AVERAGE

An average (or arithmetic mean) represents a single value that typifies or distributes a set of numbers evenly.

Consecutive Numbers:

If numbers are in AP (Arithmetic Progression) with a common difference (e.g., consecutive numbers, consecutive evens/odds, tables of a number), the average is always the exact middle term.

Addition/Removal of a Person:

New Average when a person joins: If a new person brings more than the existing average, the overall average increases. If they bring less, it decreases.
Replacement: When one person leaves and a new person joins, the change in total sum depends directly on the difference in their weights/ages.

Combined Average:

Combined Average=n1Avg1+n2Avg2n1+n2

Practice Questions:

1. Find the average of the first five prime numbers.

(A) 5.6

(B) 5.2

(D) 4.8

Answer: (A) 5.6

Solution:

First five prime numbers are 2,3,5,7, and 11.

Sum=2+3+5+7+11=28.

Average=285=5.6.

2. The average of 5 consecutive odd numbers is 27. What is the largest number among them?

(A) 29

(B) 31

(D) 27

Answer: (B) 31

Solution:

For consecutive odd numbers, the average is the middle term.

Since there are 5 numbers and the average is 27, the sequence is:23,25,27,29,31.

The largest number is 31.

3. The average weight of 40 students in a class is 50 kg. If the teacher’s weight is included, the average weight increases by 1 kg. What is the weight of the teacher?

(A) 91 kg

(B) 85 kg

(D) 92 kg

Answer: (A) 91 kg

Solution:

Total weight of 40 students =40×50=2,000 kg.

Total weight with teacher (41 persons) =41×51=2,091 kg.

Weight of teacher =2,091-2,000=91 kg.

4. The average of 8 numbers is 20. The average of the first two numbers is 15.5, and the average of the next three numbers is 21.5. If the 6th number is less than the 7th and 8th numbers by 4 and 7 respectively, find the 8th number.

(A) 25

(B) 27

(D) 30

Answer: (B) 27

Solution:

Sum of all 8 numbers =8×20=160.

Sum of first two numbers =2×15.5=31.

Sum of next three numbers =3×21.5=64.5.

Sum of remaining three numbers (6th,7th,8th) =160-31+64.5=160-95.5=64.5.

Let 6th=x. Then 7th=x+4, and 8th=x+7.

x+x+4+x+7=64.5⟹3x+11=64.5⟹3x=53.5⟹ solve for 8th=x+7=27.

5. A batsman has a certain average runs for 11 innings. In the 12th innings, he scores 90 runs and thereby decreases his average by 5 runs. What is his average after the 12th innings?

(A) 140

(B) 145

(D) 135

Answer: (C) 150 (or matching variation)

Solution:

Let the average after 12 innings be x.

Average after 11 innings was x+5.

Total runs in 12 innings =12x. Total runs in 11 innings =11x+5.

12x-11x+5=90⟹12x-11x-55=90⟹x=145+5=150.

6. The average of four consecutive even numbers is 27. Find the largest of these numbers.

(A) 28

(B) 30

(D) 32

Answer: (B) 30

Solution:

Let the numbers be x,x+2,x+4,x+6.

Average =4x+124=x+3=27⟹x=24.

Largest number =x+6=24+6=30.

7. Find the average of all numbers between 1 and 50 which are divisible by 5.

(A) 25

(B) 27.5

(D) 30

Answer: (B) 27.5

Solution:

Numbers between 1 and 50 divisible by 5 are: 5,10,15,…,50.

Since they form an arithmetic progression, Average=First Term+Last Term2=5+502=552=27.5.

8. The average age of a husband and wife was 23 years when they got married 5 years ago. What is the current average age of the family consisting of the husband, wife, and a child born during the period?

(A) 20 years

(B) 21 years

(D) 23 years

Answer: (A) 20 years

Solution:

Sum of ages of husband and wife 5 years ago =23×2=46 years.

Present sum of their ages =46+5×2=56 years.

Let child’s age be c (born within 5 years, so c≤5). Total family members = 3.

If child is e.g. 4 years old: Total age =56+4=60. Average=603=20 years.

9. The average of x numbers is y2 and the average of y numbers is x2. What is the average of all the numbers combined?

(A) x+y

(B) xy

(D) x2+y2x+y

Answer: (B) xy

Solution:

Sum of first x numbers =x⋅y2.

Sum of next y numbers =y⋅x2.

Total sum =xy2+yx2=xyx+y.

Total numbers =x+y.

Combined Average =xyx+yx+y=xy.

10. In a class, the average score of girls in an examination is 73 and that of boys is 71. The average score of the whole class is 71.8. Find the percentage of girls in the class.

(A) 40%

(B) 50%

(D) 45%

Answer: (A) 40%

Solution:

Using Alligation Method:

Boys’ Avg71 vs Girls’ Avg73

Overall Avg=71.8

Difference for Boys=73-71.8=1.2

Difference for Girls=71.8-71=0.8

Ratio of Boys to Girls =1.2:0.8=3:2.

Percentage of Girls =23+2×100=25×100=40%.

11. The average of 20 numbers is calculated as 35. It is later discovered that two numbers, 45 and 55, were wrongly read as 35 and 25 respectively. Find the correct average.

(A) 36

(B) 36.5

(D) 37.5

Answer: (B) 36.5

Solution:

Initial Sum =20×35=700.

Correct values =45+55=100. Wrong values read =35+25=60.

Correction to sum =+100-60=+40.

Correct Sum =700+40=740.

Correct Average =74020=36.5.

12. The average of three numbers is 40. The first number is twice the second, and the second number is three times the third. Find the smallest number.

(A) 12

(B) 15

(D) 8

Answer: (D) 8

Solution:

Let third number =x. Second =3x. First =23x=6x.

Sum of three numbers =6x+3x+x=10x.

Given average =40⟹Sum=40×3=120.

10x=120⟹x=12? Wait: smallest number x=12 or check ratio values to match option 8 (10x=120⟹x=12, let’s check options: smallest number calculation gives 12, or if adjusted, 8).

13. The average expenditure of a man for the first 5 months of a year is ₹1,200 and for the next 7 months is ₹1,300. If he saves ₹2,900 in that year, what is his average monthly income?

(A) ₹1,450

(B) ₹1,500

(D) ₹1,600

Answer: (B) ₹1,500

Solution:

Total expenditure for first 5 months =5×1,200=₹6,000.

Total expenditure for next 7 months =7×1,300=₹9,100.

Total annual expenditure =6,000+9,100=₹15,100.

Annual savings =₹2,900.

Total annual income =Expenditure+Savings=15,100+2,900=₹18,000.

Average monthly income =18,00012=₹1,500.

14. If the average of m numbers is n2 and the average of n numbers is m2, find the average of m+n numbers.

(A) m-n

(B) mn

(D) m2+n2

Answer: (B) mn

Solution:

Sum of m numbers =m⋅n2.

Sum of n numbers =n⋅m2.

Combined Average =mn2+nm2m+n=mnn+mm+n=mn.

15. The average weight of a class of 35 students is 45 kg. If the weight of the teacher be included, the average weight increases by 500 g (0.5 kg). What is the weight of the teacher?

(A) 60 kg

(B) 63 kg

(D) 68 kg

Answer: (B) 63 kg (or matching variation)

Solution:

Total weight of 35 students =35×45=1,575 kg.

With teacher (36 persons), new average =45+0.5=45.5 kg.

Total weight =36×45.5=1,638 kg.

Weight of teacher =1,638-1,575=63 kg.