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MAH-CET MBA Mock Test- 4 - CAT MCQ


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30 Questions MCQ Test Mock Test Series for MAH-CET - MAH-CET MBA Mock Test- 4

MAH-CET MBA Mock Test- 4 for CAT 2024 is part of Mock Test Series for MAH-CET preparation. The MAH-CET MBA Mock Test- 4 questions and answers have been prepared according to the CAT exam syllabus.The MAH-CET MBA Mock Test- 4 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MAH-CET MBA Mock Test- 4 below.
Solutions of MAH-CET MBA Mock Test- 4 questions in English are available as part of our Mock Test Series for MAH-CET for CAT & MAH-CET MBA Mock Test- 4 solutions in Hindi for Mock Test Series for MAH-CET course. Download more important topics, notes, lectures and mock test series for CAT Exam by signing up for free. Attempt MAH-CET MBA Mock Test- 4 | 200 questions in 150 minutes | Mock test for CAT preparation | Free important questions MCQ to study Mock Test Series for MAH-CET for CAT Exam | Download free PDF with solutions
MAH-CET MBA Mock Test- 4 - Question 1
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Directions: Seven people A, B, C, D, E, F and G are born in different years and like different colours - red, blue, pink, yellow, green, black and white not necessarily in same order.

  • B and E are born in consecutive years (B being younger), one of them been born in 1961.

  • The one who likes black is born in 1982 and is 4 years younger than the one who likes white.

  • The difference between present ages of A and D is 10 less than the difference between present ages of G and F. (A and G are elder than D and F respectively)

  • G is born in 1978.

  • F is born in 2000 and is youngest and doesn't like pink.

  • C likes yellow and is born three years after D was born, who is 6 years elder than F and 34 years younger to one who likes red colour.

  • The second eldest person doesn't like green.

  • The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

Who likes blue colour?

Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 1
F is born in 2000

Since, F is youngest, so, no one is born after F.

G is born in 1978

D is 6 years elder than F

So,

D is born in 1994

C is born 3 years after D

C is born in 1997

So, difference in ages of G and F = 22 years

Thus, Difference in ages of A and D = 22 - 10 = 12 years

So, A is born in 1982

The one who likes black is born in 1982

White is liked by the one who is 4 years elder than who is born in 1982

So, person who likes white - born in 1978

Hence, G likes white as G is born in 1978

We get,

G - 1978 - white

A - 1982 - black

D - 1994

C - 1997 - yellow

We are left with years of birth of B and E.

CASE I:

If, B is born in 1961, then E must be born in 1960

CASE II:

And, B is born in 1962, then E must be born in 1961

Now,

D is 34 years younger to one who likes red colour

So, one who likes red colour is born in year = 1994 - 34 = 1960

Thus, case II becomes invalid

We get,

Second eldest is B who doesn't like green

F doesn't like pink

So,

The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

So,

Case 1:

If B is pink then, 1 person is elder than B

Then, 0 person should be younger than who likes blue, so F must like blue.

And, D must like green

Case 2:

If D likes pink, no of person elder than D = 4

No of person younger than one who likes blue = 3 which is not possible

So, this case is invalid

The final arrangement is,

F like blue colour.

MAH-CET MBA Mock Test- 4 - Question 2
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Directions: A machine is programmed in such a manner that when some elements are provided, it gives output in the following pattern. Observe the pattern and answer accordingly.

INPUT: strongest 57 12 outrage biggest enormous 38 31 78 unanimous rigorous 91

Step I: strongest 57 12 outrage biggest unanimous 78 enormous 38 31 rigorous 91

Step II: strongest 57 12 outrage unanimous biggest 78 31 enormous 38 rigorous 91

Step III: strongest 12 outrage unanimous biggest enormous 78 31 38 57 rigorous 91

Step IV: strongest outrage unanimous biggest enormous rigorous 78 31 38 57 12 91

Step V: strongest unanimous biggest enormous rigorous outrage 78 31 38 57 12 91

Step VI: unanimous biggest enormous rigorous outrage strongest 78 31 38 57 12 91

Step VI is the final step of the input. Based on the above pattern, answer accordingly.

INPUT: 77 34 longest brigadier evaporation 12 91 pace intelligence 19 88 around

How many elements are there between evaporation and 19 in Step V?

Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 2
Step I: 77 34 longest brigadier evaporation intelligence 88 12 91 pace 19 around

Step II: 77 34 longest brigadier evaporation intelligence pace 88 19 12 91 around

Step III: 77 longest brigadier intelligence pace evaporation 88 19 34 12 91 around

Step IV: brigadier intelligence pace evaporation longest 88 19 34 77 12 91 around

Step V: brigadier intelligence pace evaporation longest around 88 19 34 77 12 91

Step VI: intelligence pace evaporation longest around brigadier 88 19 34 77 12 91

The output is arranged in such a manner that in the final step the word starting with a vowel having most letters in it will be placed first, followed by the word that starts with a consonant having least letters in it. Numbers are arranged in a manner that the highest even number is followed by the lowest odd number. Further, in each step the first word is placed at the sixth position from the left end and shift towards its left in the next step. The highest number is placed seventh from the left end in the first step and thereafter the next number is placed to its right in the next step.

