What are CAT LR miscellaneous questions?

CAT LR miscellaneous questions include a variety of logical reasoning problems that do not fall under fixed categories like arrangements or puzzles.

How to prepare for CAT LR miscellaneous questions?

Practice different types of LR questions regularly, focus on understanding logic, and improve problem-solving speed through mock tests.

Are miscellaneous questions important in CAT LRDI?

Yes, miscellaneous questions are important as they test adaptability and logical thinking across different problem types in CAT.

What types of questions come under LR miscellaneous?

These include puzzles, reasoning-based scenarios, pattern identification, and mixed logic problems.

How to solve CAT LR miscellaneous questions quickly?

Identify the pattern, break down the problem, and apply elimination techniques to reach the correct answer efficiently.

Where can I practice CAT LR miscellaneous questions?

You can practice from previous year CAT papers, mock tests, and online LRDI question banks.

Are CAT LR miscellaneous questions difficult?

They can range from easy to difficult depending on the logic involved, but consistent practice makes them manageable.

What is the best strategy for LR miscellaneous in CAT?

Focus on clarity of concepts, practice diverse questions, and improve time management during solving.

CAT Miscellaneous LR Questions

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.

• 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.

• 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.

• No other paper was written during the conference.

Question 1

The person having the largest Erdos number at the end of the conference must have had Erdos number (at that time):

Video Solution

Solution

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. [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

Hence the person having the largest Erdos number at the end of the conference must have had Erdos number 7 . Hence option B.

Instructions

• 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.

• 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.

• No other paper was written during the conference.

Question 2

How many participants in the conference did not change their Erdos number during the conference?

Cannot be determined

Video Solution

Solution

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.

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.

Instructions

• 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.

• 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.

• No other paper was written during the conference.

Question 3

The Erdos number of C at the end of the conference was:

Video Solution

Solution

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.

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

Erdos no. of C at the end is f+1 = 1+1 = 2. Hence option B.

Instructions

• 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.

• 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.

• No other paper was written during the conference.

Question 4

The Erdos number of E at the beginning of the conference was:

Video Solution

Solution

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.

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

Hence erdos no. of E at the beginning of conference would be f+5 = 6 .

Instructions

• 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.

• 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.

• No other paper was written during the conference.

Question 5

How many participants had the same Erdos number at the beginning of the conference?

Cannot be determined

Video Solution

Question 1

In a hospital there were 200 diabetes, 150 hyperglycaemia and 150 gastro-enteritis patients. Of these, 80 patients were treated for both diabetices and hyperglycaemia. Sixty patients were treated for gastro-enteritis and hyperglycaemia, while 70 were treated for diabetes and gastroenteritis. Some of these patients have all the three diseases. Dr. Dennis treats patients with only gastro-enteritis. Dr. Paul is a generalist. Therefore, he can treat patients with multiple diseases. Patients always prefer a specialist for their disease. If Dr. Dennis had 80 patients, then the other three doctors can be arranged in terms of the number of patients treated as:

Paul > Gerard > Hormis

Paul > Hormis > Gerard

Gerard > Paul > Hormis

Cannot be determined

Instructions

A boy is asked to put one mango in a basket when ordered 'One', one orange when ordered 'Two', one apple when ordered 'Three', and is asked to take out from the basket one mango and an orange when ordered 'Four'.

A sequence of orders is given as: 1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4

Question 2

How many total oranges were in the basket at the end of the above sequence?

Instructions

A sequence of orders is given as: 1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4

Question 3

How many total fruits will be in the basket at the end of the above order sequence?

Instructions

Directions for the next 2 questions:

A, B, C are three numbers.
Let @(A, B) = average of A and B,
/(A, B) = product of A and B, and
X(A, B) = the result of dividing A by B

Question 1

The sum of A and B is given by:

/(@ (A, B),2)

X(@(A, B), 2)

@(/(A, B), 2

@ (X(A, B), 2)

Instructions

Directions for the next 2 questions:

A, B, C are three numbers.
Let @(A, B) = average of A and B,
/(A, B) = product of A and B, and
X(A, B) = the result of dividing A by B

Question 2

Average of A, B and C is given by:

@ (/(@(/(B, A), 2), C), 3)

X(@(/(@(B, A), 3), C), 2)

/((X(@ (B, A), 2), C), 3)

/(X(@ (/(@(B, A), 2), C), 3), 2)

Question 1

Three labeled boxes containing red and white cricket balls are all mislabeled. It is known that one of the boxes contains only white balls and one only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?

White

Red

Red and White

Not possible to determine from a sample of one ball

Instructions

Answer the questions based on the following information. The following operations are defined for real numbers.

a # b=a + b, if a and b both are positive else a # b=1

a $$\triangledown$$ b=$$(a \times b)^{a+b}$$ if $$a\times\ b$$ is positive else a $$\triangledown$$b=1.

Question 1

$$\frac{(2\#1)}{(1 \triangledown 2)}$$ =

$$\frac{1}{8}$$

$$1$$

$$\frac{3}{8}$$

$$3$$

Instructions

Answer the questions based on the following information. The following operations are defined for real numbers.

a # b=a + b, if a and b both are positive else a # b=1

a $$\triangledown$$ b=$$(a \times b)^{a+b}$$ if $$a\times\ b$$ is positive else a $$\triangledown$$b=1.

Question 2

$$\frac{((1\#1) \# 2) - (10^{1.3} \triangledown \log_{10}{0.1})}{1 \triangledown 2}$$

$$\frac{3}{8}$$

$$\frac{4.\log_{10}{0.1}}{8}$$

$$\frac{(4 + 10^{13})}{8}$$

None of these

Question 1

There were 'x' pigeons and 'y' mynahs in a cage. One fine morning, 'p' birds escaped to freedom. The bird-keeper, knowing only that p = 7, was able to figure out without looking into the cage that at least one pigeon had escaped. Which of the following does not represent a possible (x,y) pair?

(10,8)

(7,2)

(25,6)

(12,4)

CAT Miscellaneous LR Questions

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