Averages is one of the important topics of Arithmetic. The questions from this topic keep repeating every year in the CAT. The funda of Average are also present in other important Quant topics such as Progressions of Algebra. Practice diverse quality questions which will strengthen your fundamentals in Averages. Averages basically incude core concepts of Averages, Weighted Averages and Alligations. 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.
Practicing questions from CAT Previous Papers will help you understand the trends, change in pattern of the exam across years. Take 3 Free CAT Mock Tests which will help you know where you currently stand, and will help you in analysing your strengths and weaknesses.
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.
Averages are some of the essential topics in the quantitative
aptitude section, and it is vital to have a clear understanding of the
formulas related to them. To help the aspirants to ace this topic, we have
made a PDF containing a comprehensive list of formulas, tips, and
tricks that you can use to solve Profit, loss and interest questions with ease
and speed. Click on the below link to download the CAT Averages formulas PDF
1. Average of n terms Formula
The average of n terms equals $$\frac{x_1+x_2+. . .+x_n}{n}$$
2. Weighted Average Formula
The weighted average of n terms equals $$\frac{w_1*x_1+ w_2*x_2+. . .+ w_n*x_n}{ w_1+ w_2+ . . . w_n}$$
The average number of copies of a book sold per day by a shopkeeper is 60 in the initial seven days and 63 in the initial eight days, after the book launch. On the ninth day, she sells 11 copies less than the eighth day, and the average number of copies sold per day from second day to ninth day becomes 66. The number of copies sold on the first day of the book launch is
The average salary of 5 managers and 25 engineers in a company is 60000 rupees. If each of the managers received 20% salary increase while the salary of the engineers remained unchanged, the average salary of all 30 employees would have increased by 5%. The average salary, in rupees, of the engineers is
Let the average salary of managers be $$x$$ and let the average salary of engineers be $$y$$. The total salary of all the employees will be $$(5+25)*60000 = 1800000$$
We have, $$5x+25y = 1800000$$ .... (1)
If the average salary of all the employees increases by $$5\%$$, the total salary of all the employees will also increase by $$5\%$$, because the total number of employees remains the same. The new total salary will be, $$1800000\times 1.05 = 1890000$$
Also, the average salary of all the managers has increased by $$20\%$$ and has become $$1.2x$$, we have
Subtracting equation (1) from equation (2), we get,
$$x = 90000$$
Which gives $$25y = 1800000 - 90000*5$$ or $$y = \dfrac{1350000}{25} = 54000$$
Therefore, the correct answer is option C.
Question 1
The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is
We are given that average of three distinct integers is 28, that means the sum of these three integers is 28x3=84
Let us write $$x+y+z=84$$ x, y, z being the three distinct integers in ascending order.
If the smallest number is increased by 7 and the largest number is reduced by 10 $$(x+7)+(y)+(z-10)=81$$
New arithmetic mean will be $$\frac{81}{3}=27$$ And this is said to be 2 more than the middle number, meaning $$27-2=y=25$$
$$x+z=59$$
We are given that difference between the largest and the smallest numbers becomes 64, $$(z-10)-(x+7)=64$$ $$z-x=81$$
Adding the two equations we get, $$2z=140$$ $$z=70$$
Question 2
There are four numbers such that average of first two numbers is 1 more than the first number, average of first three numbers is 2 more than average of first two numbers, and average of first four numbers is 3 more than average of first three numbers. Then, the difference between the largest and the smallest numbers, is
Let us assume the four numbers to be a, b, c and d in ascending order.