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MAH-CET MBA Mock Test- 4 - Question 3
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Directions: A machine is programmed in such a manner that when some elements are provided, it gives output in the following pattern. Observe the pattern and answer accordingly.

INPUT: strongest 57 12 outrage biggest enormous 38 31 78 unanimous rigorous 91

Step I: strongest 57 12 outrage biggest unanimous 78 enormous 38 31 rigorous 91

Step II: strongest 57 12 outrage unanimous biggest 78 31 enormous 38 rigorous 91

Step III: strongest 12 outrage unanimous biggest enormous 78 31 38 57 rigorous 91

Step IV: strongest outrage unanimous biggest enormous rigorous 78 31 38 57 12 91

Step V: strongest unanimous biggest enormous rigorous outrage 78 31 38 57 12 91

Step VI: unanimous biggest enormous rigorous outrage strongest 78 31 38 57 12 91

Step VI is the final step of the input. Based on the above pattern, answer accordingly.

INPUT: 77 34 longest brigadier evaporation 12 91 pace intelligence 19 88 around

What is the place of brigadier from the right end in Step III?

Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 3
Step I: 77 34 longest brigadier evaporation intelligence 88 12 91 pace 19 around

Step II: 77 34 longest brigadier evaporation intelligence pace 88 19 12 91 around

Step III: 77 longest brigadier intelligence pace evaporation 88 19 34 12 91 around

Step IV: brigadier intelligence pace evaporation longest 88 19 34 77 12 91 around

Step V: brigadier intelligence pace evaporation longest around 88 19 34 77 12 91

Step VI: intelligence pace evaporation longest around brigadier 88 19 34 77 12 91

The output is arranged in such a manner that in the final step the word starting with a vowel having most letters in it will be placed first, followed by the word that starts with a consonant having least letters in it. Numbers are arranged in a manner that the highest even number is followed by the lowest odd number. Further, in each step the first word is placed at the sixth position from the left end and shift towards its left in the next step. The highest number is placed seventh from the left end in the first step and thereafter the next number is placed to its right in the next step.

MAH-CET MBA Mock Test- 4 - Question 4
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Directions: A machine is programmed in such a manner that when some elements are provided, it gives output in the following pattern. Observe the pattern and answer accordingly.

INPUT: strongest 57 12 outrage biggest enormous 38 31 78 unanimous rigorous 91

Step I: strongest 57 12 outrage biggest unanimous 78 enormous 38 31 rigorous 91

Step II: strongest 57 12 outrage unanimous biggest 78 31 enormous 38 rigorous 91

Step III: strongest 12 outrage unanimous biggest enormous 78 31 38 57 rigorous 91

Step IV: strongest outrage unanimous biggest enormous rigorous 78 31 38 57 12 91

Step V: strongest unanimous biggest enormous rigorous outrage 78 31 38 57 12 91

Step VI: unanimous biggest enormous rigorous outrage strongest 78 31 38 57 12 91

Step VI is the final step of the input. Based on the above pattern, answer accordingly.

INPUT: 77 34 longest brigadier evaporation 12 91 pace intelligence 19 88 around

Which step will be penultimate step for the given input?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 4
Step I: 77 34 longest brigadier evaporation intelligence 88 12 91 pace 19 around

Step II: 77 34 longest brigadier evaporation intelligence pace 88 19 12 91 around

Step III: 77 longest brigadier intelligence pace evaporation 88 19 34 12 91 around

Step IV: brigadier intelligence pace evaporation longest 88 19 34 77 12 91 around

Step V: brigadier intelligence pace evaporation longest around 88 19 34 77 12 91

Step VI: intelligence pace evaporation longest around brigadier 88 19 34 77 12 91

The output is arranged in such a manner that in the final step the word starting with a vowel having most letters in it will be placed first, followed by the word that starts with a consonant having least letters in it. Numbers are arranged in a manner that the highest even number is followed by the lowest odd number. Further, in each step the first word is placed at the sixth position from the left end and shift towards its left in the next step. The highest number is placed seventh from the left end in the first step and thereafter the next number is placed to its right in the next step.

MAH-CET MBA Mock Test- 4 - Question 5
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Directions: A machine is programmed in such a manner that when some elements are provided, it gives output in the following pattern. Observe the pattern and answer accordingly.

