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Here you can download CMAT 2022 – important Averages Questions PDF by Cracku. Very Important Averages Questions for CMAT 2022 based on asked questions in previous exam papers. These questions will help your CMAT preparation. So kindly download the PDF for reference and do more practice.

Question 1:Â Students in a commerce class took a test. The average score of men is 70 and of women is 83. The class average is 76. What is the ratio of men to women in the class?

a)Â 8:7

b)Â 7:8

c)Â 7:6

d)Â 6:7

Question 2:Â Expenditures of a Company (in Lakh Rupees) per Annum Over the given Years was as under.

What is the average salary expenditure(in Lakh Rupees) per Annum during this period?

a)Â 663.6

b)Â 666.3

c)Â 636.6

d)Â 663.3

Question 3:Â A farmer got an average harvest of 25 tons of potato per acre of land. His neighbour, who had 10 acres less land than him for the cultivation of potato, got an average of 30 tons of potato per acre. If the neighbour, in total, harvested 75 tons of potato more than the farmer, what was the average harvest per acre for both the farms combined?

a)Â 25.54 tons/acre

b)Â 26.6 tons/acre

c)Â 28.95 tons/acre

d)Â 27.32 tons/acre

Question 4:Â A weigh bridge (a device which measures the weight of cargo carried by trucks) weighing the outgoing trucks of a factory developed a fault. Due to this, the weight of the last 100 trucks, in one particular lot, was measured 20% less than the actual weight. When the corrections were made, it increased the average weight of the cargo carried by the trucks of the lot by 5%. Find the total number of trucks in the lot.

a)Â 200

b)Â 500

c)Â 420

d)Â 400

Question 5:Â A travel company rents out buses and charges on the basis of distance as well as the number of passengers on a trip. It was found out that the profit was directly proportional to the distance travelled and the square of number of passengers. The ratio of the average profit (profit earned per passenger) when a bus was rented by 30 passengers for 50 kms to a bus for 150 kms carrying $\frac{1}{2}$ the number of passengers was 2:3. If the lesser of the two overall profits earned â‚¹45000 for the company, find the difference in profit when 15 people go for a trip 200 kms away and when 20 people go for a trip 300 kms away.

a)Â â‚¹80,000

b)Â â‚¹90,000

c)Â â‚¹1,00,000

d)Â â‚¹85,000

Question 6:Â Let $A$ be the average marks of the students in a class. The class has 50 students where the top 5 students scored an average of 92 marks and the students ranked 41 to 45 scored an average of 47 marks. The average of students ranked 46 to 50 is less than that of students ranked 41 to 45 and is an integer. The average of students ranked 6 to 10 is less than that of students ranked 1 to 5 and is an integer. If $60 \leq A \leq 71.1$, find the difference between the maximum and minimum average marks of students ranked 11 to 40. The passing marks are 32 and none of the students fail.

a)Â 18 marks

b)Â 25 marks

c)Â 20 marks

d)Â 23 marks

Question 7:Â In a class, there are two sections namely A and B. The average weight of students studying in section A is 85 kg whereas for section B it is 95 kg. If four students are chosen randomly from section B and transferred to section A so that the average weight of the two sections gets interchanged. It is also known that the average weight of the students who were transferred from section B is 115 kg. Find out the total number of students in the class.

Question 8:Â A fruit seller buys fruits from wholesale market to be sold at the market near his residence. The wholesale market has mangoes of 2 different varieties : alphanso and badami. Alphonso costs â‚¹120/dozen and badami costs 84/kg. In what ratio should the fruit seller mix the 2 varieties of mango to be able to earn 11.11% profit when selling the mixture at â‚¹100/kg. Assume that on an average, there are 20 badami mangoes in 2 kgs and 5 alphonso in half a kg.

a)Â 1:4

b)Â 3:5

c)Â 7:13

d)Â 14:11

Question 9:Â The average of 100 unique natural numbers is 100. If it is known that the number of numbers below the average is twice the number of numbers above it, then the largest number among the 100 numbers cannot exceed

