reason, the manager of P feels less stressed than that of Q. Also, we know that the more a manager feels in control of his work life, the lesser stress she feels. So, we can conclude that the manager of P feels more in control of his work life than manager of Q.

Which of the following is the flaw in the above reasoning?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 46 Option B is eliminated as this would have been true had the passage concluded that both these factors cause the manager of P to feel more control over her work life. Option C is eliminated as the argument makes no assumption of this sort. Option D is eliminated as the information cited stays relevant and true to the context. Option E is eliminated as the passage itself specifies that the reason why the manager of P feels less stressed than that of Q, is because of shorter work week. Option A is the right answer, as it correctly captures the flaw in the reasoning.
MAH-CET MBA Mock Test- 10 - Question 47

Each of these questions has a statements followed by two assumptions numbered I and II. An assumption is something supposed or taken for granted. Consider the statements and the assumptions following it in each of these questions and then select one of the five options based on the given statement and assumptions.

Statement: It is not always true that adoption of sophisticated technology increases production efficiency.

I. Adoption and sophisticated technology is not a difficult thing to achieve.

II. Production and efficiency can be archived by getting rid of sophisticated technology.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 47 Both the statements I and II does not discuss about the given statement. Hence, both assumptions are invalid.
MAH-CET MBA Mock Test- 10 - Question 48

Directions: Read the following information carefully and answer the question given below-

In the cricket match played between India and Pakistan, first over of 6 balls was bowled by Bhuvneshwar. Ricky and Adam were batting on the pitch. On every ball different number of runs scored among 0, 1, 2, 3, 4 and 6 not necessarily in given order. Only one player can hit a ball and score the runs. If he scores odd number of runs, then he goes to other end of pitch and other batsmen will play the next ball. Runs will be added in the account of the player who played that ball.

Most number of runs were scored on 6th ball of the over. No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over. Ricky score 10 runs more than Adam in that over.

What is the sum of runs scored on 1st and 2nd ball of the over?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 48 No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over.

Since, Adam played 5th ball of the over and no runs were scored on 4th ball of the over. So, Adam played 4th ball as well.

Since, odd number of runs scored on 5th ball, so next ball (i.e. 6th) was played by Ricky.

Most number of runs were scored on 6th ball of the over.

Ricky score 10 runs more than Adam in that over.

Total runs scored in that over = 1 + 2 + 3 + 4 + 6 = 16

So, Ricky scored 13 runs and Adam scored 3 runs in that over.

The only way Adam can score 3 runs is: - he played 3 balls and score 0, 1 and 2 runs.

He cannot play 1st or 2nd ball of the over. Because, then he will face another ball at least and will score more than 3 runs.

So, Adam faced 3rd ball and scored 2 runs on that ball.

Runs remaining to be scored are 3 and 4 which are scored by Ricky. So, Ricky scored 4 and 3 runs respectively on 1st and 2nd ball of the over.

Final Solution:

4 and 3 runs were scored on 1st and 2nd ball of the over respectively. So, require sum is 7.

MAH-CET MBA Mock Test- 10 - Question 49

Directions: Read the following information carefully and answer the question given below-

In the cricket match played between India and Pakistan, first over of 6 balls was bowled by Bhuvneshwar. Ricky and Adam were batting on the pitch. On every ball different number of runs scored among 0, 1, 2, 3, 4 and 6 not necessarily in given order. Only one player can hit a ball and score the runs. If he scores odd number of runs, then he goes to other end of pitch and other batsmen will play the next ball. Runs will be added in the account of the player who played that ball.

Most number of runs were scored on 6th ball of the over. No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over. Ricky score 10 runs more than Adam in that over.

Find the correct combination of player and run scored on any single ball:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 49 No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over.

Since, Adam played 5th ball of the over and no runs were scored on 4th ball of the over. So, Adam played 4th ball as well.

Since, odd number of runs scored on 5th ball, so next ball (i.e. 6th) was played by Ricky.

Most number of runs were scored on 6th ball of the over.

Ricky score 10 runs more than Adam in that over.

Total runs scored in that over = 1 + 2 + 3 + 4 + 6 = 16

So, Ricky scored 13 runs and Adam scored 3 runs in that over.

The only way Adam can score 3 runs is: - he played 3 balls and score 0, 1 and 2 runs.

He cannot play 1st or 2nd ball of the over. Because, then he will face another ball at-least and will score more than 3 runs.

So, Adam faced 3rd ball and scored 2 runs on that ball.

Runs remaining to be scored are 3 and 4 which are scored by Ricky. So, Ricky scored 4 and 3 runs respectively on 1st and 2nd ball of the over.

Final Solution:

None of the given combination is correct.

MAH-CET MBA Mock Test- 10 - Question 50

Directions: Read the following information carefully and answer the question given below-

In the cricket match played between India and Pakistan, first over of 6 balls was bowled by Bhuvneshwar. Ricky and Adam were batting on the pitch. On every ball different number of runs scored among 0, 1, 2, 3, 4 and 6 not necessarily in given order. Only one player can hit a ball and score the runs. If he scores odd number of runs, then he goes to other end of pitch and other batsmen will play the next ball. Runs will be added in the account of the player who played that ball.

Most number of runs were scored on 6th ball of the over. No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over. Ricky score 10 runs more than Adam in that over.

How many runs were scored on odd numbered balls? (i.e. 1st, 3rd and 5th ball of the over)

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 50 No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over.

Since, Adam played 5th ball of the over and no runs were scored on 4th ball of the over. So, Adam played 4th ball as well.

Since, odd number of runs scored on 5th ball, so next ball (i.e. 6th) was played by Ricky.

Most number of runs were scored on 6th ball of the over.

Ricky score 10 runs more than Adam in that over.

Total runs scored in that over = 1 + 2 + 3 + 4 + 6 = 16

So, Ricky scored 13 runs and Adam scored 3 runs in that over.

The only way Adam can score 3 runs is: - he played 3 balls and score 0, 1 and 2 runs.

He cannot play 1st or 2nd ball of the over. Because, then he will face another ball at-least and will score more than 3 runs.

So, Adam faced 3rd ball and scored 2 runs on that ball.

Runs remaining to be scored are 3 and 4 which are scored by Ricky. So, Ricky scored 4 and 3 runs respectively on 1st and 2nd ball of the over.

Final Solution:

4 + 2 + 1 = 7 runs scored on odd numbered balls.

MAH-CET MBA Mock Test- 10 - Question 51

Directions: Read the following information carefully and answer the question given below-

In the cricket match played between India and Pakistan, first over of 6 balls was bowled by Bhuvneshwar. Ricky and Adam were batting on the pitch. On every ball different number of runs scored among 0, 1, 2, 3, 4 and 6 not necessarily in given order. Only one player can hit a ball and score the runs. If he scores odd number of runs, then he goes to other end of pitch and other batsmen will play the next ball. Runs will be added in the account of the player who played that ball.

Most number of runs were scored on 6th ball of the over. No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over. Ricky score 10 runs more than Adam in that over.

Who played 2nd ball of the over? How many runs did he scored?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 51 No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over.

Since, Adam played 5th ball of the over and no runs were scored on 4th ball of the over. So, Adam played 4th ball as well.

