# GMAT Word Problems - Problems on Averages - 2

**IBPS PO Exams Aptitude Questions – Problems on Averages**

16.
The average age of Mr and Mrs
Sudheer and their 4 children is 20 years. If his age is excluded, the average
age of remaining members in the family would be 18 years. How old is Mr
Sudheer?

**Explanation:**

Total age of Mrs and Mrs Sudheer and their 4 children =
6 x 20 = 120 years.

When the age of Mr Sudheer excluded, the total age of
remaining 5 family members = 5x 18 = 90 years

Therefore, Sudheer’s age = 120-90 = 30 years

**SBI Bank Clerks free study material - Averages**

17.
The average age of a wife and
husband who were married 8 years ago was 25 years. The average age including a
child which was born during the interval is 23 years. Find the age of the child
now?

**Explanation:**

8 years ago, sum of the ages of W+H = 2 X 25 = 50

Including a
child boring during the interval, the number of members in the family is 3.

Present average age of the family =23

Present age of W+H+C
= 3 X 23 = 69 years

Present age of
W+H= 50+ (2 x 8) = 66 years

Age of the child
=69 -66 = 3 years

**IBPS Bank Clerks Exam study material - Averages**

18.
6 years ago, the average age
of a family of 5 members was 30 years. Including a baby, the average is increased
by 6 months today. What is the age of the child?

**Explanation:**

6 years ago, the total age of 5 members = 5 x 30 = 150 years

Present average age of 5 members and 1 child = 30.5 years

Total present age of5 members and 1 child = 6 x 30.5 = 183 year.

Total present age of 6 members = 5 x 30 + 5x6 = 180 years

Age of
the child = 183 – 180 = 3 years

**IIM-CAT Quantitative Aptitude free study material – Problems on Average**

19.
In a class of 40 students 15
are girls. The average mark of boys is 36 and average mark of girls is 40. What
is the average mark of the class?

*Explanation:*In a class of 40 students, 15 are girls. Then there are 25 boys.

The average of 25 boys is 36 and average of 15 girls is 40

Average marks of the class =

(25x36 + 15x40)
40

=
1500
40

= 37.5**Aptitude Questions and Answers – Averages Solved Examples**

20.
Of the three numbers, first
is thrice the second and the second is twice the third, If the average of three
numbers is 78. Find the first number?

**Explanation :**Let the third number be x.

2

^{ND}number is twice the third = 2x
And first number is thrice the second = 3 x 2x = 6x

Average =

(6x+2x+x)
3

=>3x= 78
ð X
=26

First number = 6x = 6 x 26 =156.

**GMAT Quantitative Aptitude – Problems on Averages**

21.
There are 4 numbers. Average
of first 3 numbers is 92 and the average of last three numbers is 100. Fourth
number is 62. Find the first number>

**Explanation:**Let four numbers be a,b,c and d.

Average of first 3 numbers=92=>
a+b+c=92 x3 =276

Average of last 3 numbers= 100
=>b+c+d= 3x100 = 300

Fourth number d= 62

First number = (a+b+c+d)-( b+c+d)= (276+62 )
– 300 =332-300= 32

**GRE Aptitude Questions - Averages ( Arithmetic Mean)**

22.
The average of 15 students is
20.Later it was found that one result was taken as 72 instead of 42. What is
the correct average?

**Explanation:**

Incorrect total marks of students = 15 x 20 = 300

Correct total marks of the students
= 300 – 72 +42 = 270

Correct average =

270
15

= 18**Quantitative Aptitude with Explanations – Arithmetic Mean Example Problems**

23.
The average marks of 25
students in a class are increased by 2 marks, when a student of marks 36 is
replaced by a new student. What are the marks obtained by new student?

**Explanation :**

Replacement of new student results in the increase of average. Means marks of new students are more than the marks of replaced student.

Total
increase in marks= Total students x Increase in Average = 25 x2 = 50 marks

Total marks are increased by 50 students.

Marks obtained by new student = marks of the
replaced student + Increase in total marks

=
36+50 = 86 marks.

**Aptitude Questions with Answers – Problems on Averages**

24.
The average weight of 30 men
is decreased by 1.5 kg, when a man of weight 95 kgs is replaced by a new man.
What is the weight of the new man?

**Explanation:**

Replacement of a man results in the decrease of the average means weight of the new man is less than the weight of the replaced man.