Question 10:Â The average marks obtained by section A of XYZ Public School in class X examinations is 85%. The number of students in sections A,B,C of class X are in the ratio 2:5:3 respectively. If the ratio of average marks of students of section B and section C is 2:1 and the overall average marks of the students of class X is 82%, find the ratio of total marks obtained by students of section A and section B.

a)Â 17:20

b)Â 11:23

c)Â 12:29

d)Â 17:50

Question 11:Â A shopkeeper mixes milk and water in the ratio 3:5. He marks up the price of the mixture and sells one litre of the mixture at a price which is equal to the average price of one litre of milk and water.Â  If it is known that water is not available free of cost and milk does not cost more than five times the cost of water, then the maximum profit earned by the shopkeeper cannot exceed

a)Â 25%

b)Â 20%

c)Â 16%

d)Â 33.33%

Question 12:Â The average earnings of 3 people go up by 15% after promotion but their individual earnings go up by 10%,4% and 15%. If the earnings of each person is a positive integer, what is the ratio of their original earnings?

a)Â 1

b)Â 2

c)Â 3

d)Â Not possible.

Question 13:Â Ankit, Beenal, Chandini, Deshna and Esha have certain amounts of money with them. The averages of the money they have taken three at a time is 121, 126, 133, 134, 139, 141, 146, 147, 152, 159
The averages of the money they spend taken three at a time is 43, 46, 50, 53, 58, 62, 65, 65, 68, 72.
It is known that the one who has the highest amount spends the least, the one who has the second highest spends the second least etc. What is the amount of money left with the one who was the richest? (type -1 if your answer is cannot be determined)

Question 14:Â A, B, C and D were the members of a team. The average runs of the team decreases by 2 if another member E is added. It is known that E scored 45 runs. No player scored less than E or more than 65 runs. If the runs scored by A and B are in the ratio 13:12 and C scored more than A, the ratio of the runs scored by B to the average runs scored by C & D is
Assume that the runs scored by all the members is a natural number.

a)Â 6:7

b)Â 4:5

c)Â 16:19

d)Â 8:9

Question 15:Â A car manufacturing company found that when their cars are moving at 60kmph without acceleration, their defective cars lose 1kmph at the end of every hour while their non defective cars maintain constant speed. When all the cars were driven for 11 hours at 60kmph without acceleration, it was found that the mean of all their average speeds was 56kmph. Find the ratio of the number of defective cars to the number of normal cars.

a)Â 2:1

b)Â 1:4

c)Â 4:1

d)Â 5:3

Question 16:Â A set R contains real numbers such that R $\epsilon$ {x: x>10} the average of R is 4 more than the average of R U {1,4,6,9}. If the average of R U {6,7,8,9,10} is equal to the total number of elements in the new set, the original average is

a)Â 40

b)Â 15

c)Â 24

d)Â 65

Question 17:Â The average age of a father, a mother and a son is 20 years. 6 years ago when the son was born, the ratio of the age of father to the age of mother was 4:3. What is the present age of father?

a)Â 40 years

b)Â 28 years

c)Â 30 years

d)Â 36 years

Question 18:Â In a class of 11 students, the average age of all the students is m years. If one student left the class, now the difference between the maximum possible average to the minimum possible average age of the class is 4. What is the difference in the age of the eldest member and the youngest member of the initial class?
Enter -1 if the answer can’t be determined

Question 19:Â The average marks of 5 students of a class is 57. The average marks of the top 3 students is 75. If the marks of the bottom 3 students are increased by 10, 4 and 9 in any order, the average of those 3 students increases by 25%. What was the third highest marks originally obtained?

a)Â 37

b)Â 28

c)Â 32

d)Â 45

Question 20:Â A batsman was having 32 runs per innings as his average after 15th innings. His average increased by 2 runs after 16th inning. Then what was his score in the 16th inning?

a)Â 64

b)Â 60

c)Â 46

d)Â 62

Let the number of men be ‘m’ and number of women be ‘w’ in the class.

