Instructions

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.

A

B

C

D

E

Text Explanation:

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.

Question 2.

A

B

C

D

E

Text Explanation:

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

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

Question 3.

A

B

C

D

E

Text Explanation:

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

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

Question 4.

A

B

C

D

E

Text Explanation:

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

Question 5.

A

B

C

D

E

Instructions

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 6.

A

B

C

D

Question 7.

A

B

C

D

Question 8.

A

B

C

D

Instructions

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 9.

A

B

C

D

Question 10.

A

B

C

D

Text Explanation:

Considering option,
In 1st we get product of a and b which is not required In 2nd we get fraction 1.5 which doesnt help in finding average In 3rd we get (a+b)/4 after 2 operations which doesn't give average in further operations. only in 4th option we get overall and exact average of A , B and C .
/(X(@ (/(@(B, A), 2), C), 3), 2) = /(X(@ (/(A+B)/2), 2), C), 3), 2)
= /(X(@ (A+B), C), 3), 2) = /(X((A+B, C)/2, 3), 2)
= /[(A+B+C)/6,2]
= (A+B+C)/3

Question 11.

A

B

C

D

Text Explanation:

All of them can be mislabeled in 2 ways:
Red box - white label
White box - Red and white label
Red and white box - Red label

Red box - red and white label
White box - Red label
Red and white box - White label
So, we would try the box with the red and white label and if it has a white ball, labelling to the boxes is done as per case 1. If it has a red ball labelling is done as per case 2.
Note: It's not a good idea to try the white label as if we get a red ball, we can't make out if we are picking from red box or red and white box. Similarly if we try from the box with red label and we get a white ball, again we can't make out if it is coming from the white box or the red and white box.

Instructions

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 12.

A

B

C

D

Question 13.

A

B

C

D

Instructions

Instructions
For the following questions answer them individually

Question 14.

A

B

C

D

Login

CAT Miscellaneous LR Questions

Instructions

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Instructions

Question 6.

Question 7.

Question 8.

Instructions

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 9.

Question 10.

Question 11.

Instructions

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 12.

Question 13.

Instructions

Instructions
For the following questions answer them individually

Question 14.

Free CAT DILR Questions

Login

CAT Miscellaneous LR Questions

Instructions

Question 1.

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

Question 2.

<div>How many participants in the conference did not change their Erdos number during the conference?</div>

Question 3.

<div>The Erdos number of C at the end of the conference was:</div>

Question 4.

<div>The Erdos number of E at the beginning of the conference was:</div>

Question 5.

<div>How many participants had the same Erdos number at the beginning of the conference?</div>

Instructions

Question 6.

<div>How many total oranges were in the basket at the end of the above sequence?</div>

Question 7.

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

Question 8.

Instructions

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, andX(A, B) = the result of dividing A by B

Question 9.

<div>The sum of A and B is given by:</div>

Question 10.

<div>Average of A, B and C is given by:</div>

Question 11.

Instructions

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

Question 12.

<div>$\frac{(2\#1)}{(1 \triangledown 2)}$ =</div>

Question 13.

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

Instructions

Instructions For the following questions answer them individually

Question 14.

Free CAT DILR Questions

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

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.

Instructions
For the following questions answer them individually