INPUT: strongest 57 12 outrage biggest enormous 38 31 78 unanimous rigorous 91

Step I: strongest 57 12 outrage biggest unanimous 78 enormous 38 31 rigorous 91

Step II: strongest 57 12 outrage unanimous biggest 78 31 enormous 38 rigorous 91

Step III: strongest 12 outrage unanimous biggest enormous 78 31 38 57 rigorous 91

Step IV: strongest outrage unanimous biggest enormous rigorous 78 31 38 57 12 91

Step V: strongest unanimous biggest enormous rigorous outrage 78 31 38 57 12 91

Step VI: unanimous biggest enormous rigorous outrage strongest 78 31 38 57 12 91

Step VI is the final step of the input. Based on the above pattern, answer accordingly.

INPUT: 77 34 longest brigadier evaporation 12 91 pace intelligence 19 88 around

Which step is the following output, 'brigadier intelligence pace evaporation longest 88 19 34 77 12 91 around'?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 5
Step I: 77 34 longest brigadier evaporation intelligence 88 12 91 pace 19 around

Step II: 77 34 longest brigadier evaporation intelligence pace 88 19 12 91 around

Step III: 77 longest brigadier intelligence pace evaporation 88 19 34 12 91 around

Step IV: brigadier intelligence pace evaporation longest 88 19 34 77 12 91 around

Step V: brigadier intelligence pace evaporation longest around 88 19 34 77 12 91

Step VI: intelligence pace evaporation longest around brigadier 88 19 34 77 12 91

The output is arranged in such a manner that in the final step the word starting with a vowel having most letters in it will be placed first, followed by the word that starts with a consonant having least letters in it. Numbers are arranged in a manner that the highest even number is followed by the lowest odd number. Further, in each step the first word is placed at the sixth position from the left end and shift towards its left in the next step. The highest number is placed seventh from the left end in the first step and thereafter the next number is placed to its right in the next step.

MAH-CET MBA Mock Test- 4 - Question 6
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Directions: A machine is programmed in such a manner that when some elements are provided, it gives output in the following pattern. Observe the pattern and answer accordingly.

INPUT: strongest 57 12 outrage biggest enormous 38 31 78 unanimous rigorous 91

Step I: strongest 57 12 outrage biggest unanimous 78 enormous 38 31 rigorous 91

Step II: strongest 57 12 outrage unanimous biggest 78 31 enormous 38 rigorous 91

Step III: strongest 12 outrage unanimous biggest enormous 78 31 38 57 rigorous 91

Step IV: strongest outrage unanimous biggest enormous rigorous 78 31 38 57 12 91

Step V: strongest unanimous biggest enormous rigorous outrage 78 31 38 57 12 91

Step VI: unanimous biggest enormous rigorous outrage strongest 78 31 38 57 12 91

Step VI is the final step of the input. Based on the above pattern, answer accordingly.

INPUT: 77 34 longest brigadier evaporation 12 91 pace intelligence 19 88 around

Which element is second to the left of the element which is third from the right end in Step V?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 6
Step I: 77 34 longest brigadier evaporation intelligence 88 12 91 pace 19 around

Step II: 77 34 longest brigadier evaporation intelligence pace 88 19 12 91 around

Step III: 77 longest brigadier intelligence pace evaporation 88 19 34 12 91 around

Step IV: brigadier intelligence pace evaporation longest 88 19 34 77 12 91 around

Step V: brigadier intelligence pace evaporation longest around 88 19 34 77 12 91

Step VI: intelligence pace evaporation longest around brigadier 88 19 34 77 12 91

The output is arranged in such a manner that in the final step the word starting with a vowel having most letters in it will be placed first, followed by the word that starts with a consonant having least letters in it. Numbers are arranged in a manner that the highest even number is followed by the lowest odd number. Further, in each step the first word is placed at the sixth position from the left end and shift towards its left in the next step. The highest number is placed seventh from the left end in the first step and thereafter the next number is placed to its right in the next step.

MAH-CET MBA Mock Test- 4 - Question 7
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Direction: 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- 4 - Question 7
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- 4 - Question 8
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Direction: In each question below is given a statement followed by two courses of action numbered I and II. You have to assume everything in the statement to be true, then decided which of the two suggested courses of action logically follows for pursuing.

Statement: The Minister said that the teachers are still not familiarized with the need, importance and meaning of population education in the higher education system. They are not even clearly aware about their role and responsibilities in the population education programme.

Courses of action:

I. Population education programme should be included in the College action Curriculum.

II. Orientation programme should be conducted for teachers on Population education.
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 8
Teachers should be familiarized with the need, importance and meaning of population education in the higher education system. Hence, instead of including the programme in the college curriculum, orientation programme will be more efficient.

Hence, only II option follows, option (b).

MAH-CET MBA Mock Test- 4 - Question 9
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Direction: In each question below is given a statement followed by two courses of action numbered I and II. You have to assume everything in the statement to be true, then decided which of the two suggested courses of action logically follows for pursuing.

Statement: Financial stringency prevented the State Government from paying salaries to its employees since April this year.