Given Average score of men = 70

$\Rightarrow$ Total score of men$\div$number of men = 70

$\Rightarrow$ Total score of men = 70$\times$m

Also, Average score of women = 83

$\Rightarrow$ Total score of women$\div$number of women = 83

$\Rightarrow$ Total score of women = 83$\times$w

Given the class average = 76

$\Rightarrow$ (Total score of men + Total score of women)$\div$(number of men + number of women) = 76

$\Rightarrow \frac{70\times m + 83\times w}{m + w} = 76$

Let the ratio of number of men to women $\frac{m}{w} = k$

$\Rightarrow$ 70k + 83 = 76k + 76

$\Rightarrow$ 6k = 7

$\Rightarrow$ k = 7 : 6.

Hence the ratio of men to women in the class is 7 : 6.

Average Salary Expenditure (in Lakh Rupees) per annum = Total salary expenditure in all these years$\div$Total number of years

= $\frac{576+682+648+672+740}{5}$

= 663.6

Let the area under cultivation in the first farm be $x$ hectares.

Therefore, the total potato harvested by first farmer = $25x$ tons

Total potato harvested by the neighbour = $30(x-10) = 30x-300$

Given, $30x-300 – 25x = 75$

$\Rightarrow 5x = 375$

$\Rightarrow x = \dfrac{375}{5} = 75$

Thus, total harvest by the first farmer = 25 x 75 = 1875 tons

Total harvest by the second farmer = 30(75-10) = 30 x 65 = 1950 tons

Thus, the average harvest for both the fars = $\dfrac{1875+1950}{75+65} =$ 27.32 tons/acre

Let us assume that there are ‘n’ trucks in the lot and each truck weigh ‘x’ kg. We are given that 100 trucks were weigh 20% less than their true weight and remaining (n-100) are weigh correctly.
Average weight of the entire lot in this case = $\dfrac{(n-100)*x+100*0.8x}{n}$

It is given that when corrections were made the average weight rose by 5%. We know that actual average weight is “x” kg.

$\Rightarrow$ $\dfrac{(n-100)*x+100*0.8x}{n}$*1.05 = x

$\Rightarrow$Â n = 420.

Therefore, we can say that there are 420 trucks in the lot.

Let the profit earned from a trip by the company be â‚¹$x$, the number of passengers be $p$ and distance travelled be $d$.

Given, $x \propto p^2$ and $x \propto d$

Combining the two proportionalities, $x \propto p^2d \Rightarrow x = kp^2d$ where k is the constant of proportionality.

We are given the ratio of profits earned per passenger. We need to find the trip which earned less overall ie we need to find the ratio of the total profit earned from the trips.

Let the average profit be $2a$ and $3a$ for the first and second trip respectively.

Thus, the ratio of overall profits of the two trips = $\dfrac{2a \times 30}{3a \times 15} = \dfrac{4}{3}$

Thus the second trip generated less overall profit. Therefore, the total profit in the second trip is â‚¹45000

Putting this in the expression for profit : $45000 = k \times (15)^2 \times 150$

$\Rightarrow k = \dfrac{45000}{225 \times 150} = \dfrac{4}{3}$

Now, profit for the first trip $p_{1} = k \times 15^2 \times 200 = 45000k$

Profit for the third trip $p_{2} = k \times 20^2 \times 300 = 120000k$

Thus, difference in profit = $p_{2} – p_{1}$

= $(120000-45000)k = 75000 \times \dfrac{4}{3} = â‚¹100000$

Let us address the students based on the rank they have in their class.

So, sum of marks of students ranked 1 to 5, $a$ : 92 x 5 = 460 marks

Let sum of marks of students ranked 6 to 10 be $b$

Sum of marks of students ranked 41 to 45, $c$ : 47 x 5 = 235 marks

Let sum of marks of students ranked 46 to 50 be $d$

Let the average marks of students ranked 11 to 40 be $x$.

For minimum value of $x_{min}$:

For the minimum value of $x$, we will maximize all the other values. Also the overall average score of the class will be at its minimum at 60.

