# Averages Important Questions for MAH-CET

Here you can download a free Averages questions PDF with answers for MAH MBA CET by Cracku. These are some tricky questions in the MAH MBA CET exam that you need to find Averages Tables answers for the given questions. These questions will help you to make practice and solve the **Averages**Â questions in the MAH MBA CET exam. Utilize this best **PDF practice set** which is included answers in detail. Click on the below link to download the **Averages Tables MCQ** PDF for MBA-CET 2022 for free.

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**Question 1:Â **A group of 20 girls has average age of 12 years. Average of first 12 from the same group is 13 years and what is the average age of other 8 girls in the group?

a)Â 10

b)Â 11

c)Â 11.5

d)Â Cannot be determined

e)Â None of these

**1)Â AnswerÂ (E)**

**Solution:**

Let the age of each of the girl in the group be $x_1,x_2,x_3,…..,x_{20}$ years

Average age of 20 girls = 12

=> $\frac{(x_1+x_2+x_3+…..+x_{20})}{20}=12$

=>Â $(x_1+x_2+x_3+…..+x_{20})=12 \times 20=240$ ————(i)

Average of first 12 girls = 13

=> $\frac{(x_1+x_2+x_2+….+x_{12})}{12}=13$

=>Â $(x_1+x_2+x_3+…..+x_{12})=13 \times 12=156$ ———–(ii)

Subtracting equation (ii) from (i)

=>Â $(x_1+x_2+x_3+…..+x_{20})$ $-$ $(x_1+x_2+x_3+…..+x_{12}) = (240-156)$

=> $(x_{13}+x_{14}+…..+x_{20})=84$

Dividing above equation by 8

=> $\frac{(x_{13}+x_{14}+……+x_{20})}{8}=\frac{84}{8} = 10.5$

=> Ans – (E)

**Question 2:Â **The average of the 5 consecutive even numbers A,B,C,D ,E is 52.what is the product of B & E

a)Â 2912

b)Â 2688

c)Â 3024

d)Â 2800

e)Â NONE OF THESE

**2)Â AnswerÂ (D)**

**Solution:**

Let the five consecutive even numbersÂ A,B,C,D ,E = $(x-4) , (x-2) , (x) , (x+2) , (x+4)$ respectively.

Average = $\frac{A+B+C+D+E}{5} = 52$

=> $(x-4) + (x-2) + (x) + (x+2) + (x+4) = 52 \times 5$

=> $5x = 52 \times 5$

=> $x = \frac{52 \times 5}{5} = 52$

=> $B = 52 – 2 = 50$ and $E = 52 + 4 = 56$

$\therefore$ Product of B & E = $50 \times 56 = 2800$

**Question 3:Â **find the average of set scores? 221,231,441,359,665,525

a)Â 399

b)Â 428

c)Â 407

d)Â 415

e)Â None of these

**3)Â AnswerÂ (C)**

**Solution:**

SetÂ :Â 221,231,441,359,665,525

Sum =Â 221 + 231 + 441 + 359 + 665 + 525 = 2442

=> Required average = $\frac{2442}{6}$

= 407

**Question 4:Â **There are three positive numbers, ${1 \over 3}$rd of average of all the three numbers is 8 less than the value of the highest number. Average of the lowest and the second lowest number is 8. Which is the highest number?

a)Â 11

b)Â 14

c)Â 10

d)Â 9

e)Â 13

**4)Â AnswerÂ (A)**

**Solution:**

Let the three positive numbers be $x,y,z$ Â Â (where $x < y < z$)

Average of the three numbers = $\frac{x + y + z}{3}$

Acc. to ques, => $\frac{1}{3} \times (\frac{x + y + z}{3}) = z – 8$

=> $x + y + z = 9z – 72$

=> $x + y = 8z – 72$

Dividing both sides by 2, we getÂ :

=> $\frac{x + y}{2} = 4z – 36$

Also, average of the lowest and the second lowest number is 8, => $\frac{x + y}{2} = 8$

=> $4z – 36 = 8$

=> $4z = 8 + 36 = 44$

=> $z = \frac{44}{4} = 11$

**Question 5:Â **The average age of some males and 15 females is 18 years. The sum of the ages of 15 females is 240 years and average age of males is 20 years. Find the number of males.

a)Â 8

b)Â 7

c)Â 10

d)Â 15

e)Â None of these

**5)Â AnswerÂ (D)**

**Solution:**

Let number of males = $x$

Average age of males = 20 years

=> Sum of age of males = $20 \times x = 20x$ years

Sum of age of females = 240 years

Acc. to ques, => $\frac{(20 x) + (240)}{15 + x} = 18$

=> $20x + 240 = 270 + 18x$

=> $20x – 18x = 2x = 270 – 240 = 30$

=> $x = \frac{30}{2} = 15$

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**Question 6:Â **Average score of a class of 50 students, in an exam is 34. Average score of the students who have passed is 52 and the average score of students who have failed is 16. How many students have failed in the exam?