Since, odd number of runs scored on 5th ball, so next ball (i.e. 6th) was played by Ricky.

Most number of runs were scored on 6th ball of the over.

Ricky score 10 runs more than Adam in that over.

Total runs scored in that over = 1 + 2 + 3 + 4 + 6 = 16

So, Ricky scored 13 runs and Adam scored 3 runs in that over.

The only way Adam can score 3 runs is: - he played 3 balls and score 0, 1 and 2 runs.

He cannot play 1st or 2nd ball of the over. Because, then he will face another ball at-least and will score more than 3 runs.

So, Adam faced 3rd ball and scored 2 runs on that ball.

Runs remaining to be scored are 3 and 4 which are scored by Ricky. So, Ricky scored 4 and 3 runs respectively on 1st and 2nd ball of the over.

Final Solution:

Ricky played 2nd ball of the over and he scored 3 runs.

MAH-CET MBA Mock Test- 10 - Question 52

Directions: Read the following information carefully and answer the question given below-

In the cricket match played between India and Pakistan, first over of 6 balls was bowled by Bhuvneshwar. Ricky and Adam were batting on the pitch. On every ball different number of runs scored among 0, 1, 2, 3, 4 and 6 not necessarily in given order. Only one player can hit a ball and score the runs. If he scores odd number of runs, then he goes to other end of pitch and other batsmen will play the next ball. Runs will be added in the account of the player who played that ball.

Most number of runs were scored on 6th ball of the over. No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over. Ricky score 10 runs more than Adam in that over.

How many balls does Adam played?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 52 No runs were scored on 4th ball of the over. Adam scored 1 run on 5th ball of the over.

Since, Adam played 5th ball of the over and no runs were scored on 4th ball of the over. So, Adam played 4th ball as well.

Since, odd number of runs scored on 5th ball, so next ball (i.e. 6th) was played by Ricky.

Most number of runs were scored on 6th ball of the over.

Ricky score 10 runs more than Adam in that over.

Total runs scored in that over = 1 + 2 + 3 + 4 + 6 = 16

So, Ricky scored 13 runs and Adam scored 3 runs in that over.

The only way Adam can score 3 runs is: - he played 3 balls and score 0, 1 and 2 runs.

He cannot play 1st or 2nd ball of the over. Because, then he will face another ball at-least and will score more than 3 runs.

So, Adam faced 3rd ball and scored 2 runs on that ball.

Runs remaining to be scored are 3 and 4 which are scored by Ricky. So, Ricky scored 4 and 3 runs respectively on 1st and 2nd ball of the over.

Final Solution:

Both Ricky and Adams played 3 balls each.

MAH-CET MBA Mock Test- 10 - Question 53

Directions: Read the following information carefully and answer the question given below-

In these questions the symbols @, #, Ú, $ and © are used with different meanings as follow:

‘A @ B’ means ‘A is not greater than B’.

‘A # B’ means ‘A is neither greater than nor equal to B’.

‘A ÚB’ means ‘A is not smaller than B’.

‘A $ B’ means ‘A is neither smaller than nor equal to B’.

‘A © B’ means ‘A is neither greater than nor smaller than B’.

Now in each of the following questions assuming the given statements to be true, find out which of the conclusions I, II, III given below them is/are definitely true and mark your answer accordingly.

Statements: J Ú E, D @ E, E $ K, K © T

Conclusions: I. J $ D II. J ÚD III. E $ T

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 53 From the given definitions,

A is not greater than B i.e @ means A≤B

A is neither greater than nor equal to B i.e # means A < B

A is not smaller than B i.e Ú means A≥B

A is neither smaller than nor equal to B, i.e $ means A > B

A is neither smaller than nor greater than B i.e © means A=B

Statement : J ≥ E ≥ D , E > K = T

Conclusion: I. J > D False

II. J ≥ D True

III. E > T True

MAH-CET MBA Mock Test- 10 - Question 54

Directions: Read the following information carefully and answer the question given below-

In these questions the symbols @, #, Ú, $ and © are used with different meanings as follow:

‘A @ B’ means ‘A is not greater than B’.

‘A # B’ means ‘A is neither greater than nor equal to B’.

‘A ÚB’ means ‘A is not smaller than B’.

‘A $ B’ means ‘A is neither smaller than nor equal to B’.

‘A © B’ means ‘A is neither greater than nor smaller than B’.

Now in each of the following questions assuming the given statements to be true, find out which of the conclusions I, II, III given below them is/are definitely true and mark your answer accordingly.

Statements: H @ I, I # L, L Ú A, A $ Q

Conclusions: I. H # L II. H ÚL III. Q # H

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 54 From the given definitions,

A is not greater than B i.e @ means A≤B

A is neither greater than nor equal to B i.e # means A < B

A is not smaller than B i.e Ú means A≥B

A is neither smaller than nor equal to B, i.e $ means A > B

A is neither smaller than nor greater than B i.e © means A = B

Statement: H ≤ I < L ≥ A > Q

Conclusion: I. H < L True

II. H ≥ L False

III. Q ≤ H Cannot be determined

MAH-CET MBA Mock Test- 10 - Question 55

Directions: Read the following information carefully and answer the question given below-

In these questions the symbols @, #, Ú, $ and © are used with different meanings as follow:

‘A @ B’ means ‘A is not greater than B’.

‘A # B’ means ‘A is neither greater than nor equal to B’.

‘A ÚB’ means ‘A is not smaller than B’.

‘A $ B’ means ‘A is neither smaller than nor equal to B’.

‘A © B’ means ‘A is neither greater than nor smaller than B’.

Now in each of the following questions assuming the given statements to be true, find out which of the conclusions I, II, III given below them is/are definitely true and mark your answer accordingly.

Statements: V # W, W ÚT, T © K, K @ F

Conclusions: I. T @ V II. T $ V III. F ÚT

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 55 From the given definitions,

A is not greater than B i.e @ means A≤B

A is neither greater than nor equal to B i.e # means A < B

A is not smaller than B i.e Ú means A≥B

A is neither smaller than nor equal to B, i.e $ means A > B

A is neither smaller than nor greater than B i.e © means A=B

Statement: V < W ≥ T = K ≤ F

Conclusion: I. T ≤ V Cannot be determined

II. T > V Cannot be determined

III. F ≥ T True

MAH-CET MBA Mock Test- 10 - Question 56

Sonia travels 7 km eastwards and then turns right and travels 3 km and further turns right and travels 13 km. What is the horizontal distance between the final destination and starting point?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 56

Hence, the horizontal distance is 6 km.

MAH-CET MBA Mock Test- 10 - Question 57

In the given question, a set of conclusions is given. There are five options comprising of three or more statements. You need to choose the option that contains the set of statements from which the given conclusions logically follow.

All bats being keys is not a possibility. No key is a sweet.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 57 The statements given in option 2 lead to the required conclusions.

(1)

We see that in this scenario both the conclusions are false.

(2)

Some bats are not drills hence all bats can never be keys because key is a subset of the larger set 'drill'. No sweet is a drill hence no key is a sweet as well. In this case both the conclusions are true.

(3)

In this scenario conclusion I is false

(4)

In this scenario neither of the conclusions is true.

