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# Must-Practice Averages CAT Questions With Video Solutions PDF

To ace the quantitative aptitude section in the CAT exam, one must focus on key concepts such as Arithmetic, Geometry, Mensuration, Algebra, etc. Among these topics, Arithmetic has the maximum weightage in the quant section of CAT. The questions from Averages, Percentages, Ratios and Proportions, Mixtures and Allegations, Time Speed & Distance, and Time & Work come under the Arithmetic concept. Almost 30% of the question might be asked based on averages from the Arithmetic concepts. The average questions in CAT are a little tricky and not very difficult to solve. So ensure that you practice more questions from the concept of the averages to ace these questions in the quant section of CAT.

In this article, we will provide you with the 20 most important CAT questions on the Averages concept will help you prepare and get a better understanding of the concept. After practising these questions, one can gain more confidence to attempt the averages questions in the actual CAT examination.

Question 1: 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

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 = $n^2 + 2n$/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: 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) 23 years

b) 22 years

c) 21 years

d) 25 years

e) 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 = 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) one of the member of age 60 dies,
therefore, 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*60 = 191
Average age = 191/8 = 23.875 $\simeq$ 24

Question 3: 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

Solution:

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

Question 4: 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)

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: 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

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

a) 550

b) 580

c) 540

d) 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: 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) 9.5

b) 10

c) 4.5

d) 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 $20\times (A) + 30\times (A+5) = 50\times A + 150$

The average score 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 term $= 30\times(A+2) = 30A + 60$

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

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: 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

Solution:

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

Question 9: A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is

Solution:

Let the total number of tests be ‘n’ and the average by ‘A’
Total score = n*A
When 1st 10 tests are excluded, decrease in total value of scores = (nA – 20 * 10) = (nA – 200)
Also, (n – 10)(A + 1) = (nA – 200)
On solving, we get 10A – n = 190……….(i)
When last 10 tests are excluded, decrease in total value of scores = (nA – 30 * 10) = (nA – 300)
Also, (n – 10)(A – 1) = (nA – 300)
On solving, we get 10A + n = 310……….(ii)
From (i) and (ii), we get n = 60
Hence, 60 is the correct answer.

Question 10: 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) 3.5

b) 5

c) 4.5

d) 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 = $\ \frac{\ 90}{20}$ = 4.5

Question 11: 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) 53

b) 51

c) 48

d) 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 of 21 students other than Ramesh = 22(a+1)-21a=a+22  = 82.5   (Given)

=> a = 60.5

Hence, 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: Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to

a) 26

b) 18

c) 16

d) 20

Solution:

Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.

Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.

Total amount bought = 22kg.

Total amount spent = 100+100+100+50+50 = 400.

Average expense = $\frac{400}{22}=Rs.18.18\approx\ 18$

Question 13: In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is

Solution:

Let Total matches played be n and in initial n-10 matches his goals be x
so we get $\frac{\left(x+1\right)}{n}=0.15$
we get x+1 =0.15n                (1)
From condition (2) we get :
$\frac{\left(x+2\right)}{n}=0.2$
we get x+2 = 0.2n                  (2)
Subtracting (1) and (2)
we get 1 =0.05n
n =20
So initially he played n-10 =10 matches

Question 14: The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of the number of original students to the number of new students is

a) 1 : 4

b) 1 : 2

c) 4 : 1

d) 3 : 1

Solution:

Let the original number of students be ‘n’ whose average weight is ‘x’

Let the number of students added be ‘m’ and the average weight will be x + 3

We need to find the value of n : m

It is given, average weight of students in a class increased by 0.6 after new students are added.

Therefore,

$\ \frac{\ nx+m\left(x+3\right)}{n+m}=x+0.6$

$\ \ nx+mx+3m=mx+nx+0.6n+0.6m$

$2.4m=0.6n$

$4m=n$

$\frac{n}{m}=\frac{4}{1}$

Question 15: If a and b are non-negative real numbers such that a+ 2b = 6, then the average of the maximum and minimum possible values of (a+ b) is

a) 3

b) 4

c) 3.5

d) 4.5

Solution:

a + 2b = 6

From the above equation, we can say that maximum value b can take is 3 and minimum value b can take is 0.

a + b + b = 6

a + b = 6 – b

a + b is maximum when b is minimum, i.e. b = 0

Maximum value of a + b = 6 – 0 = 6

a + b is minimum when b is maximum, i.e. b = 3

Minimum value of a + b = 6 – 3 = 3

Average = $\ \frac{\ 6+3}{2}$ = 4.5

Question 16: Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is

a) 22

b) 21

c) 24

d) 20

Solution:

The average marks for all the students is 38.