a)Â 25

b)Â 20

c)Â 15

d)Â 18

e)Â 30

**6)Â AnswerÂ (A)**

**Solution:**

Let the number of students who failed = $x$

=> Number of students who passed = $(50 – x)$

Acc. to ques,

=> $\frac{(52 \times (50 – x)) + (16 \times x)}{50} = 34$

=> $52 \times 50 – 52x + 16x = 34 \times 50$

=> $52x – 16x = 50 \times (52 – 34)$

=> $36x = 50 \times 18$

=> $x = \frac{50 \times 18}{36} = 25$

**Question 7:Â **In a class, the average weight of 80 boys is 64 kg and that of 75 girls is 70 kg. After a few days, 60% of the girls and 30% of the boys leave. What would be the new average weight of the class (in kg)? Assume that the average weight of the boys and the girls remain constant throughout.

a)Â 63

b)Â 66.09

c)Â 68.5

d)Â 65.5

e)Â 57.5

**7)Â AnswerÂ (B)**

**Solution:**

Initially, number of boys = 80 and number of girls = 75

Average weight of boys = 64 kg and average weight of girls = 70 kg

Now, 60% of the girls and 30% of the boys leave

=> Boys left = $\frac{100 – 30}{100} \times 80 = 56$

Girls left = $\frac{100 – 60}{100} \times 75 = 30$

Since,Â average weight of the boys and the girls remains constant throughout

$\therefore$ New average weight of the class

= $\frac{(56 \times 64) + (30 \times 70)}{56 + 30} = \frac{3584 + 2100}{86}$

= $\frac{5684}{86} = 66.09$ kg

**Question 8:Â **In a class, the average weight of 40 boys is 65 kg and that of 50 girls is 60 kg. After a few days, 40% of the girls and 50% of the boys leave. What would be the new average weight of the class (in kg)? Assume that the average weight of the boys and the girls remains constant throughout.

a)Â 65

b)Â 62

c)Â 68

d)Â 55

e)Â 58

**8)Â AnswerÂ (B)**

**Solution:**

Initially, number of boys = 40 and number of girls = 50

Average weight of boys = 65 kg and average weight of girls = 60 kg

Now,Â 40% of the girls and 50% of the boys leave

=> Boys left = $\frac{100 – 50}{100} \times 40 = 20$

Girls left = $\frac{100 – 40}{100} \times 50 = 30$

Since,Â average weight of the boys and the girls remains constant throughout

$\therefore$ New average weight of the class

= $\frac{(20 \times 65) + (30 \times 60)}{20 + 30} = \frac{1300 + 1800}{50}$

= $\frac{3100}{50} = 62$ kg

**Question 9:Â **When 9 is subtracted from a two digit number, the number so formed is reverse of the original number. Also, the average of the digits of the original number is 7.5. What is definitely the original number ?

a)Â 87

b)Â 92

c)Â 90

d)Â 69

e)Â 96

**9)Â AnswerÂ (A)**

**Solution:**

Let the ten’s digit and unit’s digit of the original number be $x$ and $y$ respectively.

=> original number = $10x + y$

Average of digits = $\frac{x + y}{2} = 7.5$

=> $x + y = 7.5 \times 2 = 15$ ————-(i)

When 9 is subtracted from it, => Reverse number = $10y + x$

=> $(10x + y) – 9 = 10y + x$

=> $9x – 9y = 9$

=> $x – y = \frac{9}{9} = 1$ —————-(ii)

Adding equations (i) & (ii), we getÂ :

=> $2x = 16$ => $x = \frac{16}{2} = 8$

Putting it in eqn(i), => $y = 15 – 8 = 7$

$\therefore$ Original number = $87$

**Question 10:Â **Average weight of three boys P, T and R is 54 1/3 kgs while the average weight of three boys, T, F and G is 53 kgs. What is the average weight of P, T, R, F and H?

a)Â 53.8 kgs

b)Â 52.4 kgs

c)Â 53.2 kgs

d)Â Canâ€™t be determined

e)Â None of these

**10)Â AnswerÂ (D)**

**Solution:**

Clearly, there is no information about the weight of H

Thus, we cannot determine their average weights.