MAH-CET MBA Mock Test- 10 - Question 58

Directions: Read the following information carefully and answer the question given below-

A certain number of persons are sitting in a circular table and are facing towards the center. Only a few persons of the information are known. More than 13 persons were seated in a table.

Five persons sit between A and B. F sits third to the right of B. A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B. Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

How many persons were seated in a table?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 58 Five persons sit between A and B. F sits third to the right of B.

A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B.

Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

Case 1 is invalid because the number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B).

Case 2b is invalid because Not more than six persons sit between F and C when counted from the right of F.

Final arrangement is

17 persons were seated in a table.

MAH-CET MBA Mock Test- 10 - Question 59

Directions: Read the following information carefully and answer the question given below-

A certain number of persons are sitting in a circular table and are facing towards the center. Only a few persons of the information are known. More than 13 persons were seated in a table.

Five persons sit between A and B. F sits third to the right of B. A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B. Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

If J sits second to the right of A, then which of the following combination is true?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 59 Five persons sit between A and B. F sits third to the right of B.

A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B.

Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

Case 1 is invalid because the number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B).

Case 2b is invalid because Not more than six persons sit between F and C when counted from the right of F.

Final arrangement is

C sits second to the right of G.

MAH-CET MBA Mock Test- 10 - Question 60

Directions: Read the following information carefully and answer the question given below-

A certain number of persons are sitting in a circular table and are facing towards the center. Only a few persons of the information are known. More than 13 persons were seated in a table.

Five persons sit between A and B. F sits third to the right of B. A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B. Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

If P sits to the immediate right of G, then what is the position of F with respect to P?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 60 Five persons sit between A and B. F sits third to the right of B.

A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B.

Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

Case 1 is invalid because the number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B).

Case 2b is invalid because Not more than six persons sit between F and C when counted from the right of F.

Final arrangement is

F sits third to the left of P.

MAH-CET MBA Mock Test- 10 - Question 61

Directions: Read the following information carefully and answer the question given below-

A certain number of persons are sitting in a circular table and are facing towards the center. Only a few persons of the information are known. More than 13 persons were seated in a table.

Five persons sit between A and B. F sits third to the right of B. A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B. Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

How many persons sit between A and H when counted from the right of A?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 61 Five persons sit between A and B. F sits third to the right of B.

A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B.

Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

Case 1 is invalid because the number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B).

Case 2b is invalid because Not more than six persons sit between F and C when counted from the right of F.

Final arrangement is

Seven persons sit between A and H when counted from the right of A.

MAH-CET MBA Mock Test- 10 - Question 62

Directions: Read the following information carefully and answer the question given below-

A certain number of persons are sitting in a circular table and are facing towards the center. Only a few persons of the information are known. More than 13 persons were seated in a table.

Five persons sit between A and B. F sits third to the right of B. A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B. Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

Who sits exactly in between C and D when counted from left of C?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 62 Five persons sit between A and B. F sits third to the right of B.

A and E are immediate neighbors. Three persons sit between D and E. Not more than two persons sit between B and D. Neither D nor H is an immediate neighbor of B.

Two persons sit between C and E. The number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B). G sits third to the right of H. Not more than six persons sit between F and C when counted from the right of F. The number of persons sit between C and H is same as between D and H.

Case 1 is invalid because the number of persons sit between A and C (when counted from the left of A) is one less than the number of persons sit between B and G (when counted from the right of B).

Case 2b is invalid because Not more than six persons sit between F and C when counted from the right of F.

Final arrangement is

H sits exactly in between C and D when counted from left of C.

MAH-CET MBA Mock Test- 10 - Question 63

Directions: Read the following information carefully and answer the question given below-

6 bikes were parked in a parking lot and were facing the North. The bikes were named A, B, C, D, E and F not necessarily in the same order. The distance between any two neighbouring bikes was a consecutive integral multiple of 4m. The distance increased left to right. Assume that there was no space after the extreme ends.

  • C was 76m to the right of bike A

  • Bike B was parked 3rd to the left of bike C

  • Bike E was parked 68m away from the left end.

  • D was 48m away from one of the ends.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X

  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y

  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

Four of the following bears a similar relationship and hence form a group, which of the following is not a part of that group?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 63
  • Bike E was parked 68m away from the left end.

    Case 1:

    Case 2:

  • C was 76m to the right of bike A
  • Bike B was parked 3rd to the left of bike C

  • D was 48m away from one of the ends. Thus, case 1 is invalid.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X
  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y
  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

Apart from DF, others have the 1st bike to the right of the 2nd bike (initially)

MAH-CET MBA Mock Test- 10 - Question 64

Directions: Read the following information carefully and answer the question given below-

6 bikes were parked in a parking lot and were facing the North. The bikes were named A, B, C, D, E and F not necessarily in the same order. The distance between any two neighbouring bikes was a consecutive integral multiple of 4m. The distance increased left to right. Assume that there was no space after the extreme ends.

  • C was 76m to the right of bike A

  • Bike B was parked 3rd to the left of bike C

  • Bike E was parked 68m away from the left end.

  • D was 48m away from one of the ends.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X

  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y

  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

What was the shortest distance between points X and Y?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 64
  • Bike E was parked 68m away from the left end.

    Case 1:

    Case 2:

  • C was 76m to the right of bike A
  • Bike B was parked 3rd to the left of bike C

  • D was 48m away from one of the ends. Thus, case 1 is invalid.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X
  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y
  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

The shortest distance between points X and Y was 63m

MAH-CET MBA Mock Test- 10 - Question 65

Directions: Read the following information carefully and answer the question given below-

6 bikes were parked in a parking lot and were facing the North. The bikes were named A, B, C, D, E and F not necessarily in the same order. The distance between any two neighbouring bikes was a consecutive integral multiple of 4m. The distance increased left to right. Assume that there was no space after the extreme ends.

  • C was 76m to the right of bike A

  • Bike B was parked 3rd to the left of bike C

  • Bike E was parked 68m away from the left end.

  • D was 48m away from one of the ends.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X

  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y

  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

How much bike M must travel to reach point X from point Z?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 65
  • Bike E was parked 68m away from the left end.

    Case 1:

    Case 2:

  • C was 76m to the right of bike A
  • Bike B was parked 3rd to the left of bike C

  • D was 48m away from one of the ends. Thus, case 1 is invalid.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X
  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y
  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

M must travel 16m north, 20m west to reach point X

MAH-CET MBA Mock Test- 10 - Question 66

Directions: Read the following information carefully and answer the question given below-

6 bikes were parked in a parking lot and were facing the North. The bikes were named A, B, C, D, E and F not necessarily in the same order. The distance between any two neighbouring bikes was a consecutive integral multiple of 4m. The distance increased left to right. Assume that there was no space after the extreme ends.

  • C was 76m to the right of bike A

  • Bike B was parked 3rd to the left of bike C

  • Bike E was parked 68m away from the left end.

  • D was 48m away from one of the ends.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X

  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y

  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

What was the initial distance between bikes D and C?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 66
  • Bike E was parked 68m away from the left end.

    Case 1:

    Case 2:

  • C was 76m to the right of bike A
  • Bike B was parked 3rd to the left of bike C

  • D was 48m away from one of the ends. Thus, case 1 is invalid.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X
  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y
  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

D was 44m to the right of C (initially).