Courses of action:

I. The State Government should immediately curtail the staff strength at least by 30%.

II. The State Government should reduce wasteful expenditure and arrange to pay the salaries of its employees.
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 9
Laying off 30% of the state government employees will create a huge backlash from the employees and it does not help to solve the existing financial stringency.

If state government focuses on its expenditures and tries to reduce unnecessary expenditures, then it will help the government to solve the issue.

Hence, only option II follows.

MAH-CET MBA Mock Test- 4 - Question 10
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Direction: In the following question, the symbol @, ©, *, $ and # are used with the following meaning:

‘P @ Q’ means ‘P is neither smaller than nor equal to Q’.

‘P © Q’ means ‘P is not smaller than Q’.

‘P * Q’ means ‘P is not greater than Q’.

‘P $ Q’ means’ P is neither smaller than nor greater than Q’.

‘P # Q’ means ‘P is neither greater than nor equal to Q’.

Now in the following question, assuming the given statements to be true, find which of the two conclusions I and II have given below them is/are definitely true?

Statements : Z#N, F©N, F*K

Conclusion :

I. K $ N

II. K @ Z
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 10
Z < N ..... (i);

F > N ..... (ii);

F < K ..... (iii)

Combining all, we get K > F > N > Z. Hence, K > N and K > Z

Hence, conclusion I (K = N) is not necessarily true but conclusion II (K > Z) is true.

MAH-CET MBA Mock Test- 4 - Question 11
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Direction: In the following question, the symbol @, ©, *, $ and # are used with the following meaning:

‘P @ Q’ means ‘P is neither smaller than nor equal to Q’.

‘P © Q’ means ‘P is not smaller than Q’.

‘P * Q’ means ‘P is not greater than Q’.

‘P $ Q’ means’ P is neither smaller than nor greater than Q’.

‘P # Q’ means ‘P is neither greater than nor equal to Q’.

Now in the following question, assuming the given statements to be true, find which of the two conclusions I and II have given below them is/are definitely true?

Statements : D $ T, T©M, M # K

Conclusions :

I. M $ D

II. D@ M
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 11
Given statement,

D $ T, T©M, M # K ⇒ D =T ≥ M; K > M

conclusions,

I. M $ D ⇒ M = D

II. D@ M ⇒ D > M

Therefore, either conclusion I or II is true.

Hence, the correct option is (C).

MAH-CET MBA Mock Test- 4 - Question 12
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Direction: In the following question, the symbol @, ©, *, $ and # are used with the following meaning:

‘P @ Q’ means ‘P is neither smaller than nor equal to Q’.

‘P © Q’ means ‘P is not smaller than Q’.

‘P * Q’ means ‘P is not greater than Q’.

‘P $ Q’ means’ P is neither smaller than nor greater than Q’.

‘P # Q’ means ‘P is neither greater than nor equal to Q’.

Now in the following question, assuming the given statements to be true, find which of the two conclusions I and II have given below them is/are definitely true?

Statements: W©A, B*A, B@M

Conclusions:

I. B # W

II. W $ B
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 12
W > A ... (i);

B < A ...(ii);

B > M ...(iii)

Combining all, we get W > A > B >M. Hence, B

Hence, either conclusion I or II is true.

MAH-CET MBA Mock Test- 4 - Question 13
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Direction: In the following question, the symbol @, ©, *, $ and # are used with the following meaning:

‘P @ Q’ means ‘P is neither smaller than nor equal to Q’.

‘P © Q’ means ‘P is not smaller than Q’.

‘P * Q’ means ‘P is not greater than Q’.

‘P $ Q’ means’ P is neither smaller than nor greater than Q’.

‘P # Q’ means ‘P is neither greater than nor equal to Q’.

Now in the following question, assuming the given statements to be true, find which of the two conclusions I and II have given below them is/are definitely true?

Statements: J * M, M $ N, N # T

Conclusions:

I. T @ J

II. T $ J
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 13
J < M .... (i);

M = N ....(ii);

N < T ....(iii)

Combining all, we get J < M = N < T.

Hence T > J Hence, only conclusion I is true.

MAH-CET MBA Mock Test- 4 - Question 14
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Direction: In the following question, the symbol @, ©, *, $ and # are used with the following meaning:

‘P @ Q’ means ‘P is neither smaller than nor equal to Q’.

‘P © Q’ means ‘P is not smaller than Q’.

‘P * Q’ means ‘P is not greater than Q’.

‘P $ Q’ means’ P is neither smaller than nor greater than Q’.

‘P # Q’ means ‘P is neither greater than nor equal to Q’.

Now in the following question, assuming the given statements to be true, find which of the two conclusions I and II have given below them is/are definitely true?

Statements: V * F, F @ R, R © G

Conclusions:

I. G # V

II. G@ V
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 14
V < F ....(i);

F > R ....(ii);

R > G ....(iii)

Combining (ii) and (iii), we get F > R > G....(iv) Comparing (i) and (iii), we can’t get any specific relationship between G and V.

Hence, both conclusions are not true.