Average of first two numbers is 1 more than the first number $$\frac{\left(a+b\right)}{2}=a+1$$ $$b-a=2$$ $$b=a+2$$
Average of first three numbers is 2 more than average of first two numbers $$\frac{\left(a+b+c\right)}{3}=\frac{\left(a+b\right)}{2}+2$$ $$2c=a+b+12$$ Substituting the value for b $$2c=a+a+2+12$$ $$2c=2a+14$$ $$c=a+7$$
Average of first four numbers is 3 more than average of first three numbers. $$\frac{\left(a+b+c+d\right)}{4}=\frac{\left(a+b+c\right)}{3}+3$$ $$3d=a+b+c+36$$ Substituting the value of b and c $$3d=a+a+2+a+7+36$$ $$3d=3a+45$$ $$d=a+15$$
d is the largest and a is the smallest and we know that d=a+15
Hence the difference between the smallest and the largest values is 15.
Question 1
There are three persons A, B and C in a room. If a person D joins the room, the average weight of the persons in the room reduces by x kg. Instead of D, if person E joins the room, the average weight of the persons in the room increases by 2x kg. If the weight of E is 12 kg more than that of D, then the value of x is
=> $$\frac{\left(e-d\right)}{4}=3x$$ => e - d = 12x
Given that 12x = 12 => x = 1.
Question 2
In a company, 20% of the employees work in the manufacturing department. If the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company, then the ratio of the average salary obtained by the manufacturing employees to the average salary obtained by the nonmanufacturing employees is
Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.
It is given that 20% of the employees work in the manufacturing department, and the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.
Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)
Average salary in the manufacturing department = (100y/6*20x) = 5y/6x
Similarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)
Hence, the average salary in the nonmanufacturing department = (500y/6*80x) = 25y/24x
Hence, the ratio is:- (5y/6x): (25y/24x)
=> 120: 150 = 4:5
The correct option is B
Question 1
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
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,
$$\ \frac{\ nx+m\left(x+3\right)}{n+m}=x+0.6$$
$$\ \ nx+mx+3m=mx+nx+0.6n+0.6m$$
$$2.4m=0.6n$$
$$4m=n$$
$$\frac{n}{m}=\frac{4}{1}$$
The answer is option C.
Question 2
Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onward, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his monthly earnings, in rupees, from the tenth month onward should be
Saving for remaining 3 months should be 6600-4500, i.e. Rs 2100
Savings for each month in last 3 months = $$\frac{2100}{3}$$ = Rs 700
It is given, monthly expenses in last 3 months = Rs 3700
This implies, his monthly earnings from 10th month should be 3700+700, i.e. Rs 4400
The answer is option A.
Question 3
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
The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is
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
The answer is option D.
Question 6
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
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
The answer is option D.
Question 1
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
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 2
The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is
Let sum of marks of students be x Now therefore x = 25*50 =1250 Now to maximize the marks of the toppers We will minimize the marks of 20 students so their scores will be (30,31,32.....49 ) let score of toppers be y so we get 5y +$$\frac{20}{2}\left(79\right)$$=1250 we get 5y +790=1250 5y=460 y=92 So scores of toppers = 92
Question 3
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
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 1
A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is
Given, $$\frac{\text{sum of scores in n matches+38+15}}{n+2}=29$$
Given, $$\frac{\text{sum of scores in n matches}}{n}=30$$
=> 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 of 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
Question 1
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?
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 2
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
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 1
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?
The possible average age of people whose ages are below 51 years will be maximum if the average age of the number of people aged 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 2
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
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 1
An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?
It is given that the maximum weight limit is 630. The lightest person's weight is 53 Kg and the heaviest person's weight is 57 Kg.
In order to have maximum people in the lift, all the remaining people should be of the lightest weight possible, which is 53 Kg.
Let there be n people.
53 + n(53) + 57 = 630
n is approximately equal to 9.8. Hence, 9 people are possible.
Therefore, a total of 9 + 2 = 11 people can use the elevator.
Question 2
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
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 3
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
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 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 ?
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
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 = 8x4 + 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 1
Amol was asked to calculate the arithmetic mean of 10 positive integers, each of which had 2 digits. By mistake, he interchanged the 2 digits, say a and b, in one of these 10 integers. As a result, his answer for the arithmetic mean was 1.8 more than what it should have been. Then |b - a| equals
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 1
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?
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 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.
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
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