Sum = 5*38 = 190

To find the minimum marks scored by Amit, we need to maximise the score of remaining students.

Maximum scores sum of remaining students = 50 + 49 + 48 + 32 = 179

Minimum possible score of Amit = 190 – 179 = 11

It is given, Amit scored least. This implies maximum possible score of Amit is 31.

Difference = 31 – 11 = 20

Question 17: In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is

Solution:

Let the number of students in section A and B be a and b, respectively.

It is given, a = b – 10

$\ \ \frac{\ 32a+60b}{a+b}$ is an integer

$\ \ \frac{\ 32a+60\left(a+10\right)}{a+a+10}=k$

$\ \ \frac{\ 46a+300}{a+5}=k$

$k=\ \frac{\ 46\left(a+5\right)}{a+5}+\frac{70}{a+5}$

$k=\ \ 46+\frac{70}{a+5}$

a can take values 2, 5, 9, 30, 65

Difference = 65 – 2 = 63

Question 18: A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
[CAT 2004]

a) 2 : 3

b) 1 : 2

c) 1 : 3

d) 3 : 4

Solution:

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

Question 19: Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?

a) 2 : 3

b) 4 : 3

c) 3 : 2

d) 3 : 4

Solution:

Fraction of A in contained 1 = $\frac{5}{6}$

Fraction of A in contained 2 = $\frac{1}{4}$

Let the ratio of liquid required from containers 1 and 2 be x:1-x

x($\frac{5}{6}$) + (1-x)($\frac{1}{4}$) = $\frac{1}{2}$

$\frac{7x}{12}$ = $\frac{1}{4}$

=> x = $\frac{3}{7}$

=> Ratio = 3:4

Question 20: There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then,

a) A > B

b) A < B

c) A = B

d) Cannot be determined

Solution:

Let the volume of the cup be V.
Hence, after removing three cups of alcohol from the first container,

Volume of alcohol in the first container is 500-3V
Volume of water in the second container is 500 and volume of alcohol in the second container is 3V.
So, in each cup, the amount of water contained is $\frac{500}{500+3V}*V$

Hence, after adding back 3 cups of the mixture, amount of water in the first container is $0+\frac{1500V}{500+3V}$
Amount of alcohol contained in the second container is $3V – \frac{9V^2}{500+3V} = \frac{1500V}{500+3V}$

So, the required proportion of water in the first container and alcohol in the second container are equal.

## Tips and tricks to solve average questions in CAT

As mentioned earlier, the average questions in CAT are a little tricky and often, its applications are used to solve other quants questions. One can solve these questions quickly with some simple alternative methods. Here are some tips and tricks to calculate the average with a detailed example. Firstly, let us look at the definition of average.

Definition of Average: Average is simply defined as the ratio of the sum of observations divided by the number of observations. The formula to calculate the average is given below.

Average  = Sum of the observations/Number of the Observations

Let us take one simple example from the concept of averages and solve it in the standard method and the deviation method.

Example Question: The average weight of a class of 50 students is 60 kg. Ten students join the class, and the average weight of the class becomes 62 kg. What is the average weight of the newly joined students?

The given question can be solved in two approaches. Let’s see them in detail.

• The solution in Standard Method:
Total weight of the class = 60*50 = 3000
New total weight = 62*60 = 3720
Total weight of the newly joined students = 3720-3000 = 720
Average weight = 720/10 = 72 kg
• The solution in Deviation Method:
Initial average = 60 kg
New average = 62 kg
The average increases by 2 kg and the number of people increase to 60 students. Therefore, total change for 10 students added = 60(2) = 120 kg
Average change per person = 120/10 = 12 kg
Average weight of newly added students = 60 + 12 = 72 kg

## CAT Averages Questions – FAQs

##### How do you calculate the Average?
• Average is defined as the ratio of the sum of observations divided by the number of observations. The formula to calculate the average is given below.

Average  = Sum of the observations/Number of the Observations

##### Do Average Questions repeat in the CAT exam?
• The exact questions will not repeat in the CAT, but similar questions will be asked. The questions based on the Averges appear every year in the CAT exam.
##### Is arithmetic enough for CAT?
• Arithmetic is the most important concept in the CAT Quant section. However, Arithmetic is one part of the Quantitative aptitude section. One must also practice the other key concepts such as Geometry & Mensuration, Algebra, etc.