Total
decrease in weight= Total men x Decrease in Average = 30 x 1.5 = 45 kgs

Weight of new man = Weight of replaced
man – Total decrease in weight

= 95-45= 50 kgs

**Quantitative Aptitude Questions with Answers – Averages Solved Question**

25. A batsman has a certain average for runs for 20 innings. In the
21

^{st}innings he makes a score of 142 runs thereby increasing his average by 5 runs. What is the average of 21 innings?**Explanation :**

Let average for 20 innings be x

Total score in 20 innings = 20x

In 21

^{st}innings, he scored 142 runs
Total score in 21 innings = 20x +142

Average increased by 5 after 21 innings->Average = x +5

Total score in 21 innings = 21 (x+5)

Average increased by 5 after 21 innings->Average = x +5

Total score in 21 innings = 21 (x+5)

20x+142 = 21 (x+5)

ð 20x+142
= 21x+105

ð x=
37

Average
after 21 innings = 37 +5 = 42

**Aptitude Questions for Campus Placements – Averages**

26. The batting average of 40 matches of a cricket player is 60
runs. His highest score exceeds his lowest score by 132 runs. If these two
matches are excluded, the average of the remaining 38 matches 58. Find his
highest score?

**Explanation :**

Let his highest score be x then his lowest score= x-132

Total score in 38 matches = score in 40
matches – (Highest score + Lowest score)

=
(40 x 60)-(x+x-132)

= 2400 – 2x+132 => 2532-2x

After excluding highest and lowest scores,
the average becomes 58.

Therefore, total score in 38 matches = 38
x 59 =2242

2532- 2x = 2242 => 2x =
2532-2242 =>2x = 290 => x=145

His highest score= 145

**Placement Papers Aptitude Questions – Problems on Arithmetic Mean**

27. The average weight of 20 students in a class is 60 kgs. When 30
new students are admitted, the average weight reduced by 3 kg. What is the
average weight of newly admitted students?

**Explanation:**

Total weight of 20 students = 20 x 60 = 1200 kg

Total weight of (20+30=) 50 students = 50
x 57 = 2850 kg

Total weight of the 30 students = Weight
of 50 students – Weight of 20 students

= 2850 – 1200 =1650

Average weight of newly admitted 30
students =

1650
30

=55 kg**GRE Mathematics – Averages and Applications of Averages**

28. The average weight of students of a class is 40 kg. 12 new
students with an average weight of 32 kg join the class, thereby decreasing the
average by 4 kgs. Find the original strength of the class?

**Explanation:**Let the original strength of the class be x.

Total weight of original strength of the
class = 40x

Total weight of newLY joined students =
12 x 32 =384 kg

Total weight of original and new
students = 40x + 384 – (1)

Average is decreased by 4 kg =>New
Average =36

Total Number of students = (x+12)

Total weight of original and new
students = 36 (x+12) --- (2)

Equating (1) and
(2)

36(x+12) = 40x+384 => 36x +432= 40x+384

ð 12x
= 48 =>x= 48

**Problems on Averages – GMAT Word Problems**

29. The average weight of A,B,C is 80 kgs. A fourth man D joins them
and the average weight of four becomes 85 kgs. If E whose weight 5 kg less than
D , replaces A , the average weight of four men B,C,D and E becomes 82 kgs. Find the weight of A?

**Explanation:**Total weight of A+B+C = 3X 80 = 240

Weight of
A+B+C+D= 4x 85 =340.

Weight of
D = 340-240 = 100 kg

Weight of E =
100-5= 95 kg

Weight of (B+C+D+E) = 4 x 82 =328

Weight of (B+C+D) = 328-95= 233

Weight of A
=(A+B+C+D)-(B+C+D)= 340-233= 107

**Aptitude Questions and Answers for Competitive Examinations - Averages**

30. In a cricket match, 6 players had a certain average of theri
runs. 7

^{th}player makes a score of 56 runs, thereby increasing the average of their runs by 5. Find the average of 6 players?**Explanation:**

Let the average runs of 6 players be x.

Total runs of 6 players= 6x

Runs scored by 7

^{th}player= 56
Total score of 7 players= 6x+56 -(1)

Increase in
average = 5

Average of 7 players = (x+5)

Total score of 7 players = 7(x+5) -(2)

Equating (1) and (2) , 6x +56 = 7(x+5)

ð 6x+56
= 7x+35

ð X=21

New average = x+5 = 21+5 = 25