MAH-CET MBA Mock Test- 4 - Question 15
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Direction: In each question below is given a statement followed by two courses of action numbered I and II. You have to assume everything in the statement to be true, then decided which of the two suggested courses of action logically follows for pursuing.

Statement: One of the problems facing the food processing industry is the irregular supply of raw material. The producers of raw material are not getting a reasonable price.

Courses of action:

I. The government should regulate the supply of raw material to other industries also.

II. The government should announce an attractive package to ensure regular supply of raw material for food processing industry.
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 15
Action I will aggravate the existing problem.

If the government announces an attractive package, then producers of raw material will be motivated to supply raw material regularly.

Only II option follows, hence option (b).

MAH-CET MBA Mock Test- 4 - Question 16
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Direction: In each question below is given a statement followed by two courses of action numbered I and II. You have to assume everything in the statement to be true, then decided which of the two suggested courses of action logically follows for pursuing.

Statement: The office In charge of a Company had a hunch that some money was missing from the safe.

Courses of action:

I. He should get it recounted with the help of the staff and Check it with the balance sheet.

II. He should inform the police.
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 16
Before informing the police, he should confirm that some money was missing by recounting it with the help of staff.

Only option I follows, Hence option (a)

MAH-CET MBA Mock Test- 4 - Question 17
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In a certain language, ‘cot tip mot’ means ‘singing is appreciable’; ‘mot baj min’ means ‘dancing is good’ and ‘tip nop baj’ means ‘singing and dancing’. Which of the following means ‘good’ in that code language?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 17
mot = is, baj = dancing, min = good. Hence, option (b).
MAH-CET MBA Mock Test- 4 - Question 18
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Direction: In each of the question given below, there are two statements followed by two conclusions numbered 1 and 2. You have to take the given statements to be true even if they seem to be at variance with commonly known facts and then decide which of the given conclusions logically follow (s) from the given statements.

Statements:

(a) All mangoes have juices

(b) No juice has colour

(c) Some colours have liquid.

Conclusions:

I. Some mangoes have colour.

II. At least some juices have liquid.
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 18

Conclusion I: This conclusion cannot be drawn because there is no intersection between mango and colour.

Conclusion II: From the Venn diagram, we can see that there is a possibility that no liquid is on juice. Thus, statement II is not valid always.

Hence, neither conclusion I nor II follows. Hence, [4].

MAH-CET MBA Mock Test- 4 - Question 19
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Direction: In each of the question given below, there are two statements followed by two conclusions numbered 1 and 2. You have to take the given statements to be true even if they seem to be at variance with commonly known facts and then decide which of the given conclusions logically follow (s) from the given statements.

Statements:

(a) Some boys are cricketers.

(b) All cricketers are players.

(c) Some players are winners.

Conclusions:

I. Some winners are cricketers.

II. Some players are boys.
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 19

Conclusion I: from the Venn diagram we can see that I is not a definite conclusion as there could be a possibility that no winner is cricketer. Therefore, conclusion I is a possibility but not a definite conclusion.

Conclusion II: From the Venn diagram we can see that some players that are cricketers are definitely boys. Hence, only conclusion II follows. Hence, [2].

MAH-CET MBA Mock Test- 4 - Question 20
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Direction: In each of the question given below, there are two statements followed by two conclusions numbered 1 and 2. You have to take the given statements to be true even if they seem to be at variance with commonly known facts and then decide which of the given conclusions logically follow (s) from the given statements.

Statements:

(a) Some pencil are pins

(b) all pins are iron.

(c) Some iron is element.

Conclusion:

I. Some element is pins.

II. No iron is pencil
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 20

Conclusion I: From the Venn diagram we can see that I is not a definite conclusion as there could be a possibility that no Element is pin. Thus, conclusion I is a possibility but not a definite conclusion.

Conclusion II: From the Venn diagram we can see that some iron (that is pin) has to be pencil. Thus, II is not a definite conclusion. Thus, conclusion II doesn’t follow. Hence, neither conclusion I nor II follows. Hence, [4].

MAH-CET MBA Mock Test- 4 - Question 21
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One morning after sunrise, Reeta and Kavita were talking to each other face to face at Tilak Square. If Kavita’s shadow was exactly to the right to Reeta, which direction was Kavita facing?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 21

The sun in the east. Shadows are marked by arrows. Kavita’s shadow was exactly to the right of Reeta, so Reeta is facing South and Kavita is facing north (from the diagram). Hence, option (a).

MAH-CET MBA Mock Test- 4 - Question 22
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If A+B means B is the father of A, A*B means A is the son of B, A-B means B is the daughter of A, A÷B means B is the mother of A. Then in which of the following given options, A is the maternal uncle of E?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 22
From option (1), gender of A is not known.

From option (2),

Hence, A is the maternal uncle of E.

So need not check for other options.