Total marks of the class = 50 x 60 = 3000 marks

We know that a=460 and c=235.

For students ranked 6 to 10, since we are maximizing the marks, b = 91 x 5 = 455 marks

For the students ranked 46 to 50, the maximum average marks possible = 46

Thus, total marks of students ranked 46 to 50, $d$= 5 x 46 = 230 marks

Thus, $(30 \times x)+a+b+c+d = 3000 \Rightarrow 30x=3000-a-b-c-d$

$\Rightarrow 30x = 3000 – 460-235-230-455 = 1620 \Rightarrow x_{min} = \dfrac{1620}{30}$ = 54 marks

For maximum value of $x_{max}$:

For the maximum value of $x$, we will minimize all the other values. Also the overall average score of the class will be at its maximum at 70.

Total marks of the class = 50 x 71.1 = 3555 marks

We know that a=460 and c=235.

For the students ranked 6 to 10, the average marks will be same as the average marks of students 11 to 40.

Thus, $b = 5x$

Thus, total marks of students ranked 46 to 50, $c$= 5 x 32 = 160 marks

Thus, $(30 \times x)+a+b+c+d= 3555 \Rightarrow 30x+5x=3555-460-235-160$

$\Rightarrow 35x = 2695 \Rightarrow x_{max} = \dfrac{2700}{35}$ = 77.14 marks

This is not possible as we know that the value of $b$ has to be an integer. Therefore we take the next smallest integer as the average value of marks of students 6 to 10.

Thus, $b$ = 78 x 5 = 390 marks

Thus, $(30 \times x)+a+b+c+d= 3555 \Rightarrow 30x=3555-460-390-235-160$

$\Rightarrow 30x = 2310 \Rightarrow x_{max} = \dfrac{2310}{30}$ = 77 marks

Thus, the difference between the two values = $x_{max} – x_{min}$ = 77-54 = 23 marks

Let ‘a’ and ‘b’ be the number of students in section A and B respectively.
Before transfer, the total weight of all the students in section A = 85a
It is given that after transfer the average weight of both the sections gets interchanged.
$\Rightarrow$ 85a+4*115 = 95*(a+4)
$\Rightarrow$ a = 8
Similarly, for section B we can say that,
$\Rightarrow$ 95b-4*115 = 85*(b-4)
$\Rightarrow$ b=12.
Hence, the total number of students in the class = 8+12 = 20.

Let the cost price of the mixture of the mangoes be $â‚¹x$

Now, $x \times (1+\dfrac{1}{9}) = 100$

$\Rightarrow x \times \dfrac{9}{10}$ = â‚¹90/kg

Now, cost of alphonso = $\dfrac{120}{12}$ = â‚¹10/piece

Weight of one alphonso mango = $\dfrac{500}{5}$ = 100 gms

Cost of alphonso per kg = $\dfrac{1000}{100} \times 10Â$ = â‚¹100/kg

Let the ratio of mixing the two types of mangoes be $k$

Let the weight of badami mangoes in the mixture be 1 kg

Thus, weight of alphonso = $k \times$ weight of badami = $k$ kgs

Total cost of the mixture = $90 \times (k+1)$

Cost of badami mangoes = â‚¹84

Cost of alphonso mangoes = $k$ x 100 = â‚¹100k

Since we know the total amount, we can equate it.

So, $100k + 84 = 90k+90$

$\Rightarrow 10k = 6 \Rightarrow k = \dfrac{3}{5}$

Thus the ratio of mixing the mangoes is 3:5

Alternately,

We know that the CP of alphonso is 100/kg and CP of Badami is 84 / kg.

The seller makes a profit of 11.11% if he sells it at 100/kg.

So without profit , he must sell it at, 100Â * 100/111.11 = 90/kg.

Applying alligation,

100Â  Â  Â  Â  Â  Â  Â  Â  84

90

6Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  10

Therefore, the required ratio is 3:5.

The average of 100 unique natural numbers is 100. The number of numbers below the average is twice the numbers above it.

