Monday, 28 March 2016

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.
2ND 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 21st 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 21st 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)
      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. 7th 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 7th 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