# Averages Questions for CAT Set-2 PDF

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## Averages Questions for CAT Set-2 PDF

Download important CAT Averages  Questions Set-2 with Solutions PDF based on previously asked questions in CAT exam. Practice Averages  Questions Set-2 with Solutions for CAT exam.

Question 1: The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.

a) 55

b) 60

c) 62

d) Cannot be determined

Question 2: Prof. Suman takes a number of quizzes for a course. All the quizzes are out of 100. A student
can get an A grade in the course if the average of her scores is more than or equal to 90.Grade B is awarded to a student if the average of her scores is between 87 and 89 (both included). If the average is below 87, the student gets a C grade. Ramesh is preparing for the last quiz and he realizes that he will score a minimum of 97 to get an A grade. After the quiz, he realizes that he will score 70, and he will just manage a B. How many quizzes did Prof. Suman take?

a) 6

b) 7

c) 8

d) 9

e) None of these

Question 3: 2 years ago, one-fifth of Amita’s age was equal to one-fourth of the age of Sumita, and the average of their age was 27 years. If the age of Paramita is also considered, the average age of three of them declines to 24. What will be the average age of Sumita and Paramita 3 years from now?

a) 25 years

b) 26 years

c) 27 years

d) cannot be determined

Question 4: The average of 7 consecutive numbers is P. If the next three numbers are also added, the average shall

a) remain unchanged

b) increase by 1

c) increase by 1.5

d) increase by 2

Question 5: The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

a) 30

b) 28

c) 32

d) 26

Question 6: Consider the set S = {2, 3, 4, …., 2n+1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y ?

a) 0

b) 1

c) (1/2)*n

d) (n+1)/2n

e) 2008

Question 7: A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people to be solicited?

a) Rs. 300

b) Rs. 250

c) Rs. 400

d) 500

Question 8: Three classes X, Y and Z take an algebra test.
The average score in class X is 83.
The average score in class Y is 76.
The average score in class Z is 85.
The average score of all students in classes X and Y together is 79.
The average score of all students in classes Y and Z together is 81.
What is the average for all the three classes?

a) 81

b) 81.5

c) 82

d) 84.5

Question 9: Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is:
[CAT 2000]

a) n

b) n+1

c) kn, where k is a function of n

d) n+(2/7)

InstructionsThere are 60 students in a class. These students are divided into three groups A, B and C of 15, 20 and 25 students each. The groups A and C are combined to form group D.

Question 10: What is the average weight of the students in group D?

a) More than the average weight of A

b) More than the average weight of C

c) Less than the average weight of C

d) Cannot be determined

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Total marks = 80 x 10 = 800
Total marks except highest and lowest marks = 81 x 8 = 648
So Summation of highest marks and lowest marks will be = 800 – 648 = 152
When highest marks is 92, lowest marks will be = 152-92 = 60

Grade A $\geq$ 90 and Grade B = 87 to 89

If Ramesh scores 70 instead of 97, => Change of marks = 97 – 70 = 27

It creates a change from grade A to B, this means an overall change in average by

= Minimum marks for grade A – Minimum marks for Grade B = 90 – 87 = 3

$\therefore$ Number of subjects = $\frac{27}{3} = 9$

Let ‘A’, ‘S’ and ‘P’ be Amita’s, Sumita’s and Paramita’s present age.

It is given that 2 years ago, one-fifth of Amita’s age was equal to one-fourth of the age of Sumita, and the average of their age was 27 years.

$\dfrac{(A-2)+(S-2)}{2} = 27$

$A+S = 58$  … (1)

Also, $\dfrac{A-2}{5} = \dfrac{S-2}{4}$

$4A-8 = 5S-10$

$5S – 4A = 2$ … (2)

From equation (1) and (2) we can say that S = 26, A = 32.

Average age of Amita, Sumita and Paramita before 2 years = 24.

$\dfrac{(A-2)+(S-2)+(P-2)}{3} = 24$

$A+S+P = 78$. Hence, P = 20.

Therefore, the average age of Sumita and Paramita 3 years from now? = $\dfrac{(S+3)+(P+3)}{2}$ = $\dfrac{(26+3)+(20+3)}{2}$ = 26 years.

Hence, option B is the correct answer.

Let the 7 consecutive numbers be a-3, a-2, a-1, a, a+1, a+2 and a+3.
Sum of the numbers = 7a and the average of these numbers = a
If next 3 numbers a+4, 4+5 and a+6 are also added then the average of these 10  numbers = $\dfrac{7a+a+4+a+5+a+6}{10} = a+1.5$
Thus, the average increases by 1.5
Hence, option C is the correct answer.

Let the average height of 22 toddlers be 3x.
Sum of the height of 22 toddlers = 66x
Hence average height of the two toddlers who left the group = x
Sum of the height of the remaining 20 toddlers = 66x – 2x = 64x
Average height of the remaining 20 toddlers = 64x/20 = 3.2x
Difference = 0.2x = 2 inches => x = 10 inches
Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

The odd numbers in the set are 3, 5, 7, …2n+1

Sum of the odd numbers = 3+5+7+…+(2n+1) = $n^2 + 2n$

Average of odd numbers = $n^2 + 2n$/n = n+2

Sum of even numbers = 2 + 4 + 6 + … + 2n = 2(1+2+3+…+n) = 2*n*(n+1)/2 = n(n+1)

Average of even numbers = n(n+1)/n = n+1

So, difference between the averages of even and odd numbers = 1

Let there be total 100 people whom the college will ask for donation. Out of these 60 people have already given average donation of 600 Rs. Thus total amount generated by 60 people is 36000. This is 75% of total amount required . so the amount remaining is 12000 which should be generated from remaining 40 people. So average amount needed is 12000/40 = 300

Let x , y and z be no. of students in class X, Y ,Z respectively.

From 1st condition we have

83*x+76*y = 79*x+79*y which give 4x = 3y.

Next we have 76*y + 85*z = 81(y+z) which give 4z = 5y .

Now overall average of all the classes can be given as $\frac{83x+76y+85z}{x+y+z}$

Substitute the relations in above equation we get,

$\frac{83x+76y+85z}{x+y+z}$  = (83*3/4 + 76 + 85*5/4)/(3/4 + 1 + 5/4) = 978/12 = 81.5

The first five numbers could be n-2, n-1, n, n+1, n+2. The next two number would then be, n+3 and n+4, in which case, the average of all the 7 numbers would be $\frac{(5n+2n+7)}{7}$ = n+1