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# CAT Averages Questions PDF [Important Questions]

Averages is one of the most important topics in the Quantitative Ability section of CAT. It is an easy topic and so one must not avoid this topic. You can check out these Averages CAT Previous year questions. Practice a good number of questions on CAT Averages questions. In this article, we will look into some important Averages Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT Averages Questions PDF below, which is completely Free.

Question 1:Â 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)Â 0

b)Â 1

c)Â (1/2)*n

d)Â (n+1)/2n

e)Â 2008

Solution:

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 =Â $n^2 + 2n$/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

Question 2:Â Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight-member joint family is nearest to
[CAT 2007]

a)Â 23 years

b)Â 22 years

c)Â 21 years

d)Â 25 years

e)Â 24 years

Solution:

Ten years ago, the total age of the family is 231 years.

Seven years ago, (Just before the death of the first person), the total age of the family would have been 231+8*3 = 231+24 = 255.
This is because, in 3 years, every person in the family would have aged by 3 years,
Total change in age = 231+24 = 255

After the death of one member, the total age is 255-60 =Â 195 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8

Four years ago, (i.e. 6 years after start date)Â one of the member of age 60 dies,
therefore,Â total age of the family is 195+24-60 = 159 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8

After 4 more years, the current total age of the family is = 8×4 + 159 = 191 years

The average age is 191/8 = 23.875 years = 24 years (approx)

Alternatively,
Since the number of members is always the same throughout
The 2 older members dropped their age by 60
So, after 10yrs, total age = 231 + 8*10 – 2*60Â = 191
Average age = 191/8 = 23.875 $\simeq$ 24

Question 3:Â 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)Â Rs. 300

b)Â Rs. 250

c)Â Rs. 400

d)Â 500

Solution:

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

Question 4:Â 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)Â n

b)Â n+1

c)Â kn, where k is a function of n

d)Â n+(2/7)

Solution:

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

Question 5:Â 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)Â 55

b)Â 60

c)Â 62

d)Â Cannot be determined

Solution:

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

Question 6:Â Total expenses of a boarding house are partly fixed and partly varying linearly with the number ofÂ boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600Â when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

a)Â 550

b)Â 580

c)Â 540

d)Â 560

Solution:

Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

=> y = 500 and x = 5000

x + 100y = 5000 + 50000 = 55000

Average expense = $\frac{55000}{100}$ = Rs.550.

Question 7:Â A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is

a)Â 9.5

b)Â 10

c)Â 4.5

d)Â 6

Solution:

Let, the average score of boys in the mid semester exam is A.
Therefore, the average score of girls in the mid semester exam be A+5.
Hence, the total marks scored by the class is $20\times (A) +Â 30\times (A+5)Â = 50\times AÂ + 150$

The average scoreÂ of the entire class is $\dfrac{(50\times A + 150)}{50} = A + 3$

wkt, class average increased by 2, class average in final term $= (A+3) + 2 = A + 5$

Given, that score of girls dropped by 3, i.e $(A+5)-3 = A+2$
Total score of girls in final termÂ $= 30\times(A+2) = 30A + 60$

Total class score in final term $= (A + 5)\times50 = 50A +Â 250$

the total marks scored by the boys is $(50A + 250) – (30A – 60) = 20A + 190$
Hence, the average of the boys in the final exam is $\dfrac{(20G + 190)}{20} = A + 9.5$

Hence, the increase in the average marks of the boys is $(A+9.5) – A = 9.5$

Question 8:Â In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

a)Â 27

b)Â 25

c)Â 26

d)Â 28

Solution:

The possible average age of peopleÂ whose ages are below 51 years will be maximum if the average age of the number of people agedÂ 51 years and above is minimum. Hence, we can say that that there are 30 people having same age 51 years.

Let ‘x’ be the maximum average age of people whose ages are below 51.

Then we can say that,

$\dfrac{51*30+39*x}{30+39} = 38$

$\Rightarrow$ $1530+39x = 2622$

$\Rightarrow$ $x = 1092/39 = 28$

Hence, we can say that option D is the correct answer.

Question 9:Â The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

a)Â 3.5

b)Â 5

c)Â 4.5

d)Â 4

Solution:

It is given that the average of the 30 integers = 5

Sum of the 30 integers = 30*5=150

There are exactly 20 integers whose value is less than 5.

To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers

So the sum of 10 integers = 10*6=60

The sum of the 20 integers = 150-60= 90

Average of 20 integers =Â $\ \frac{\ 90}{20}$ = 4.5

Question 10:Â The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

a)Â 3.5

b)Â 5

c)Â 4.5

d)Â 4

Solution:

It is given that the average of the 30 integers = 5

Sum of the 30 integers = 30*5=150

There are exactly 20 integers whose value is less than 5.

To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers

So the sum of 10 integers = 10*6=60

The sum of the 20 integers = 150-60= 90

Average of 20 integers =Â $\ \frac{\ 90}{20}$ = 4.5

Question 11:Â Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is

a)Â 53

b)Â 51

c)Â 48

d)Â 49

Solution:

Assume the average of 21 students other than Ramesh = a

Sum of the scores of 21 students other than Ramesh = 21a

Hence the average of 22 students = a+1

Sum of the scores of all 22 students = 22(a+1)

The score of Ramesh = Sum of scores of all 22 students – Sum of the scores ofÂ 21 students other than Ramesh = 22(a+1)-21a=a+22Â  = 82.5Â  Â (Given)

=> a = 60.5

Hence,Â sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353

Now the sum of the scores of students other than Gautam = 21*62 = 1302

Hence the score of Gautam = 1353-1302=51

Question 12:Â A batsman played n + 2 innings and got out on all occasions. HisÂ average score in these n + 2 innings was 29 runs and he scored 38 andÂ 15 runs in the last two innings. The batsman scored less than 38 runsÂ in each of the first n innings. In these n innings, his average score wasÂ 30 runs and lowest score was x runs. The smallest possible value of xÂ is

a)Â 4

b)Â 3

c)Â 2

d)Â 1

Solution:

Given, $\frac{\text{sum of scoresÂ in n matches+38+15}}{n+2}=29$

Given,Â $\frac{\text{sum of scoresÂ in n matches}}{n}=30$

=> 30n + 53 = 29(n+2) => n=5

Sum of the scores in 5 matches = 29*7 – 38-15 = 150

Since the batsmen scored less than 38, in each ofÂ the first 5 innings. The value of x will be minimum when remaining four values are highest

=> 37+37+37+37 + x = 150

=> x = 2