Let x be the number of numbers below the average.
=> Number of numbers above the average = 2x
Total number of numbers = 3x.
100 is not of the form 3x.
Therefore, 100 should be one of the 100 numbers and the remaining 99 numbers should be distributed above and below it.

66 numbers are below 100 and 33 numbers are above 100.
Now, for the highest number to be maximized, the sum of the remaining numbers should be minimized.

Let us assume that the numbers below 100 are 1, 2, 3, 4, …,66 (Since the sum has to be minimized).

Sum of the numbers below 100 = 1+2+3+4+….+66 = (66*67)/2 = 2211.

Let us calculate the sum of the values above 100 (excluding the largest value).
Again, for the highest value to be maximized, every other value should be minimized.
Let us assume that the numbers above 100 are 100, 101, 102,….,132.
Sum of the numbers greater than 100 = 101+102+103+…..+132
= 1+2+3+…+132 – (1+2+….100)
= (132*133)/2 – (100*101)/2
= 8778 – 5050
= 3728

We know that the average of the 100 numbers is 100.
=> Sum of the 100 numbers = 100*100 = 10000.

Let the largest number be X.
2211 +Â 100 + 3728 +Â X = 10000
6039 + X = 10000
=> X = 3961

Therefore, 3961 is the right answer.

Let the number of students in sections A,B,C be $2x$,$5x$ and $3x$ respectively.

Let the maximum marks be 100.

Total marks obtained by students of section A = $85 \times 2x = 170x$ marks

Total marks obtained by students of class X = $82 \times (2x+5x+3x) = 820x$

Total marks obtained by students of section B and section C = $820x – 170x = 650x$

Let the average marks of students of section B and section C be $2y$ and $y$ respectively.

Now, total marks of students of section B = $2y \times 5x = 10xy$

Total marks of students of section C = $y \times 3x = 3xy$

Total marks of students of section B and section C = $13xy$

$\Rightarrow 650x = 13xy$

$\Rightarrow y = \dfrac{650x}{13x} = 50$

Total marks obtained by students of section B = $10x \times 50 = 500x$

Required ratio = $\dfrac{170x}{500x} = 17:50$

Let the price of milk be $x$ and the price of water be $y$.
Now, the shopkeeper mixes milk and water in the ratio $3:5$.

The cost of 1 litre of the mixture for the milkman = $\dfrac{3x+5y}{5+3}$ = $\dfrac{3x+5y}{8}$
Average price of 1 litre of milk and water = $\dfrac{x+y}{2}$

We have to find the maximum possible profit that the shopkeeper could have made provided milk does not cost more than 5 times the cost of water. Therefore, x=5y will be the limiting condition.

Substituting x = 5y and calculating the profit percentage, we get,

Profit percentage =$\dfrac{\frac{6y}{2}-\frac{20y}{8}}{\frac{20y}{8}}$
=> Profit percentage = $\dfrac{\frac{4y}{8}}{\frac{20y}{8}}$
=> Profit percentage = $\frac{4}{20}$ = $20$%.

The maximum possible profit percentage cannot exceed $20 %$ and hence, option B is the right answer.

Here, the average earning goes up by 15%, and the individual earnings go up by 10%, 4%, and 15%.

If we consider only the third friend, the earning goes up by 15%. But since the increase in the earnings of the other two friends is less than 15%, and their earnings are positive, the average increase will always be lower than 15%. Hence, the case is not possible.