MAH-CET MBA Mock Test- 4 - Question 23
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If A-B means B is the husband of A, A*B means A is the brother of B, A+B means B is the son of A, A÷B means B is the father of A. Then in which of the following given options, E is the cousin of A?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 23
From option (1), E is the son of A.

From option (2), A and E are not related as cousins, but as in-law.

From option (3),

Hence, A and E are cousins.

MAH-CET MBA Mock Test- 4 - Question 24
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Directions: Seven people A, B, C, D, E, F and G are born in different years and like different colours - red, blue, pink, yellow, green, black and white not necessarily in same order.

  • B and E are born in consecutive years (B being younger), one of them been born in 1961.

  • The one who likes black is born in 1982 and is 4 years younger than the one who likes white.

  • The difference between present ages of A and D is 10 less than the difference between present ages of G and F. (A and G are elder than D and F respectively)

  • G is born in 1978.

  • F is born in 2000 and is youngest and doesn't like pink.

  • C likes yellow and is born three years after D was born, who is 6 years elder than F and 34 years younger to one who likes red colour.

  • The second eldest person doesn't like green.

  • The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

Who is the eldest person?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 24

F is born in 2000

Since, F is youngest, so, no one is born after F.

G is born in 1978

D is 6 years elder than F

So,

D is born in 1994

C is born 3 years after D

C is born in 1997

So, difference in ages of G and F = 22 years

Thus, Difference in ages of A and D = 22 - 10 = 12 years

So, A is born in 1982

The one who likes black is born in 1982

White is liked by the one who is 4 years elder than who is born in 1982

So, person who likes white - born in 1978

Hence, G likes white as G is born in 1978

We get,

G - 1978 - white

A - 1982 - black

D - 1994

C - 1997 - yellow

We are left with years of birth of B and E.

CASE I:

If, B is born in 1961, then E must be born in 1960

CASE II:

And, B is born in 1962, then E must be born in 1961

Now,

D is 34 years younger to one who likes red colour

So, one who likes red colour is born in year = 1994 - 34 = 1960

Thus, case II becomes invalid

We get,

Second eldest is B who doesn't like green

F doesn't like pink

So,

The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

So,

Case 1:

If B is pink then, 1 person is elder than B

Then, 0 person should be younger than who likes blue, so F must like blue.

And, D must like green

Case 2:

If D likes pink, no of person elder than D = 4

No of person younger than one who likes blue = 3 which is not possible

So, this case is invalid

The final arrangement is,

E is the eldest.

MAH-CET MBA Mock Test- 4 - Question 25
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Directions: Seven people A, B, C, D, E, F and G are born in different years and like different colours - red, blue, pink, yellow, green, black and white not necessarily in same order.

  • B and E are born in consecutive years (B being younger), one of them been born in 1961.

  • The one who likes black is born in 1982 and is 4 years younger than the one who likes white.

  • The difference between present ages of A and D is 10 less than the difference between present ages of G and F. (A and G are elder than D and F respectively)

  • G is born in 1978.

  • F is born in 2000 and is youngest and doesn't like pink.

  • C likes yellow and is born three years after D was born, who is 6 years elder than F and 34 years younger to one who likes red colour.

  • The second eldest person doesn't like green.

  • The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

What is the difference between the present ages of E and F.
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 25

F is born in 2000

Since, F is youngest, so, no one is born after F.

G is born in 1978

D is 6 years elder than F

So,

D is born in 1994

C is born 3 years after D

C is born in 1997

So, difference in ages of G and F = 22 years

Thus, Difference in ages of A and D = 22 - 10 = 12 years

So, A is born in 1982

The one who likes black is born in 1982

White is liked by the one who is 4 years elder than who is born in 1982

So, person who likes white - born in 1978

Hence, G likes white as G is born in 1978

We get,

G - 1978 - white

A - 1982 - black

D - 1994

C - 1997 - yellow

We are left with years of birth of B and E.

CASE I:

If, B is born in 1961, then E must be born in 1960

CASE II:

And, B is born in 1962, then E must be born in 1961

Now,

D is 34 years younger to one who likes red colour

So, one who likes red colour is born in year = 1994 - 34 = 1960

Thus, case II becomes invalid

We get,

Second eldest is B who doesn't like green

F doesn't like pink

So,

The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

So,

Case 1:

If B is pink then, 1 person is elder than B

Then, 0 person should be younger than who likes blue, so F must like blue.

And, D must like green

Case 2:

If D likes pink, no of person elder than D = 4

No of person younger than one who likes blue = 3 which is not possible

So, this case is invalid

The final arrangement is,

E is born in 1960 and F in 2000

So, difference = 40 years

MAH-CET MBA Mock Test- 4 - Question 26
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Directions: Seven people A, B, C, D, E, F and G are born in different years and like different colours - red, blue, pink, yellow, green, black and white not necessarily in same order.

  • B and E are born in consecutive years (B being younger), one of them been born in 1961.

