# Logical Reasoning Questions for CMAT Set-2

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## Logical Reasoning Questions for CMAT Set-2

Download important Logical Reasoning Questions for CMAT Set-2 PDF based on previously asked questions in the CMAT exam. Practice Logical Reasoning QuestionsÂ  Set-2 PDF for the CMAT exam.

Instructions

DIRECTIONS for the following three questions: Answer the questions on the basis of the information given below.

Seven varsity basketball players (A, B, C, D, E, F, and G) are to be honoured at a special luncheon. The players will be seated on the dais in a row. A and G have to leave the luncheon early and so must be seated at the extreme right. B will receive the most valuable player’s trophy and so must be in the centre to facilitate presentation. C and D are bitter rivals and therefore must be seated as far apart as possible.

Question 1:Â Which of the following cannot be seated at either end?

a)Â C

b)Â D

c)Â F

d)Â G

Instructions

DIRECTIONS for the following three questions: Answer the questions on the basis of the information given below.

Five women decided to go shopping to M.G. Road, Bangalore. They arrived at the designated meeting place in the following order:

1. Archana,
2. Chellamma,
3. Dhenuka,
4. Helen, and
5. Shahnaz.

Each woman spent at least Rs. 1000. Below are some additional facts about how much they spent during their shopping spree.

i. The woman who spent Rs. 2234 arrived before the lady who spent Rs. 1193.

ii. One woman spent Rs. 1340 and she was not Dhenuka.

iii. One woman spent Rs. 1378 more than Chellamma.

iv. One woman spent Rs. 2517 and she was not Archana.

v. Helen spent more than Dhenuka.

vi. Shahnaz spent the largest amount and Chellamma the smallest.

Question 2:Â Which of the following amounts was spent by one of them?

a)Â Rs. 1139

b)Â Rs. 1378

c)Â Rs. 2571

d)Â Rs. 2718

Instructions

Mathematicians are assigned a number called Erdos number (named after the famous mathematician, Paul Erdos). Only Paul Erdos himself has an Erdos number of zero. Any mathematician who has written a research paper with Erdos has an Erdos number of 1.For other mathematicians, the calculation of his/her Erdos number is illustrated below:

Suppose that a mathematician X has co-authored papers with several other mathematicians. ‘From among them, mathematician Y has the smallest Erdos number. Let the Erdos number of Y be y. Then X has an Erdos number of y+1. Hence any mathematician with no co-authorship chain connected to Erdos has an Erdos number of infinity. :

In a seven day long mini-conference organized in memory of Paul Erdos, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdos number. Nobody had an Erdos number less than that of F.

On the third day of the conference F co-authored a paper jointly with A and C. This reduced the average Erdos number of the group of eight mathematicians to 3. The Erdos numbers of B, D, E, G and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdos number of the group of eight to as low as 3.

â€˘ At the end of the third day, five members of this group had identical Erdos numbers while the other three had Erdos numbers distinct from each other.

â€˘ On the fifth day, E co-authored a paper with F which reduced the group’s average Erdos number by 0.5. The Erdos numbers of the remaining six were unchanged with the writing of this paper.

â€˘ No other paper was written during the conference.

Question 3:Â How many participants in the conference did not change their Erdos number during the conference?

a)Â 2

b)Â 3

c)Â 4

d)Â 5

e)Â Cannot be determined

Instructions

Directions for the following three questions: Answer the questions based on the passage below.

A group of three or four has to be selected from seven persons. Among the seven are two women: Fiza and Kavita, and five men: Ram, Shyam, David, Peter and Rahim. Ram would not like to be in the group If Shyam is also selected. Shyam and Rahim want to be selected together in the group. Kavita would like to be in the group only if David is also there. David, if selected, would not like Peter in the group. Ram would like to be in the group only if Peter is also there. David insists that Fiza be selected in case he is there in the group.