Ans – (D)

**Question 11:Â **The average age of a man and his son is 48 years. The ratio of their ages is 11 : 5 respectively. What will be ratio of their ages after 6 years ?

a)Â 6 : 5

b)Â 5 : 3

c)Â 4 : 3

d)Â 2 : 1

e)Â None of these

**11)Â AnswerÂ (D)**

**Solution:**

Let age of man = $11x$ and age of son = $5x$

Sum of ages of man and son = 48 * 2 = 96 years

=> $11x + 5x = 96$

=> $x = \frac{96}{16} = 6$

=> Age of man = 11 * 6 = 66 years

Age of son = 5 * 6 = 30 years

=> Age of man after 6 years = 66 + 6 = 72 years

Age of son after 6 years = 30 + 6 = 36 years

$\therefore$ Required ratio = 72Â : 36 =Â **2Â : 1**

**Question 12:Â **The average age of a man and his son is 18 years. The ratio of their ages is 5 : 1 respectively. What will be the ratio of their ages after 6 years ?

a)Â 10 : 3

b)Â 5 : 2

c)Â 4 : 3

d)Â 3 : 1

e)Â None of these

**12)Â AnswerÂ (D)**

**Solution:**

Let age of man = $5x$ and age of son = $x$

Sum of ages of man and son = 18 * 2 = 36 years

=> $5x + x = 36$

=> $x = \frac{36}{6} = 6$

=> Age of man = 5 * 6 = 30 years

Age of son = 1 * 6 = 6 years

=> Age of man after 6 years = 30 + 6 = 36 years

Age of son after 6 years = 6 + 6 = 12 years

$\therefore$ Required ratio = 36 : 12 =Â **3 : 1**

**Question 13:Â **The average age of a man and his son is 54 years. The ratio of their ages is 23 : 13 respectively. What will be ratio of their ages after 6 years?

a)Â 10 : 7

b)Â 5 : 3

c)Â 4 : 3

d)Â 3 : 2

e)Â None of these

**13)Â AnswerÂ (B)**

**Solution:**

Let age of man = $23x$ and age of son = $13x$

Sum of ages of man and son = 54 * 2 = 108 years

=> $23x + 13x = 108$

=> $x = \frac{108}{36} = 3$

=> Age of man = 23 * 3 = 69 years

Age of son = 13 * 3 = 39 years

=> Age of man after 6 years = 69 + 6 = 75 years

Age of son after 6 years = 39 + 6 = 45 years

$\therefore$ Required ratio = 75 : 45 =Â **5Â : 3**

**Question 14:Â **The average of 5 consecutive even numbers A, B, C. D and E is 52. What is the product of B & E?

a)Â 2916

b)Â 2988

c)Â 3000

d)Â 2800

e)Â None of these

**14)Â AnswerÂ (D)**

**Solution:**

Let the 5 consecutive even numbers be = $(x) , (x+2) , (x+4) , (x+6) , (x+8)$

Sum of these 5 consecutive numbers = 52 * 5 = 260

=> $(x) + (x+2) + (x+4) + (x+6) + (x+8) = 260$

=> $5x = 260 – 20 = 240$

=> $x = 48$

=> B = $x + 2$ = 48 + 2 = 50

E = $x + 8$ = 48 + 8 = 56

$\therefore$ B $\times$ E = 50 * 56 = 2800

**Question 15:Â **Find the average of the following set of scores :

221, 231, 441, 359, 665, 525

a)Â 399

b)Â 428

c)Â 407

d)Â 415

e)Â None of these

**15)Â AnswerÂ (C)**

**Solution:**

Sum of scores = 221 + 231 + 441 + 359 + 665 + 525

= 2442

=> Average score = 2442/6 = 407

**Question 16:Â **The ratio of roses and lillies in a garden is 3 : 2 respectively. The average number of roses and lillies is 180. What is the number of lillies in the garden?

a)Â 144

b)Â 182

c)Â 216

d)Â 360

e)Â None of these

**16)Â AnswerÂ (A)**

**Solution:**

Let the number of roses and lillies in the garden be $3x$ and $2x$ resp

Acc to ques,

=> $\frac{3x + 2x}{2} = 180$

=> $5x = 360$

=> $x = 360/5 = 72$

$\therefore$ Number of lillies = 2 * 72 = 144

**Question 17:Â **The average of five positive numbers is 308. The average of first two numbers is 482.5 and the average of last two numbers is 258.5. What is the third number?