MAH-CET MBA Mock Test- 10 - Question 67

Directions: Read the following information carefully and answer the question given below-

6 bikes were parked in a parking lot and were facing the North. The bikes were named A, B, C, D, E and F not necessarily in the same order. The distance between any two neighbouring bikes was a consecutive integral multiple of 4m. The distance increased left to right. Assume that there was no space after the extreme ends.

  • C was 76m to the right of bike A

  • Bike B was parked 3rd to the left of bike C

  • Bike E was parked 68m away from the left end.

  • D was 48m away from one of the ends.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X

  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y

  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

How many bikes were parked to the right of F (initially)?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 67
  • Bike E was parked 68m away from the left end.

    Case 1:

    Case 2:

  • C was 76m to the right of bike A
  • Bike B was parked 3rd to the left of bike C

  • D was 48m away from one of the ends. Thus, case 1 is invalid.

  • Bike F went 60m towards west, turned right and moved for 40m. It then turned left and moved 20m and stopped at point X
  • Bike E moved 52m towards east, took a right turn and moved 23m and stopped at point Y
  • Bike M was parked 20m to the east of point Y. It moved 47m towards North and stopped at point Z.

F was parked the rightmost

MAH-CET MBA Mock Test- 10 - Question 68

In the given question, a set of conclusions is given. There are five options comprising of three or more statements. You need to choose the option that contains the set of statements from which the given conclusions logically follow.

Some tracks are not bars. Some shoots are definitely not bars.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 68 (1)

In this scenario all tracks are bars hence conclusion 1 does not follow.

(2)

In this scenario neither of the conclusions is true.

(3)

In this scenario both the conclusions are true. All tracks are mangoes and no mango is bar hence no tracks will be bars. Also whatever part of shoots is tracks will never be bars.

(4)

It is given that all bars is tracks; hence the circles for bars and tracks can overlap as well. In this scenario all tracks will be bars hence conclusion 1 does not follow.

The statements given in option 3 lead to the required conclusion.

MAH-CET MBA Mock Test- 10 - Question 69

Directions: Read the following information carefully and answer the question given below-

Eight persons B, D, F, G, K, L, M and P were seated around a square table facing the centre. They were born in the month of February of the same year. They were born on different dates and were seated such that the persons born on dates which were multiples of 7 were seated along the 4 sides and the persons born on dates which were multiples of 6 were seated at the 4 corners.

  • P and D were born at a difference of 10 days

  • K was 3rd to the right of L who was opposite to the one born on 18th

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F

  • K was born just after D and was 2nd to the left of G's neighbour

  • M was to the immediate left of the one born on 14th

  • L was younger to the one who was to the immediate left of B.

How many persons were born after K?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 69 The persons were born in February and on dates which were multiples of 6 and 7. The dates will be: 6, 7, 12, 14, 18, 21, 24 and 28
  • P and D were born at a difference of 10 days. Combination can be (14, 24) or (18, 28)
  • K was 3rd to the right of L who was opposite to the one born on 18th
  • K was born just after D and was 2nd to the left of G's neighbour. Thus, K could be born on 18th, 21st or 28th.

Thus, we see that K could not be born on 18th, hence he was born on 21st or 28th. Thus, D was born either on 18th or 24th. P was born either on 14th or 28th.

  • M was to the immediate left of the one born on 14th

In both cases D is not the one born on 18th, thus, D must have been born on 24th and P must have been born on 14th. K was thus born on 28th.

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F. Dates left: 6, 7, 12, 21. Only combination will be 12 and 21.

  • L was younger to the one who was to the immediate left of B. Thus, case 2 and 3 become invalid.

Thus, the final arrangement is as follows:

K was the youngest person hence nobody was born after him.

MAH-CET MBA Mock Test- 10 - Question 70

Directions: Read the following information carefully and answer the question given below-

Eight persons B, D, F, G, K, L, M and P were seated around a square table facing the centre. They were born in the month of February of the same year. They were born on different dates and were seated such that the persons born on dates which were multiples of 7 were seated along the 4 sides and the persons born on dates which were multiples of 6 were seated at the 4 corners.

  • P and D were born at a difference of 10 days

  • K was 3rd to the right of L who was opposite to the one born on 18th

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F

  • K was born just after D and was 2nd to the left of G's neighbour

  • M was to the immediate left of the one born on 14th

  • L was younger to the one who was to the immediate left of B.

How many persons were born between B and M?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 70 The persons were born in February and on dates which were multiples of 6 and 7. The dates will be: 6, 7, 12, 14, 18, 21, 24 and 28
  • P and D were born at a difference of 10 days. Combination can be (14, 24) or (18, 28)
  • K was 3rd to the right of L who was opposite to the one born on 18th
  • K was born just after D and was 2nd to the left of G's neighbour. Thus, K could be born on 18th, 21st or 28th.

Thus, we see that K could not be born on 18th, hence he was born on 21st or 28th. Thus, D was born either on 18th or 24th. P was born either on 14th or 28th.

  • M was to the immediate left of the one born on 14th

In both cases D is not the one born on 18th, thus, D must have been born on 24th and P must have been born on 14th. K was thus born on 28th.

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F. Dates left: 6, 7, 12, 21. Only combination will be 12 and 21.

  • L was younger to the one who was to the immediate left of B. Thus, case 2 and 3 become invalid.

Thus, the final arrangement is as follows:

L and P were born between B and M

MAH-CET MBA Mock Test- 10 - Question 71

Directions: Read the following information carefully and answer the question given below-

Eight persons B, D, F, G, K, L, M and P were seated around a square table facing the centre. They were born in the month of February of the same year. They were born on different dates and were seated such that the persons born on dates which were multiples of 7 were seated along the 4 sides and the persons born on dates which were multiples of 6 were seated at the 4 corners.

  • P and D were born at a difference of 10 days

  • K was 3rd to the right of L who was opposite to the one born on 18th

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F

  • K was born just after D and was 2nd to the left of G's neighbour

  • M was to the immediate left of the one born on 14th

  • L was younger to the one who was to the immediate left of B.

Who among the following was the third eldest?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 71 The persons were born in February and on dates which were multiples of 6 and 7. The dates will be: 6, 7, 12, 14, 18, 21, 24 and 28
  • P and D were born at a difference of 10 days. Combination can be (14, 24) or (18, 28)
  • K was 3rd to the right of L who was opposite to the one born on 18th
  • K was born just after D and was 2nd to the left of G's neighbour. Thus, K could be born on 18th, 21st or 28th.

Thus, we see that K could not be born on 18th, hence he was born on 21st or 28th. Thus, D was born either on 18th or 24th. P was born either on 14th or 28th.

  • M was to the immediate left of the one born on 14th

In both cases D is not the one born on 18th, thus, D must have been born on 24th and P must have been born on 14th. K was thus born on 28th.

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F. Dates left: 6, 7, 12, 21. Only combination will be 12 and 21.

  • L was younger to the one who was to the immediate left of B. Thus, case 2 and 3 become invalid.

Thus, the final arrangement is as follows:

L was the 3rd eldest person

MAH-CET MBA Mock Test- 10 - Question 72

Directions: Read the following information carefully and answer the question given below-

Eight persons B, D, F, G, K, L, M and P were seated around a square table facing the centre. They were born in the month of February of the same year. They were born on different dates and were seated such that the persons born on dates which were multiples of 7 were seated along the 4 sides and the persons born on dates which were multiples of 6 were seated at the 4 corners.