Question 4:Â Which of the following is a feasible group in four?

a)Â Ram, Peter, Fiza and Rahim

b)Â Shyam, Rahim, Kavita and David

c)Â Shyam, Rahim, Fiza and David

d)Â Fiza, David, Ram and Peter

Question 5:Â A king has unflinching loyalty from eight of his ministers M1 to M8, but he has to select only four to make a cabinet committee. He decides to choose these four such that each selected person shares a liking with at least one of the other three selected. The selected persons must also hate at least one of the likings of any of the other three persons selected.
M1 likes fishing and smoking, but hates gambling.
M2 likes smoking and drinking, but hates fishing.
M3 likes gambling, but hates smoking,
M4 likes mountaineering, but hates drinking,
M5 likes drinking, but hates smoking and mountaineering.
M6 likes fishing, but hates smoking and mountaineering.
M7 likes gambling and mountaineering, but hates fishing.
M8 likes smoking and gambling, but hates mountaineering.
Who are the four people selected by the king?

a)Â M1, M2, M5 and M6

b)Â M3, M4, M5 and M6

c)Â M4, M5, M6 and M8

d)Â M1, M2, M4 and M7

Instructions

Directions for the next 5 questions:

Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament is conducted in two stages. In the first stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of the first stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprises of several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated, The team that remains undefeated in the second stage is declared the winner and claims the Gold Cup.

The tournament rules are such that each match results in a winner and a loser with no possibility of a tie. In the first stage a team earns one point for each win and no points for a loss. At the end of the first stage teams in each group are ranked on the basis of total points to determine the qualifiers advancing to the next stage. Ties are resolved by a series of complex tie-breaking rules so that exactly four teams from each group advance to the next stage.

Question 6:Â What is the total number of matches played in the tournament?

a)Â 28

b)Â 55

c)Â 63

d)Â 35

Instructions

Four students â€” Ashish, Dhanraj, Felix and Sameer sat for the Common Entrance Exam for Management (CEEM). One student got admission offers from three NIMs (National Institutes of Management), another from two NIMs, the third from one NIM, while the fourth got none. Below are some of the facts about who got admission offers from how many NIMs and what is their educational background.

I. The one who is an engineer didnâ€™t get as many admissions as Ashish.

II. The one who got offer for admissions in two NIMs isnâ€™t Dhanraj nor is he a chartered accountant.

III. Sameer is an economist.

IV. Dhanraj isnâ€™t an engineer and received more admission offers than Ashish.

V. The doctor got the most number of admission offers.

Question 7:Â Which one of the following statements is necessarily true?

a)Â Ashish is a chartered accountant and got offer for admission in three NIMs.

b)Â Dhanraj is a doctor and got admission offer in one NIM.

c)Â Sameer is an economist who got admission offers in two NIMs.

d)Â Felix who is not an engineer did not get any offer for admission.

Question 8:Â Six persons are playing a card game sitting around a circular table. Suresh is facing Raghubir who is to the left of Ajay and to the right of Pramod. Ajay is to the left of Dhiraj. Yogendra is to the left of Pramod. If Dhiraj exchanges his seat with Yogendra and Pramod exchanges with Raghubir, who will be sitting to the left of Dhiraj?

a)Â Yogendra

b)Â Raghubir

c)Â Suresh

d)Â Ajay

Question 9:Â Choose the set in which the statements are most logically related.
A. First-year students of this college like to enter for the prize.
B. All students of this college rank as University students.
C. First-year students of this college are entitled to enter for he prize.
D. Some who rank as University students are First-year students.
E. All University students are eligible to enter for the prize.
F. All those who like to are entitled to enter for the prize.

a)Â AEF

b)Â ABC

c)Â BEC

d)Â CDF

Instructions

The following questions relate to a game to be played by you and your friend. The game consists of a 4 x 4 board (see below) where each cell contains a positive integer. You and your friend make moves alternately. A move by any of the players consists of splitting the current board configuration into two equal halves and retaining one of them. In your moves you are allowed to split the board only vertically and to decide to retain either the left or the right half. Your friend, in his/her moves, can split the board only horizontally and can retain either the lower or the upper half. After two moves by each player a single cell will remain which can no longer be split and the number in that cell will be treated as the gain (in rupees) of the person who has started the game. A sample game is shown below. So your gain is Re.1. With the same initial board configuration as above and assuming that you have to make the first move, answer the following questions.
Initial Board

After your friends move (retain upper)

After your friends move (retain lower)

Question 10:Â If your first move is (retain right), then whatever moves your friend may select you can always force a gain of no less than

a)Â Rs.3

b)Â Rs.6

c)Â Rs.4

d)Â None of these

According to given conditions we can conclude ,

Hence , F cannot be seated at either end.

From the given statements, we can infer that the amounts spent by the women are Rs.2234, Rs.1193, Rs.1340, and Rs.2517. Also, we know that one of the 5 women spent Rs.1378 more than Chellamma. Also, we know that each woman spent at least Rs.1000.