  • The one who likes black is born in 1982 and is 4 years younger than the one who likes white.

  • The difference between present ages of A and D is 10 less than the difference between present ages of G and F. (A and G are elder than D and F respectively)

  • G is born in 1978.

  • F is born in 2000 and is youngest and doesn't like pink.

  • C likes yellow and is born three years after D was born, who is 6 years elder than F and 34 years younger to one who likes red colour.

  • The second eldest person doesn't like green.

  • The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

A is born in _____ and likes _____ colour.
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 26
F is born in 2000

Since, F is youngest, so, no one is born after F.

G is born in 1978

D is 6 years elder than F

So,

D is born in 1994

C is born 3 years after D

C is born in 1997

So, difference in ages of G and F = 22 years

Thus, Difference in ages of A and D = 22 - 10 = 12 years

So, A is born in 1982

The one who likes black is born in 1982

White is liked by the one who is 4 years elder than who is born in 1982

So, person who likes white - born in 1978

Hence, G likes white as G is born in 1978

We get,

G - 1978 - white

A - 1982 - black

D - 1994

C - 1997 - yellow

We are left with years of birth of B and E.

CASE I:

If, B is born in 1961, then E must be born in 1960

CASE II:

And, B is born in 1962, then E must be born in 1961

Now,

D is 34 years younger to one who likes red colour

So, one who likes red colour is born in year = 1994 - 34 = 1960

Thus, case II becomes invalid

We get,

Second eldest is B who doesn't like green

F doesn't like pink

So,

The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

So,

Case 1:

If B is pink then, 1 person is elder than B

Then, 0 person should be younger than who likes blue, so F must like blue.

And, D must like green

Case 2:

If D likes pink, no of person elder than D = 4

No of person younger than one who likes blue = 3 which is not possible

So, this case is invalid

The final arrangement is,

A is born in 1982 and likes black.

MAH-CET MBA Mock Test- 4 - Question 27
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Directions: Seven people A, B, C, D, E, F and G are born in different years and like different colours - red, blue, pink, yellow, green, black and white not necessarily in same order.

  • B and E are born in consecutive years (B being younger), one of them been born in 1961.

  • The one who likes black is born in 1982 and is 4 years younger than the one who likes white.

  • The difference between present ages of A and D is 10 less than the difference between present ages of G and F. (A and G are elder than D and F respectively)

  • G is born in 1978.

  • F is born in 2000 and is youngest and doesn't like pink.

  • C likes yellow and is born three years after D was born, who is 6 years elder than F and 34 years younger to one who likes red colour.

  • The second eldest person doesn't like green.

  • The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

When is B born?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 27
F is born in 2000

Since, F is youngest, so, no one is born after F.

G is born in 1978

D is 6 years elder than F

So,

D is born in 1994

C is born 3 years after D

C is born in 1997

So, difference in ages of G and F = 22 years

Thus, Difference in ages of A and D = 22 - 10 = 12 years

So, A is born in 1982

The one who likes black is born in 1982

White is liked by the one who is 4 years elder than who is born in 1982

So, person who likes white - born in 1978

Hence, G likes white as G is born in 1978

We get,

G - 1978 - white

A - 1982 - black

D - 1994

C - 1997 - yellow

We are left with years of birth of B and E.

CASE I:

If, B is born in 1961, then E must be born in 1960

CASE II:

And, B is born in 1962, then E must be born in 1961

Now,

D is 34 years younger to one who likes red colour

So, one who likes red colour is born in year = 1994 - 34 = 1960

Thus, case II becomes invalid

We get,

Second eldest is B who doesn't like green

F doesn't like pink

So,

The no. of people elder than the one who likes pink are 1 more than the no. of people younger than one who likes blue.

So,

Case 1:

If B is pink then, 1 person is elder than B

Then, 0 person should be younger than who likes blue, so F must like blue.

And, D must like green

Case 2:

If D likes pink, no of person elder than D = 4

No of person younger than one who likes blue = 3 which is not possible

So, this case is invalid

The final arrangement is,

B is born in 1961.

MAH-CET MBA Mock Test- 4 - Question 28
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Directions : Study the following information carefully and answer the questions given below:

8 persons A, B, C, D, E, F, G and H live in 2 different buildings Building X and Building Y. Both buildings have 5 floors each with ground floor numbered 1; floor above it numbered 2 and so on until topmost is numbered 5. One of the floors are vacant in each building. Only one person lives on each floor of the building X and building Y. No same number of floors are vacant. Only one person lives below E. At least one person lives between F and G. Number of persons above D is one more than the person below D. D lives in building Y. No one lives between F and B. G lives one of the floors below B. D lives immediately above G. No one lives between E and H and they both lives on odd numbered floor. F and A lives on the same numbered floor. C lives below A. 1st floor is not vacant in any of the building. B's and E's floor numbers are not same.