  • P and D were born at a difference of 10 days

  • K was 3rd to the right of L who was opposite to the one born on 18th

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F

  • K was born just after D and was 2nd to the left of G's neighbour

  • M was to the immediate left of the one born on 14th

  • L was younger to the one who was to the immediate left of B.

What was the position of B with respect to the person seated opposite to D?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 72 The persons were born in February and on dates which were multiples of 6 and 7. The dates will be: 6, 7, 12, 14, 18, 21, 24 and 28
  • P and D were born at a difference of 10 days. Combination can be (14, 24) or (18, 28)
  • K was 3rd to the right of L who was opposite to the one born on 18th
  • K was born just after D and was 2nd to the left of G's neighbour. Thus, K could be born on 18th, 21st or 28th.

Thus, we see that K could not be born on 18th, hence he was born on 21st or 28th. Thus, D was born either on 18th or 24th. P was born either on 14th or 28th.

  • M was to the immediate left of the one born on 14th

In both cases D is not the one born on 18th, thus, D must have been born on 24th and P must have been born on 14th. K was thus born on 28th.

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F. Dates left: 6, 7, 12, 21. Only combination will be 12 and 21.

  • L was younger to the one who was to the immediate left of B. Thus, case 2 and 3 become invalid.

Thus, the final arrangement is as follows:

B was to the immediate right of G who was seated opposite to D

MAH-CET MBA Mock Test- 10 - Question 73

Directions: Read the following information carefully and answer the question given below-

Eight persons B, D, F, G, K, L, M and P were seated around a square table facing the centre. They were born in the month of February of the same year. They were born on different dates and were seated such that the persons born on dates which were multiples of 7 were seated along the 4 sides and the persons born on dates which were multiples of 6 were seated at the 4 corners.

  • P and D were born at a difference of 10 days

  • K was 3rd to the right of L who was opposite to the one born on 18th

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F

  • K was born just after D and was 2nd to the left of G's neighbour

  • M was to the immediate left of the one born on 14th

  • L was younger to the one who was to the immediate left of B.

Four of the following bear a similar relationship and hence form a group, who among the following is not a part of that group?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 73 The persons were born in February and on dates which were multiples of 6 and 7. The dates will be: 6, 7, 12, 14, 18, 21, 24 and 28
  • P and D were born at a difference of 10 days. Combination can be (14, 24) or (18, 28)
  • K was 3rd to the right of L who was opposite to the one born on 18th
  • K was born just after D and was 2nd to the left of G's neighbour. Thus, K could be born on 18th, 21st or 28th.

Thus, we see that K could not be born on 18th, hence he was born on 21st or 28th. Thus, D was born either on 18th or 24th. P was born either on 14th or 28th.

  • M was to the immediate left of the one born on 14th

In both cases D is not the one born on 18th, thus, D must have been born on 24th and P must have been born on 14th. K was thus born on 28th.

  • Number of persons who were younger to F was equal to the number of persons who were elder to the one who was to the immediate left of F. Dates left: 6, 7, 12, 21. Only combination will be 12 and 21.

  • L was younger to the one who was to the immediate left of B. Thus, case 2 and 3 become invalid.

Thus, the final arrangement is as follows:

Apart from D others were seated along the sides

MAH-CET MBA Mock Test- 10 - Question 74

In each question below is given a statement followed by three courses of action numbered (A), (B) and (C). A course of action is a step or administrative decision to be taken for improvement, follow-up or further action in regard to the problem, policy, etc. On the basis of the information given in the statement, you have to assume everything in the statement to be true, then decide which of the suggested courses of action logically follow(s) for pursuing.

Statement: A large number of students lost a considerable amount of their valuable time taking the on-line admission test of prestigious technical schools in the country due to technical glitches and were not compensated for the lost time.

Courses of action:

(A) The technical school authorities should cancel the entire on-line admission testing process and conduct the examination again.

(B) The technical school authorities should identify all those who were disadvantaged due to technical glitches and call them for a reexamination on a separate date.

(C) The Government should ask the technical school authorities to conduct the admission test offline from next year onwards.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 74 There is no point in conducting re-exam for everyone. Only those who were affected by the technical problems should be allowed to sit for a re exam.

The reason for making the admission test online is to improve the efficiency of the whole process. So instead of scrapping the online test, technical schools should take steps to avoid any technical glitches in the future.

Hence, option (b).

MAH-CET MBA Mock Test- 10 - Question 75

In the question below are given few statements depicting an issue or a problem. You have to assume everything in the statements to be true and on the basis of the information given in the statements, decide which of the suggested courses of action logically follow(s).

The meteorological department forecasts that floods may play havoc in the city of F this year.

1. The government should take precautionary measures to protect people residing in areas likely to be worst hit by the flood.

2. The government should assure people not to be afraid of floods.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 75 1 follows as it directly aims to solve the problem. 2 is incorrect, as it is unrealistic and it does nothing to avert or improve the situation. A is the right answer.
MAH-CET MBA Mock Test- 10 - Question 76

Choose the figure which is different from the rest.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 76 Figure (3) is different from the others.

Only in figure (3), the two shaded portions lie on the same side of the mainline.

Hence, the correct option is (C).

MAH-CET MBA Mock Test- 10 - Question 77

Direction: Select a figure from amongst the Answer Figures which will continue the same series as established by the five Problem Figures.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 77 In one step, a black circle is added to the figure at the CW-end of the existing circles and a line segment is added on the upper side. In the next step, a white circle is added to the figure at the ACW-end of the existing circles and a line segment is added on the lower side.

Hence, the correct option is (B).

MAH-CET MBA Mock Test- 10 - Question 78

Select the suitable figure from the Answer figures that would replace the question mark (?).

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 78 The element on the left is rotated 90 degrees clockwise and that on the right is rotated 90 degrees anticlockwise.

So, option 1 is the correct answer.

MAH-CET MBA Mock Test- 10 - Question 79

Find the figure number which is different from the other four:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 79 All the given figures are mirror images of each other except figure 4, so the figure which is different from the other four is figure 4.

Hence, the correct option is (D).

MAH-CET MBA Mock Test- 10 - Question 80

Find the odd one out.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 80 The number of figures in right box increases by 1 with respect to left box except in figure 2.
MAH-CET MBA Mock Test- 10 - Question 81

Find the odd one out.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 81 Each figure rotates once in the clockwise direction(90 degrees) except figure 4.

Hence, the correct option is (D).

MAH-CET MBA Mock Test- 10 - Question 82

Find the odd one out.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 82 In all figures, the part of shade in the central figure is shaded once inside and once outside but this does not happen in figure 3.
MAH-CET MBA Mock Test- 10 - Question 83

Find the odd one out.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 83 The first two letters in left box are interchanged in the right box and the last two letters in left box are interchanged in the right box. Also if there is any vowel, it is replaced by immediate next vowel. Only figure 3 does not follow this.
MAH-CET MBA Mock Test- 10 - Question 84

Find the odd one out.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 84 Only in fig. (2), while moving in an ACW direction, the numbers do not form a sequence.

Hence, the correct option is (B).