Let us assume that the money Ankit, Beenal, Chandini, Deshna and Esha have is in ascending order and take it as a, b, c, d, e respectively.
Then the amount each person has appears in the averages 6 times (One person is fixed and the other two spots have to be filled from the remaining 4 people which is 4C2)
So, $\frac{a+b+c+d+e}{5}=\frac{121+126+133+134+139+141+146+147+152+159}{10}$
Or $\frac{a+b+c+d+e}{5}=\frac{1398}{10}$
So, a+b+c+d+e=699â€¦(equation 1)
Also, a+b+c=363 (121*3)â€¦(equation 2)
And a+b+d=378 (126*3)â€¦(equation 3)
And c+d+e=477 (159*3)â€¦(equation 4)
And b+d+e=456 (152*3)â€¦(equation 5)
Using equation 1 and 2, we get d+e=336. Putting in equation 4, we get c=141.
Using equation 1 and 3, we get c+e=321. We know c=141 so e=180.
E is the richest. We do not need to calculate the amount with the others although they can be calculated as we have 5 equations with 5 variables.
Similarly, we must calculate their spending. Let us take the amount they spend as a, b, c, d, e respectively. They must be in descending order since it is given in the question that the one who has the most spends the least, the one who has the second most spends the second least etc.
$\frac{a+b+c+d+e}{5}=\frac{43+46+50+53+58+62+65+65+68+72}{10}$
So, a+b+c+d+e=291â€¦(equation 1)
Also, e+d+c=129 (43*3) â€¦(equation 2)
And e+d+b=138 (46*3) â€¦(equation 3)
And a+b+c=216 (72*3) â€¦(equation 4)
And a+b+d=204 (68*3) â€¦(equation 5)
From these we can calculate e=33.
So, the amount left with the richest person is 180-33=147 rupees.

Consider average of 4 players = x and scores of A,B,C and D be a,b,c & d.

Total score = 4*x

(4*x+45)/5 = x-2

x= 55

45=<a,b,c,d<=65

a:b = 13:12 Possible values (a,b) = (52,48) or (65,60)

If C scored more than A, A cannot be 65.

Hence a=52, b=48

Average runs scored by C&D = (4x-(52+48))/2 = (220-100)/2 =60

The ratio of runs scored by B to the average runs of C & D = 48:60 = 4:5

Average speed of a defective car over 11 hours = (60+50)/2 = 55kmph.
Let x be the number of defective cars and y be the number of non-defective cars.

X(55) +Y(60)=56(X+Y)
4Y=X.

Hence the required ratio is 4:1.

Assume the average is $a$ and total number of elements in original set = n.

Total of all the elements of the set R= an

If 4 elementsÂ {1,4,6,9} are added then average reduces by 4

Now it is given that (an+20)/(n+4)=a-4

=> an+20=(a-4)(n+4)

=> 20=4a-4n-16

=> a-n=9

The average of R U {6,7,8,9,10} is equal to the total number of elements in the new set

(an+40)/(n+5)=(n+5)

=> an+40=n$^{2}$+10n+25

=> n(n+9)+40=n$^{2}$+10n+25

=> 9n+40=10n+25

=> n=15Â  Â Hence a=24

Let the age of the father and mother 6 years back be 4x and 3x.

Present age of the father and mother will be 4x+6 and 3x+6.

Age of the son will be 6 years.

Average age of the family is ={4x+6+3x+6+6}/3=20

60= 7x+18

x=6

Thus present age of father is 4×6+6= 30 years.

Let the age of the eldest child be a and youngest be b years.

Given the averages of their ages= m

Sum of ages of all the students: 11m years

If the youngest student is the one who left the average will be maximum= (11m-b)/10

If the eldest student is the one who left the average will be minimum= (11m-a)/10

Given,

$\frac{11m-b-{(11m-a)}}{10}$=4

a-b=40

Thus difference in their ages will be 40 years.

The sum of marks of all the students = 57*5 = 285
The sum of marks of top 3 students = 75*3=225

If the average increases by 25%, the sum of marks will also increase by 25%.
The increase in sum of the marks of bottom 3 students= 25% of x =Â  (10+4+9) = 23Â  where x is the sum of marks of bottom 3 students.
=> 25x/100=23Â  Â => x=92

The third highest marks = Sum of top 3 marks+ Sum of bottom 3 marks – Sum marks of all students = 225+92-285= 32

Alligations :$p=\frac{\left(p1q1+p2q2\right)}{q1+q2}$Â  where q1 and q2 are the number of innings in two group’s , p1 and p2 are there respective averages(average runs)Â , p is the overall average.