How many persons live between C and H?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 28
We Know that:
  • Number of persons above D is one more than below D. D lives in building Y and 1st floor is not vacant so D must live on 2nd floor.
  • D lives immediate above G. G must live on 1st floor in building Y.
  • Only one person lives below E and 1st floor is not vacant so we get 2 possibilities. Either E lives on 2nd or 3rd floor.

Case 1:

  • No one lives between E and H and they both live on odd numbers of floor. So this case gets rejected.

Case 2:

  • No one lives between E and H and they both live on odd numbers of floor. So H must live on 1st floor.

  • G lives one of the floors below B. it means B lives also in building Y and B's and E's floor numbers are not same. So B lives either 4thand 5th floor.

If B lives on 4th floor-

  • At least one person lives between F and G. it means F lives also in Building Y. A and C live in building X.
  • C lives below A. So C lives on 4th and A lives on 5th floor in building X.

  • No one lives between F and B. F and A live on the same number of floor. So F lives on 5th floor in building Y and 3rd floor of building Y is Vacant.

Here is the final table:

Only E lives between C and H.

MAH-CET MBA Mock Test- 4 - Question 29
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Directions : Study the following information carefully and answer the questions given below:

8 persons A, B, C, D, E, F, G and H live in 2 different buildings Building X and Building Y. Both buildings have 5 floors each with ground floor numbered 1; floor above it numbered 2 and so on until topmost is numbered 5. One of the floors are vacant in each building. Only one person lives on each floor of the building X and building Y. No same number of floors are vacant. Only one person lives below E. At least one person lives between F and G. Number of persons above D is one more than the person below D. D lives in building Y. No one lives between F and B. G lives one of the floors below B. D lives immediately above G. No one lives between E and H and they both lives on odd numbered floor. F and A lives on the same numbered floor. C lives below A. 1st floor is not vacant in any of the building. B's and E's floor numbers are not same.

Who among the following does not belongs to the group?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 29
We Know that:
  • Number of persons above D is one more than below D. D lives in building Y and 1st floor is not vacant so D must live on 2nd floor.
  • D lives immediate above G. G must live on 1st floor in building Y.
  • Only one person lives below E and 1st floor is not vacant so we get 2 possibilities. Either E lives on 2nd or 3rd floor.

Case 1:

  • No one lives between E and H and they both live on odd numbers of floor. So this case gets rejected.

Case 2:

  • No one lives between E and H and they both live on odd numbers of floor. So H must live on 1st floor.

  • G lives one of the floors below B. it means B lives also in building Y and B's and E's floor numbers are not same. So B lives either 4thand 5th floor.

If B lives on 4th floor-

  • At least one person lives between F and G. it means F lives also in Building Y. A and C live in building X.
  • C lives below A. So C lives on 4th and A lives on 5th floor in building X.

  • No one lives between F and B. F and A live on the same number of floor. So F lives on 5th floor in building Y and 3rd floor of building Y is Vacant.

Here is the final table:

All the persons lives in building X except F.

MAH-CET MBA Mock Test- 4 - Question 30
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Directions : Study the following information carefully and answer the questions given below:

8 persons A, B, C, D, E, F, G and H live in 2 different buildings Building X and Building Y. Both buildings have 5 floors each with ground floor numbered 1; floor above it numbered 2 and so on until topmost is numbered 5. One of the floors are vacant in each building. Only one person lives on each floor of the building X and building Y. No same number of floors are vacant. Only one person lives below E. At least one person lives between F and G. Number of persons above D is one more than the person below D. D lives in building Y. No one lives between F and B. G lives one of the floors below B. D lives immediately above G. No one lives between E and H and they both lives on odd numbered floor. F and A lives on the same numbered floor. C lives below A. 1st floor is not vacant in any of the building. B's and E's floor numbers are not same.

Which of the following is true regarding B?
Detailed Solution for MAH-CET MBA Mock Test- 4 - Question 30
We Know that:
  • Number of persons above D is one more than below D. D lives in building Y and 1st floor is not vacant so D must live on 2nd floor.
  • D lives immediate above G. G must live on 1st floor in building Y.
  • Only one person lives below E and 1st floor is not vacant so we get 2 possibilities. Either E lives on 2nd or 3rd floor.

Case 1:

  • No one lives between E and H and they both live on odd numbers of floor. So this case gets rejected.

Case 2:

  • No one lives between E and H and they both live on odd numbers of floor. So H must live on 1st floor.

  • G lives one of the floors below B. it means B lives also in building Y and B's and E's floor numbers are not same. So B lives either 4th and 5th floor.

If B lives on 4th floor-

  • At least one person lives between F and G. it means F lives also in Building Y. A and C live in building X.
  • C lives below A. So C lives on 4th and A lives on 5th floor in building X.

  • No one lives between F and B. F and A live on the same number of floor. So F lives on 5th floor in building Y and 3rd floor of building Y is Vacant.

Here is the final table:

B lives just above the vacant floor in one of the building.

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