MAH-CET MBA Mock Test- 10 - Question 85

Select a suitable figure from the given figures that will replace the question mark (?) :

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 85 Figure 2 replaces the ? mark.

The two figures are the water images and mirror images of their respective figures in every successive figure. Also the rotation takes place simultaneously.

MAH-CET MBA Mock Test- 10 - Question 86

Direction: Select a suitable answer figure from the given question figures that will replace the question mark (?).

Each one of the upper element is replaced by an element similar to the lower element(s) and each one of the lower elements is replaced by an element similar to the upper element(s).

Hence, the correct option is (B).

MAH-CET MBA Mock Test- 10 - Question 87

Select a suitable figure from the given figures that will replace the question mark (?) :

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 87 Figure 5 replaces the ? mark.

The symbols keep increasing by 1 and also rotate anti clockwise in next step.

MAH-CET MBA Mock Test- 10 - Question 88

Select a suitable figure from the given figures that will replace the question mark (?) :

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 88 Figure 2 replaces the ? mark.

The figure on top right corner comes to bottom left corner and then inverts(changes its direction from top to bottom) . The figure on bottom right corner goes to top left corner and inverts (changes its direction from left to right). The remaining two figures move to their respective right corners and their shading inverts.

MAH-CET MBA Mock Test- 10 - Question 89

Select a suitable figure from the given figures that will replace the question mark (?):

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 89 Figure 3 replaces the ? mark.

The output is

First the water image of the given image (C) in the series is found then the mirror image of that gives the final output.

MAH-CET MBA Mock Test- 10 - Question 90

Select a figure from the given option figures which will continue the same series as the figures in the question:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 90 Figure 3 will follow the series.

The new figure added in every consecutive step and then in the next step the figure added in previous step comes to the centre. The remaining figures move in anticlockwise direction.

MAH-CET MBA Mock Test- 10 - Question 91

Select a figure from amongst the answer figures which will continue the same series as established by the five.

Question Figure:

Answer Figure:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 91 The next figure will be

Logic: One of the pins gets inverted in each step. The pins get inverted sequentially from right to left.

Hence, the correct option is (E).

MAH-CET MBA Mock Test- 10 - Question 92

Select a figure from the given option figures which will continue the same series as the figures in the question:

Question Figure:

Answer Figure:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 92 Figure 2 will follow the series.

The consecutive arrows and the line with a circle are facing in opposite direction. There is an addition of one line(either arrow line or line with a circle) in each step and the figure orientation is changed (like vertical and horizontal) in every alternate step.

Hence, the correct option is (C).

MAH-CET MBA Mock Test- 10 - Question 93

Select a figure from the given option figures which will continue the same series as the figures in the question:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 93 The black dot and the separated line in triangle move in clockwise direction and then after completion of one round, it moves in anticlockwise direction. Similarly, the separated line in square moves in anticlockwise direction and then after completion of one round, it moves in clockwise direction.
MAH-CET MBA Mock Test- 10 - Question 94

Select a figure from the given option figures which will continue the same series as the figures in the question:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 94 Figure 5 will follow the series. The pair of signs opposite to each other, increase by 1 in alternate steps. Also, the figure rotates once in anticlockwise direction.
MAH-CET MBA Mock Test- 10 - Question 95

Select a figure from the given option figures which will continue the same series as the figures in the question:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 95 Figure 3 will follow the series. The shaded portion rotates in clockwise direction. The shaded dots at the boundary rotate in anticlockwise direction.
MAH-CET MBA Mock Test- 10 - Question 96

Select a figure from the given option figures which will continue the same series as the figures in the question:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 96 Figure 4 will follow the series. The arrows attached to their outer sides are open and closed in alternate steps. The letters inside rotate in the clockwise direction. First, there is no gap with rotation. Then one gap and rotation. Then two-gap and rotation and so on.

Hence, the correct option is (D).

MAH-CET MBA Mock Test- 10 - Question 97

Select a figure from the given option figures which will continue the same series as the figures in the question:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 97 Figure 5 will follow the series. One line is deleted from bottom and one from top in alternate steps. The line at odd position from bottom are deleted and line from even position from top are deleted.
MAH-CET MBA Mock Test- 10 - Question 98

Select a figure from the given option figures which will continue the same series as the figures in the question:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 98 Figure 5 will follow the series. The letter at top right corner comes to north sector of the cross. The letter at north comes to west section. The letter at west section goes to bottom left corner. The letter at bottom left corner goes to south section of cross. The letter at south section goes to east section and the letter at east section goes to top right corner.
MAH-CET MBA Mock Test- 10 - Question 99

Select a figure from the given option figures which will continue the same series as the figures in the question:

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 99 Figure 5 will follow the series. Each section of the square rotates clockwise. The arrow rotates in anticlockwise direction and after one full rotation, it rotates clockwise. The rest symbols alternatively come at top and bottom in their respective sections.
MAH-CET MBA Mock Test- 10 - Question 100

Select a suitable figure from the given figures that will replace the question mark (?):

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 100 Figure 2 will replace the? mark. Each symbol is changed by changing its face to opposite direction. Also, the figures at top get interchanged and the figures at bottom also get interchanged.
MAH-CET MBA Mock Test- 10 - Question 101

Directions : Study the table carefully and answer the questions that follow:

Distribution of Malnutrition in children in Percentage in Various Years

In which of the following years there was highest percentage increase in general malnutrition cases in comparison to previous year?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 101 In 2001, there were 698 cases (10.6% of 6775) compared to 58 (3.4% of 1721) which was the highest percentage improvement over previous year.
MAH-CET MBA Mock Test- 10 - Question 102

Direction: Study the table carefully and answer the questions that follow:

Distribution of Malnutrition in children in Percentage in Various Years

How many percent approximately was mild malnutrition in all the years together ?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 102 Required percentage = (14% of 18000+21.9% of 2410+22.1% of 1721+41.1% of 6775+42.4% of 4713+47.9% of 4008+53.4% of 9180)/46807 = 15033/46807 = ~32%
MAH-CET MBA Mock Test- 10 - Question 103

Direction: Study the table carefully and answer the questions that follow:

Distribution of Malnutrition in children in Percentage in Various Years

How many were normal in the number of the surveying children in all the years?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 103

MAH-CET MBA Mock Test- 10 - Question 104

Direction: Study the table carefully and answer the questions that follow:

Distribution of Malnutrition in children in Percentage in Various Years

How many cases were moderate and serious malnutrition together in the surveying in 1998?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 104

MAH-CET MBA Mock Test- 10 - Question 105

Direction: Study the table carefully and answer the questions that follow:

Distribution of Malnutrition in children in Percentage in Various Years

What was the percentage decrease in moderate cases from 1998 to 2004?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 105 Required percentage = = [(0.65*18000)-(0.28*9180)]/(0.28*9180) = 78%
MAH-CET MBA Mock Test- 10 - Question 106

Direction: Study the pie- charts carefully and answer the questions that follow.

The chart on left shows Percentage of Employees working in Different Departments

The chart on the right shows Percentage of Women Working in Different Departments

Also, Total number of employees = 8450 and Total number of women = 3500

Which department has the lowest number of men working in it?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 106 Computation

HR Department:

Number of men in Finance department is 308.