The person who spent Rs.2234 cannot be the person who spent Rs.1378 more than Chellamma. Also, we know that Chellamma spent the least among the five women.

Case 1:
Chellamma spent Rs.1193.
=> One of the 5 women spent 1193+1378 = Rs.2571.

We know that Shahnaz spent the most. Therefore, Shahnaz should have spent Rs.2571.

Archana didn’t spend Rs.2517. Also, Helen spent more than Dhenuka.
Therefore, Helen should have spent Rs.2517.

Dhenuka didn’t spend Rs.1340.
Therefore, Dhenuka should have spent Rs. 2234 and Archana should have spent Rs.1340.

However, it has been given that the woman who spent Rs. 2234 arrived before the lady who spent Rs. 1193. According to the given order, Archana arrived before Dhenuka. Therefore, we can eliminate this case.

Case 2:

Rs.2517 is the highest amount spent. Shahnaz spent Rs.2517.
=> Amount spent by Chellamma = 2517 – 1378 = Rs.1139

Dhenuka didn’t spend Rs.1340. Helen spent more than Dhenuka. Therefore, Dhenuka should have spent Rs.1193.

Now, we know that theÂ woman who spent Rs. 2234 arrived before the lady who spent Rs. 1193. Helen did not arrive before Dhenuka but Archana did. Therefore, Archana should have spent Rs.2234 and Helen should have spent Rs.1340.

According to given conditions amount spent by everyone is,

Hence, option A is the right answer.

Let us consider the Erdos number of A,B,C,D,E,F,G,H be a,b,c,d,e,f,g,h where f is the min, a is infinity.

At the end of 3rd day, F co authored with A and C. Since F has min Erdos number ,the values of c,a will change to f+1 and the Erdos number of F will remain the same.Â Â  [Because according to Erdos principle if a person co-authors with some one who has higher Erdos number then the Erdos number of co-authors will be min Erdos value + 1]

Average of the mathematicians is 3
Sum of the Erdos number of eight mathematicians=24

Erdos number at the third day:f+1,b,f+1,d,e,f,g,h

At the end of the fifth day, F co-authors with E thereby changing the average to 2.5 and the Erdos number of rest of the mathematicians remain unchanged.
Sum of the Erdos numbers of eight mathematicians=20
So here the difference of 4[24-20] arose, which means e will be f+5 initially and changed to f+1 after co-authoring with F.

So the Erdos number at the third day:f+1,b,f+1,d,f+5,f,g,h

At the end of the third day, five mathematicians had the same Erdos number and the rest had distinct Erdos number from each other.

It cannot be f+5 because then there will be two mathematicians with the same Erdos number f+1.

So five mathematicians will have f+1, one with f+5, one with f, one with some different value say x

5(f+1)+f+5+f+x=24

7f+x=14

The only value which satisfies the above equation is f=1,x=7

Erdos number at the end of fifth day,f+1,b,f+1,d,f+1,f,g,h
On tabulating, we get

So B,D ,F,G,H are 5 participants in the conference who did not change their Erdos number during the conference.

David and Peter cannot be selected together. So, option d) is not possible.
David and Fiza have to be selected together. So, option b) is ruled out.
Shyam and Rahim have to be selected together. So, option a) is also ruled out.
Option c) is the correct answer

Looking at each option and the table. Only Option D satisfies the given conditionÂ Â that each selected person shares a liking with at least one of the other three selected. The selected persons must also hate at least one of the likings of any of the other three persons selected.

CMAT Free Solved Previous Papers.

In 1st stage there were 2 groups each group played 28 matches as $^8C_2$ = 28. So total 56 matches after 1st stage . IN 1st round of stage 2, 4 matches are played and in the next round we have 4 teams .The winner is one who defeats every team out of remaining 3Â matchesÂ . so 3 matches more needed. In total 56+4+3=63.

According to given information, we finally know that ashish is aÂ CA and will get 1 or no offers. dhanraj is doctor and gets 3 offers felix is an engineer and can get 2 offers.

Sameer is an economist and can get 2 offers.

The correct original circular arrangement sequence in clockwise manner is Suresh, Dhiraj, Ajay, Raghuveer, Pramod, Yogendra.

So after the changes, Suresh is to the left of Dhiraj.