a)Â 224

b)Â 58

c)Â 121

d)Â Cannot be determined

e)Â None of these

**17)Â AnswerÂ (B)**

**Solution:**

Let the numbers be $A_1 , A_2 , A_3 , A_4 , A_5$

=> $A_1 + A_2 + A_3 + A_4 + A_5 = 308 * 5 = 1540$

Also, $A_1 + A_2 = 482.5 * 2 = 965$

and $A_4 + A_5 = 258.5 * 2 = 517$

=> $A_3 = 1540 – 965 – 517 = 58$

**Question 18:Â **What will be the average of the following set of scores (Rounded off to the nearest integer) ?

46, 54, 62, 68, 56, 29, 58

a)Â 45

b)Â 59

c)Â 62

d)Â 48

e)Â 53

**18)Â AnswerÂ (E)**

**Solution:**

Sum of the scores = 46 + 54 + 62 + 68 + 56 + 29 + 58 = 373

=> Required Average = $\frac{373}{7}$

= $53.28 \approx 53$

**Question 19:Â **What will be the average of the followings set of scores?

59, 84, 44, 98, 30, 40, 58

a)Â 62

b)Â 66

c)Â 75

d)Â 52

e)Â 59

**19)Â AnswerÂ (E)**

**Solution:**

Sum of the scores

= 59 + 84 + 44 + 98 + 30 + 40 + 58 = 413

=> Average = $\frac{413}{7}$ = 59

**Question 20:Â **The average of five positive numbers is 213. The average of the first two numbers is 233.5 and the average of last two numbers is 271. What is the third number?

a)Â 64

b)Â 56

c)Â 106

d)Â Cannot be determined

e)Â None of these

**20)Â AnswerÂ (B)**

**Solution:**

Let the numbers be $A_1 , A_2 , A_3 , A_4 , A_5$

=> $A_1 + A_2 + A_3 + A_4 + A_5 = 213 * 5 = 1065$

Also, $A_1 + A_2 = 233.5 * 2 = 467$

and $A_4 + A_5 = 271 * 2 = 542$

=> $A_3 = 1065 – 467 – 542 = 56$

**Question 21:Â **The ratio of ducks and frogs in a pond is 37 : 39 respectively. The average number of ducks and frogs in the pond is 152. What is the number of frogs in the pond?

a)Â 148

b)Â 152

c)Â 156

d)Â 144

e)Â None of these

**21)Â AnswerÂ (C)**

**Solution:**

Let the number of ducks and frogs in the pond be $37x$ and $39x$ resp

Acc to ques,

=> $\frac{37x + 39x}{2} = 152$

=> $38x = 152$

=> $x = \frac{152}{38} = 4$

$\therefore$ Number of frogs = 4 * 39 = 156

**Question 22:Â **The average of five numbers is 49. The average of the first and the second numbers is 48 and the average of the fourth and fifth numbers is 28. What is the third number?

a)Â 92

b)Â 91

c)Â 95

d)Â Cannot be determined

e)Â None of these

**22)Â AnswerÂ (E)**

**Solution:**

Let the numbers be $A_1 , A_2 , A_3 , A_4 , A_5$

=> $A_1 + A_2 + A_3 + A_4 + A_5 = 49 * 5 = 245$

Also, $A_1 + A_2 = 48 * 2 = 96$

and $A_4 + A_5 = 28 * 2 = 56$

=> $A_3 = 245 – 96 – 56 = 93$

**Question 23:Â **The average of five numbers is 26. The average of first two numbers is 30 and that of the last two numbers is 7. What is the third number ?

a)Â 23

b)Â 75

c)Â 56

d)Â Cannot be determined

e)Â None of these

**23)Â AnswerÂ (C)**

**Solution:**

Let the numbers be $A_1 , A_2 , A_3 , A_4 , A_5$

=> $A_1 + A_2 + A_3 + A_4 + A_5 = 26 * 5 = 130$

Also, $A_1 + A_2 = 30 * 2 = 60$

and $A_4 + A_5 = 7 * 2 = 14$

=> $A_3 = 130 – 60 – 14 = 56$

**Question 24:Â **The respective ratio between the number of rose and lily flowers in a garden is 3Â : 2. The average number of rose and lily is 180. What is the number of lily flowers in the garden ?

a)Â 144

b)Â 360

c)Â 181

d)Â 216

e)Â None of these

**24)Â AnswerÂ (A)**

**Solution:**

Let the number of roses and lillies in the garden be $3x$ and $2x$ resp

Acc to ques,

=> $\frac{3x + 2x}{2} = 180$

=> $5x = 360$

=> $x = 360/5 = 72$

$\therefore$ Number of lillies = 2 * 72 = 144

**Question 25:Â **The average weight of 40 students in a class is 55 kg. Six students of them whose average weight is 52 kg leave the class and another group of 6 students whose average weight is 42 kg joins the class. What is the new average of the class ? (in kg)

a)Â 54.25

b)Â 52.75

c)Â 53.5

d)Â 54

e)Â 53

**25)Â AnswerÂ (C)**

**Solution:**

Total weight of original 40 students = $40 \times 55 = 2200$ kg

Weight of 6 students who left = $6 \times 52 = 312$ kg

Weight of 6 new students who joined = $6 \times 42 = 252$ kg

=> New weight of new group of 40 students = $(2200 – 312 + 252)$ kg

= $2140$ kg

$\therefore$ Required Average = $\frac{2140}{40}$

= $53.5$ kg