MAH-CET MBA Mock Test- 10 - Question 107

Direction: Study the pie- charts carefully and answer the questions that follow.

The chart on left shows Percentage of Employees working in Different Departments

The chart on the right shows Percentage of Women Working in Different Departments

Also, Total number of employees = 8450 and Total number of women = 3500

What is the average number of men working in different departments in the organization?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 107

Total Employees = 8450, Total Women = 3500

So, Total Men = 8450 - 3500 = 4950

Average number of men working in different departments = 4950/5 = 990

MAH-CET MBA Mock Test- 10 - Question 108

Direction: Study the pie- charts carefully and answer the questions that follow.

The chart on left shows Percentage of Employees working in Different Departments

The chart on the right shows Percentage of Women Working in Different Departments

Also, Total number of employees = 8450 and Total number of women = 3500

What is the respective ratio of number of women working in Administration department to the total number of employees working in that department?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 108

Required ratio = 490 : 1014 = 245 : 507.

MAH-CET MBA Mock Test- 10 - Question 109

Direction: Study the pie- charts carefully and answer the questions that follow.

The chart on left shows Percentage of Employees working in Different Departments

The chart on the right shows Percentage of Women Working in Different Departments

Also, Total number of employees = 8450 and Total number of women = 3500

Number of women working in the Finance and It department together form what percent of total number of employees in the organization? (rounded off to 2 digits after decimal)

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 109 Computation

HR Department:

Number of women working in the Finance and It department

= (875 +735) = 1610

Total number of employees = 8450.

MAH-CET MBA Mock Test- 10 - Question 110

Direction: Study the pie- charts carefully and answer the questions that follow.

The chart on left shows Percentage of Employees working in Different Departments

The chart on the right shows Percentage of Women Working in Different Departments

Also, Total number of employees = 8450 and Total number of women = 3500

What is the number of men working in HR department?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 110 Computation

HR Department:

Number of men working in HR department = 926

MAH-CET MBA Mock Test- 10 - Question 111

Direction: Study the following graph carefully and answer the questions below it.

Number of students (Males & Females) passed out from various Colleges in a year.

(Number in thousands)

What is the average number of students (Males & Females) passed out from all the colleges together?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 111 Total number of boys = 15 + 17.5 + 27.5 + 25 + 10 = 95 = 95000

Total number of girls = 22.5 + 20 + 35 + 30 + 7.5 = 115 = 115000

Average = (95000 + 115000)/5 = 42000

MAH-CET MBA Mock Test- 10 - Question 112

Direction: Study the following graph carefully and answer the questions below it.

Number of students (Males & Females) passed out from various Colleges in a year.

(Number in thousands)

The number of Females passed out from college C is approximately what percent the total number of Females passed out from all the colleges together?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 112 the number of female students passed out from college C = 35000

Total number of girls = 22.5 + 20 + 35 + 30 + 7.5 = 115 = 115000

Required percentage = 35000 X 100/115000 = 30.43%

MAH-CET MBA Mock Test- 10 - Question 113

Direction: Study the following graph carefully and answer the questions below it.

Number of students (Males & Females) passed out from various Colleges in a year.

(Number in thousands)

What is the difference between the total number of students passing out from college A and the total number of students passing out from college E?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 113 Total number of students passing out from college A = 15 + 22.5 = 37.5

Total number of students passing out from college E = 10 + 7.5 = 17.5

Difference = 37.5 – 17.5 = 20 = 20000

MAH-CET MBA Mock Test- 10 - Question 114

Direction: Study the following graph carefully and answer the questions below it.

Number of students (Males & Females) passed out from various Colleges in a year.

(Number in thousands)

What is the respective ratio of the total number of Males to the total number of Females passed out from all the colleges together?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 114 Total number of boys = 15 + 17.5 + 27.5 + 25 + 10 = 85 = 95000

Total number of girls = 22.5 + 20 + 35 + 30 + 7.5 = 115 = 115000

Required ratio = 95 : 115 = 19 : 23

MAH-CET MBA Mock Test- 10 - Question 115

Direction: Study the following graph carefully and answer the questions below it.

Number of students (Males & Females) passed out from various Colleges in a year.

(Number in thousands)

The number of Males passing out from colleges A and B together is what percent of the number of females passing out from colleges C and D together?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 115 The number of male students passing out of colleges A and B = 32.5 = 32500

The number of female students passing out of colleges C and D = 65 = 65000

Required percentage = 32500 X 100 / 65000 = 50%

MAH-CET MBA Mock Test- 10 - Question 116

Direction: Study the information carefully to answer the questions that follow:

In a school consisting of 2800 children, the ratio of girls to boys is 4: 3 respectively. All the children have taken different hobby classes viz. Singing, Dancing, Painting and Cooking. 20 percent of the boys take Painting Classes. The number of girls taking Dancing Classes is five fourth of the number of boys taking the same. One-fourth of the girls take cooking classes. The total number of students taking cooking classes is 700. Two-fifth of the boys take Singing Classes and the remaining boys take Dancing Classes. The girls taking Singing Classes is twice the number of boys taking the same. The remaining girls take Painting Classes.

What is the respective ratio of boys taking painting classes to the boys taking Singing Classes?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 116

Total number of children - 2800

Girls: boys = 4: 3

Total number of boys - (3/7) X 2800 -1200 Total number of girls - (4/7) X 2800 =1600

Number of boys taking painting class - 20 % of 1200 = 240

Number of girls taking cooking class - 25% of 1600 = 400

Number of boys taking cooking class = 700 - 400 - 300

Number of boys taking singing class -(2/5) X1200 = 480

Number of girls taking singing class - 480 X 960

Number of girls taking dance class = (5/4) 180 = 225

Ratio = 240: 480 = 1:2

MAH-CET MBA Mock Test- 10 - Question 117

Direction: Study the information carefully to answer the questions that follow:

In a school consisting of 2800 children, the ratio of girls to boys is 4: 3 respectively. All the children have taken different hobby classes viz. Singing, Dancing, Painting and Cooking. 20 percent of the boys take Painting Classes. The number of girls taking Dancing Classes is five fourth of the number of boys taking the same. One-fourth of the girls take cooking classes. The total number of students taking cooking classes is 700. Two-fifth of the boys take Singing Classes and the remaining boys take Dancing Classes. The girls taking Singing Classes is twice the number of boys taking the same. The remaining girls take Painting Classes.

The number of girls taking cooking classes is what percent of the total number of children in the school? (rounded off to the nearest integer)

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 117 Total number of boys - (3/7) X 2800 -1200 Total number of girls - (4/7) X 2800 =1600

Number of boys taking painting class - 20 % of 1200 = 240

Number of girls taking cooking class - 25% of 1600 = 400

Number of boys taking cooking class = 700 - 400 - 300

Number of boys taking singing class -(2/5) X1200 = 480

Number of girls taking singing class - 480 X 960

Number of girls taking dance class = (5/4) 180 = 225

Girls taking cooking class = 400

Required percentage =

MAH-CET MBA Mock Test- 10 - Question 118

Direction: Study the information carefully to answer the questions that follow:

In a school consisting of 2800 children, the ratio of girls to boys is 4: 3 respectively. All the children have taken different hobby classes viz. Singing, Dancing, Painting and Cooking. 20 percent of the boys take Painting Classes. The number of girls taking Dancing Classes is five fourth of the number of boys taking the same. One-fourth of the girls take cooking classes. The total number of students taking cooking classes is 700. Two-fifth of the boys take Singing Classes and the remaining boys take Dancing Classes. The girls taking Singing Classes is twice the number of boys taking the same. The remaining girls take Painting Classes.

The number of boys taking cooking classes is what percent of the total number of children in the school? (rounded off to two digits after decimal)

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 118 Total number of boys - (3/7) X 2800 -1200 Total number of girls - (4/7) X 2800 =1600

Number of boys taking painting class - 20 % of 1200 = 240

Number of girls taking cooking class - 25% of 1600 = 400

Number of boys taking cooking class = 700 - 400 - 300

Number of boys taking singing class -(2/5) X1200 = 480

Number of girls taking singing class - 480 X 960

Number of girls taking dance class = (5/4) 180 = 225

Number of boys taking cooking class = 300

Required percentage =

MAH-CET MBA Mock Test- 10 - Question 119

Direction: Study the information carefully to answer the questions that follow:

In a school consisting of 2800 children, the ratio of girls to boys is 4: 3 respectively. All the children have taken different hobby classes viz. Singing, Dancing, Painting and Cooking. 20 percent of the boys take Painting Classes. The number of girls taking Dancing Classes is five fourth of the number of boys taking the same. One-fourth of the girls take cooking classes. The total number of students taking cooking classes is 700. Two-fifth of the boys take Singing Classes and the remaining boys take Dancing Classes. The girls taking Singing Classes is twice the number of boys taking the same. The remaining girls take Painting Classes.

What is the total number of children taking Dancing Classes?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 119 Total number of boys - (3/7) X 2800 -1200 Total number of girls - (4/7) X 2800 =1600

Number of boys taking painting class - 20 % of 1200 = 240

Number of girls taking cooking class - 25% of 1600 = 400

Number of boys taking cooking class = 700 - 400 - 300

Number of boys taking singing class -(2/5) X1200 = 480

Number of girls taking singing class - 480 X 960

Number of girls taking dance class = (5/4) 180 = 225

Total number of children dancing class = 180 + 225 = 405

MAH-CET MBA Mock Test- 10 - Question 120

Direction: Study the information carefully to answer the questions that follow:

In a school consisting of 2800 children, the ratio of girls to boys is 4: 3 respectively. All the children have taken different hobby classes viz. Singing, Dancing, Painting and Cooking. 20 percent of the boys take Painting Classes. The number of girls taking Dancing Classes is five fourth of the number of boys taking the same. One-fourth of the girls take cooking classes. The total number of students taking cooking classes is 700. Two-fifth of the boys take Singing Classes and the remaining boys take Dancing Classes. The girls taking Singing Classes is twice the number of boys taking the same. The remaining girls take Painting Classes.

What is the number of girls taking Painting Classes?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 120 Total number of boys - (3/7) X 2800 -1200 Total number of girls - (4/7) X 2800 =1600

Number of boys taking painting class - 20 % of 1200 = 240

Number of girls taking cooking class - 25% of 1600 = 400

Number of boys taking cooking class = 700 - 400 - 300

Number of boys taking singing class -(2/5) X1200 = 480

Number of girls taking singing class - 480 X 960

Number of girls taking dance class = (5/4) 180 = 225

Total number of children dancing class = 180 + 225 = 405

MAH-CET MBA Mock Test- 10 - Question 121

In each of these questions two equations numbered I and II are given. You have to solve both the equations and choose the correct option.

I. y4 - 16y2 + 55 = 0

II. 3x2 - 23x + 14 = 0

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 121 y4 - 16y2 + 55 = 0

∴ (y2)2 - 16y2 + 55 = 0

∴ let y2 be b

∴ b2 - 16b + 55 = 0

∴ (b - 11) (b - 5) = 0

∴ b = 5 or b = 11

∴ y ≈ ±2.24 or y ≈ ±3.31

3x2 - 23x + 14 = 0

∴ 3x2 - 2x - 21x + 14 = 0

∴ (3x - 2) (x - 7) = 0

∴ x = 2/3 or x = 7

∴ x = 7 is greater than all roots of y, but x = 2/3 is lesser than the all roots of y.

Hence, no relation is obtained between x and y

Hence, option 5 is correct.

MAH-CET MBA Mock Test- 10 - Question 122

In each of these questions two equations numbered I and II are given. You have to solve both the equations and choose the correct option.

I. 4x3 + 8x2 - 60x = 0

II. 7y2 − 77y + 196 = 0

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 122 4x3 + 8x2 - 60x = 0

∴ 4x(x2 + 2x - 15) = 0

∴ 4x = 0 or x2 + 2x - 15 = 0

∴ x2 + 2x - 15 = 0

∴ x2 + 5x - 3x - 15 = 0

∴ (x + 5)(x - 3) = 0

∴ x = - 5 or x = 3

7y2 − 77y + 196 = 0

∴ y2 − 11y + 28 = 0

∴ y2 − 7y − 4y + 28 = 0

∴ (y - 7)(y - 4) = 0

∴ y = 7 or y = 4

∴ Both the values of y are greater than all three values of x.

Hence, option 2 is correct.

MAH-CET MBA Mock Test- 10 - Question 123

Find the wrong number of the series

1888, 1892, 1896, 1900, 1904, 1908

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 123 All the other years except 1900 is a leap year.

Remember a year which is a multiple of 100 must be a multiple of 400 as well to be a leap year. For examples, 1600, 2000 are leap years but 1900, 1700 are not.

MAH-CET MBA Mock Test- 10 - Question 124

Find the wrong number in the series.

18739, 2676, 445, 88, 21, 6, 3, 1

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 124 (18739-7)/7=2676

(2676-6)/6=445

(445-5)/5=88

(88-4)/4=21

(21-3)/3=6

(6-2)/2=2

(2-1)/1=1

MAH-CET MBA Mock Test- 10 - Question 125

Find the wrong number in the series.

311,316,326,337,342,358,374

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 125 Next number =previous number+digit sum of previous number

311+(3+1+1)=316

316+(3+1+6)=326

326+(3+2+6)=337

337+(3+3+7)=350

350+(3+5+0)=358

358+(3+5+8)=374

MAH-CET MBA Mock Test- 10 - Question 126

On a School’s Annual Day sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get?

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 126 Let each child supposed to get ‘x’ sweets

112x = 80(x+6)

x = 15

MAH-CET MBA Mock Test- 10 - Question 127

These questions consist of two quantities, one in column A and another in column B. compare the two quantities.

Perimeter of an isosceles triangle is 22. Largest side is 10. Other two sides of the triangle are x and y, where x and y are integers.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 127 It is given that the triangle is isosceles and the largest side is also given as 10 so the remaining two sides have to be equal.
MAH-CET MBA Mock Test- 10 - Question 128

These questions consist of two quantities, one in column A and another in column B. compare the two quantities.

ABCD is a rectangle.

Detailed Solution for MAH-CET MBA Mock Test- 10 - Question 128 If the angle between the lines of length a and b is a right angle then we can calculate DC as V square root of (a2 +b2 ). Since this information is not available, no relation can be established.
MAH-CET MBA Mock Test- 10 - Question 129